Difference between revisions of "Groups"

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==Groups==
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<p>Recall that an operation <span class="math-inline"><math>\cdot</math></span> on <span class="math-inline"><math>S</math></span> is said to be associative if for all <span class="math-inline"><math>a, b, c \in S</math></span> we have that <span class="math-inline"><math>a \cdot (b \cdot c) = (a \cdot b) \cdot c</math></span> and <span class="math-inline"><math>\cdot</math></span> is said to be commutative if for all <span class="math-inline"><math>a, b \in S</math></span> we have that <span class="math-inline"><math>a \cdot b = b \cdot a</math></span>.</p>
 
<p>Recall that an operation <span class="math-inline"><math>\cdot</math></span> on <span class="math-inline"><math>S</math></span> is said to be associative if for all <span class="math-inline"><math>a, b, c \in S</math></span> we have that <span class="math-inline"><math>a \cdot (b \cdot c) = (a \cdot b) \cdot c</math></span> and <span class="math-inline"><math>\cdot</math></span> is said to be commutative if for all <span class="math-inline"><math>a, b \in S</math></span> we have that <span class="math-inline"><math>a \cdot b = b \cdot a</math></span>.</p>
 
<p>An element <span class="math-inline"><math>e \in S</math></span> is the identity element of <span class="math-inline"><math>S</math></span> under <span class="math-inline"><math>\cdot</math></span> if for all <span class="math-inline"><math>a \in S</math></span> we have that <span class="math-inline"><math>a \cdot e = a</math></span> and <span class="math-inline"><math>e \cdot a = a</math></span>.</p>
 
<p>An element <span class="math-inline"><math>e \in S</math></span> is the identity element of <span class="math-inline"><math>S</math></span> under <span class="math-inline"><math>\cdot</math></span> if for all <span class="math-inline"><math>a \in S</math></span> we have that <span class="math-inline"><math>a \cdot e = a</math></span> and <span class="math-inline"><math>e \cdot a = a</math></span>.</p>
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<p>Furthermore, <span class="math-inline"><math>(\mathbb{Z}, +)</math></span> is also a group under the operation of standard addition since the sum of any two integers is an integer, addition is associative, the additivity identity is <span class="math-inline"><math>0 \in \mathbb{Z}</math></span>, and for all <span class="math-inline"><math>a \in \mathbb{Z}</math></span> we have <span class="math-inline"><math>-a \in \mathbb{Z}</math></span> as additive inverses.</p>
 
<p>Furthermore, <span class="math-inline"><math>(\mathbb{Z}, +)</math></span> is also a group under the operation of standard addition since the sum of any two integers is an integer, addition is associative, the additivity identity is <span class="math-inline"><math>0 \in \mathbb{Z}</math></span>, and for all <span class="math-inline"><math>a \in \mathbb{Z}</math></span> we have <span class="math-inline"><math>-a \in \mathbb{Z}</math></span> as additive inverses.</p>
 
<p>We will examine many other (more interesting) groups later on, but for now, let's look at an example of a set and a binary operation that does NOT form a group.</p>
 
<p>We will examine many other (more interesting) groups later on, but for now, let's look at an example of a set and a binary operation that does NOT form a group.</p>
<h2 id="toc1"><span>Example 1</span></h2>
+
===Example 1===
 
<p>Consider the set of integers <span class="math-inline"><math>\mathbb{Z}</math></span> and define <span class="math-inline"><math>*</math></span> for all <span class="math-inline"><math>a, b \in \mathbb{Z}</math></span> by:</p>
 
<p>Consider the set of integers <span class="math-inline"><math>\mathbb{Z}</math></span> and define <span class="math-inline"><math>*</math></span> for all <span class="math-inline"><math>a, b \in \mathbb{Z}</math></span> by:</p>
 
<div style="text-align: center;"><math>\begin{align} \quad a * b = a + 2b \end{align}</math></div>
 
<div style="text-align: center;"><math>\begin{align} \quad a * b = a + 2b \end{align}</math></div>
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<p>Clearly <span class="math-inline"><math>a * (b * c) \neq (a * b) * c</math></span> so <span class="math-inline"><math>\mathbb{Z}</math></span> does not form a group under the operation <span class="math-inline"><math>*</math></span>.</p>
 
<p>Clearly <span class="math-inline"><math>a * (b * c) \neq (a * b) * c</math></span> so <span class="math-inline"><math>\mathbb{Z}</math></span> does not form a group under the operation <span class="math-inline"><math>*</math></span>.</p>
  
 +
==Basic Theorems Regarding Groups==
 +
<p>A group <span class="math-inline"><math>(G, \cdot)</math></span> is a set <span class="math-inline"><math>G</math></span> with a binary operation <span class="math-inline"><math>\cdot</math></span> such that:</p>
 +
<ul>
 +
<li><strong>1)</strong> <span class="math-inline"><math>\cdot</math></span> is associative, i.e., for all <span class="math-inline"><math>a, b, c \in G</math></span>, <span class="math-inline"><math>a \cdot (b \cdot c) = (a \cdot b) \cdot c)</math></span>.</li>
 +
</ul>
 +
<ul>
 +
<li><strong>2)</strong> There exists an identity element <span class="math-inline"><math>e \in G</math></span> such that <span class="math-inline"><math>a \cdot e = a = e \cdot a</math></span> for all <span class="math-inline"><math>a \in G</math></span>.</li>
 +
</ul>
 +
<ul>
 +
<li><strong>3)</strong> For each <span class="math-inline"><math>a \in G</math></span> there exists an <span class="math-inline"><math>a^{-1} \in G</math></span> such that <span class="math-inline"><math>a \cdot a^{-1} = a^{-1} \cdot a = e</math></span>.</li>
 +
</ul>
 +
<p>We will now look at some rather basic results regarding groups which we can derive from the group axioms above.</p>
 +
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 +
<td><strong>Proposition 1:</strong> Let <span class="math-inline"><math>(G, \cdot)</math></span> be a group and let <span class="math-inline"><math>e</math></span> be the identity for this group. Then:<br />
 +
<strong>a)</strong> The identity element <span class="math-inline"><math>e \in G</math></span> is unique.<br />
 +
<strong>b)</strong> For each <span class="math-inline"><math>a \in G</math></span>, the corresponding inverse <span class="math-inline"><math>a^{-1} \in G</math></span> is unique.<br />
 +
<strong>c)</strong> For each <span class="math-inline"><math>a \in G</math></span>, <span class="math-inline"><math>(a^{-1})^{-1} = a</math></span>.<br />
 +
<strong>d)</strong> For all <span class="math-inline"><math>a, b \in G</math></span>, <span class="math-inline"><math>(a \cdot b)^{-1} = b^{-1} \cdot a^{-1}</math></span>.<br />
 +
<strong>e)</strong> For all <span class="math-inline"><math>a, b \in G</math></span>, if <span class="math-inline"><math>a \cdot b = e</math></span> then <span class="math-inline"><math>a = b^{-1}</math></span> and <span class="math-inline"><math>b = a^{-1}</math></span>.<br />
 +
<strong>f)</strong> If <span class="math-inline"><math>a^2 = a</math></span> then <span class="math-inline"><math>a = e</math></span>.</td>
 +
</blockquote>
 +
<ul>
 +
<li><strong>Proof of a)</strong> Suppose that <span class="math-inline"><math>e</math></span> and <span class="math-inline"><math>e'</math></span> are both identities for <span class="math-inline"><math>\cdot</math></span>. Then:</li>
 +
</ul>
 +
<div style="text-align: center;"><math>\begin{align} \quad e = e \cdot e = e \cdot e' = e' \cdot e' = e' \end{align}</math></div>
 +
<ul>
 +
<li>Therefore <span class="math-inline"><math>e = e'</math></span> so the identity for <span class="math-inline"><math>\cdot</math></span> is unique. <span class="math-inline"><math>\blacksquare</math></span></li>
 +
</ul>
 +
<ul>
 +
<li><strong>Proof of b)</strong> Suppose that <span class="math-inline"><math>a^{-1} \in G</math></span> and <span class="math-inline"><math>a^{-1'} \in G</math></span> are both inverses for <span class="math-inline"><math>a \in G</math></span> under <span class="math-inline"><math>\cdot</math></span>. Then:</li>
 +
</ul>
 +
<div style="text-align: center;"><math>\begin{align} \quad a^{-1} = a^{-1} \cdot e = a^{-1} \cdot (a \cdot a^{-1'}) = (a^{-1} \cdot a)*a^{-1} = e \cdot a^{-1'} = a^{-1'} \end{align}</math></div>
 +
<ul>
 +
<li>Therefore <span class="math-inline"><math>a^{-1} = a^{-1'}</math></span> so the inverse for <span class="math-inline"><math>a</math></span> is unique. <span class="math-inline"><math>\blacksquare</math></span></li>
 +
</ul>
 +
<ul>
 +
<li><strong>Proof of c)</strong> Let <span class="math-inline"><math>a \in G</math></span>. Then <span class="math-inline"><math>(a^{-1})^{-1}</math></span> is the inverse to <span class="math-inline"><math>a^{-1}</math></span>. However, the inverse to <span class="math-inline"><math>a^{-1}</math></span> is <span class="math-inline"><math>a</math></span> and by (b) we have shown that the inverse of each element in <span class="math-inline"><math>G</math></span> is unique. Therefore <span class="math-inline"><math>a = (a^{-1})^{-1}</math></span>. <span class="math-inline"><math>\blacksquare</math></span></li>
 +
</ul>
 +
<ul>
 +
<li><strong>Proof of d)</strong> If we apply the operation <span class="math-inline"><math>\cdot</math></span> between <span class="math-inline"><math>b^{-1} \cdot a^{-1}</math></span> and <span class="math-inline"><math>(a \cdot b)</math></span> we get:</li>
 +
</ul>
 +
<div style="text-align: center;"><math>\begin{align} \quad (a \cdot b) \cdot [b^{-1} \cdot a^{-1}] & = a \cdot [(b \cdot b^{-1}) \cdot a^{-1}] \\ \quad &= a \cdot [e \cdot a^{-1}] \\ \quad &= a \cdot a^{-1} \\ \quad &= e \end{align}</math></div>
 +
<ul>
 +
<li>Therefore the inverse of <span class="math-inline"><math>(a \cdot b)</math></span> is <span class="math-inline"><math>b^{-1} \cdot a^{-1}</math></span>. We also have that the invere of <span class="math-inline"><math>(a \cdot b)</math></span> is <span class="math-inline"><math>(a \cdot b)^{-1}</math></span>. By (b), the inverse of <span class="math-inline"><math>(a \cdot b)</math></span> is unique and so:</li>
 +
</ul>
 +
<div style="text-align: center;"><math>\begin{align} \quad (a \cdot b)^{-1} = b^{-1} \cdot a^{-1} \quad \blacksquare \end{align}</math></div>
 +
<ul>
 +
<li><strong>Proof of e)</strong> Suppose that <span class="math-inline"><math>a \cdot b = e</math></span>. Then:</li>
 +
</ul>
 +
<div style="text-align: center;"><math>\begin{align} \quad a \cdot b &= e \\ \quad (a \cdot b) \cdot b^{-1} &= e \cdot b^{-1} \\ \quad a \cdot (b \cdot b^{-1}) &= b^{-1} \\ \quad a \cdot e &= b^{-1} \\ \quad a &= b^{-1} \end{align}</math></div>
 +
<ul>
 +
<li>Similarly:</li>
 +
</ul>
 +
<div style="text-align: center;"><math>\begin{align} \quad a \cdot b &= e \\ \quad a^{-1} \cdot (a \cdot b) &= a^{-1} \cdot e \\ \quad (a^{-1} \cdot a) \cdot b &= a^{-1} \\ \quad e \cdot b &= a^{-1} \\ \quad b &= a^{-1} \quad \blacksquare \end{align}</math></div>
 +
<ul>
 +
<li><strong>Proof of f)</strong> Suppose that <span class="math-inline"><math>a^2 = a \cdot a = a</math></span>. Then:</li>
 +
</ul>
 +
<div style="text-align: center;"><math>\begin{align} \quad a^2 &= a \\ \quad a \cdot a &= a \\ \quad a^{-1} \cdot (a \cdot a) &= a^{-1} \cdot a \\ \quad (a^{-1} \cdot a) \cdot a &= e \\ \quad e \cdot a &= e \\ \quad a &= e \end{align}</math></div>
 +
<ul>
 +
<li>Hence <span class="math-inline"><math>a = e</math></span>. Alternatively we see that if <span class="math-inline"><math>a \cdot a = a</math></span> then the inverse of <span class="math-inline"><math>a</math></span> with respect to <span class="math-inline"><math>\cdot</math></span> is <span class="math-inline"><math>e</math></span>, that is <span class="math-inline"><math>a^{-1} = e</math></span>. Multiplying both sides of this equation by <span class="math-inline"><math>a</math></span> gives us that <span class="math-inline"><math>a = e</math></span>. <span class="math-inline"><math>\blacksquare</math></span></li>
 +
</ul>
 +
 +
 +
== Subgroups and Group Extensions==
 +
 +
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 +
<td><strong>Definition:</strong> Let <span class="math-inline"><math>(G, \cdot)</math></span> be a group. If <span class="math-inline"><math>S \subseteq G</math></span> and <span class="math-inline"><math>S</math></span> forms a group under the same operation <span class="math-inline"><math>\cdot</math></span> then <span class="math-inline"><math>(S, \cdot)</math></span> is said to be a <strong>Subgroup</strong> of <span class="math-inline"><math>(G, \cdot)</math></span>. If <span class="math-inline"><math>S</math></span> is a subgroup of <span class="math-inline"><math>G</math></span> then we write <span class="math-inline"><math>S \leq G</math></span>.</td>
 +
</blockquote>
 +
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 +
<td><strong>Definition:</strong> Let <span class="math-inline"><math>(G, \cdot)</math></span> be a group. If <span class="math-inline"><math>(H, \cdot)</math></span> is a group such that <span class="math-inline"><math>G \subseteq H</math></span> then <span class="math-inline"><math>(H, \cdot)</math></span> is said to be a <strong>Group Extension</strong> of <span class="math-inline"><math>(G, \cdot)</math></span>.</td>
 +
</blockquote>
 +
<p>For example, consider the group of complex numbers under the operation of standard addition, <span class="math-inline"><math>(\mathbb{C}, +)</math></span>. We know that the set of real numbers is a subset of the set of complex numbers, that is, <span class="math-inline"><math>\mathbb{R} \subset \mathbb{C}</math></span> and so the group of real numbers under the operation of standard addition, <span class="math-inline"><math>(\mathbb{R}, +)</math></span> is a subgroup of <span class="math-inline"><math>(\mathbb{C}, +)</math></span> and <span class="math-inline"><math>(\mathbb{C}, +)</math></span> is a group extension of <span class="math-inline"><math>(\mathbb{R}, +)</math></span>.</p>
 +
<p>We will now look at a nice theorem which tells us that to determine if <span class="math-inline"><math>(S, \cdot)</math></span> is a subgroup of <span class="math-inline"><math>(G, \cdot)</math></span> where <span class="math-inline"><math>S \subseteq G</math></span>, that then we only need to check two of the four group axioms for verification</p>
 +
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 +
<td><strong>Theorem 1:</strong> If <span class="math-inline"><math>(G, \cdot)</math></span> is a group with the identity <span class="math-inline"><math>e \in G</math></span> and <span class="math-inline"><math>S \subseteq G</math></span> then <span class="math-inline"><math>(S, \cdot)</math></span> is a subgroup of <span class="math-inline"><math>(G, \cdot)</math></span> if and only if <span class="math-inline"><math>S</math></span> is closed under <span class="math-inline"><math>\cdot</math></span> and for all <span class="math-inline"><math>a \in S</math></span> there exists an <span class="math-inline"><math>a^{-1} \in S</math></span> such that <span class="math-inline"><math>a \cdot a^{-1} = e</math></span> and <span class="math-inline"><math>a^{-1} \cdot a = e</math></span>.</td>
 +
</blockquote>
 +
<ul>
 +
<li><strong>Proof:</strong> Let <span class="math-inline"><math>(G, \cdot)</math></span> be a group with the identity <span class="math-inline"><math>e \in G</math></span> of <span class="math-inline"><math>\cdot</math></span> and let <span class="math-inline"><math>S \subseteq G</math></span>.</li>
 +
</ul>
 +
<ul>
 +
<li><span class="math-inline"><math>\Rightarrow</math></span> Suppose that <span class="math-inline"><math>(S, \cdot)</math></span> is a subgroup of <span class="math-inline"><math>(G, \cdot)</math></span>. Then by definition, <span class="math-inline"><math>(S, \cdot)</math></span> is a group itself and satisfies all of the group axioms - namely that <span class="math-inline"><math>S</math></span> is closed under the operation <span class="math-inline"><math>\cdot</math></span> and that for all <span class="math-inline"><math>a \in S</math></span> there exists an <span class="math-inline"><math>a^{-1} \in S</math></span> such that <span class="math-inline"><math>a \cdot a^{-1} = e</math></span> and <span class="math-inline"><math>a^{-1} \cdot a = e</math></span>.</li>
 +
</ul>
 +
<ul>
 +
<li><span class="math-inline"><math>\Leftarrow</math></span> Now suppose that <span class="math-inline"><math>S</math></span> is closed under <span class="math-inline"><math>\cdot</math></span> and that for all <span class="math-inline"><math>a \in S</math></span> there exists an <span class="math-inline"><math>a^{-1} \in S</math></span> such that <span class="math-inline"><math>a \cdot a^{-1} = e</math></span> and <span class="math-inline"><math>a^{-1} \cdot a = e</math></span>. These are precisely two of the group axioms we have looked at, and to show that <span class="math-inline"><math>(S, \cdot)</math></span> is a subgroup of <span class="math-inline"><math>(G, \cdot)</math></span> we only need to show that the other two axioms hold.</li>
 +
</ul>
 +
<ul>
 +
<li>First suppose that <span class="math-inline"><math>a, b, c \in S</math></span> and that <span class="math-inline"><math>a \cdot (b \cdot c) \neq (a \cdot b) \cdot c</math></span>, that is, suppose that <span class="math-inline"><math>\cdot</math></span> is not associative on <span class="math-inline"><math>S</math></span>. Since <span class="math-inline"><math>S \subseteq G</math></span> we must have that <span class="math-inline"><math>a \cdot (b \cdot c) \neq (a \cdot b) \cdot c</math></span> for this particular <span class="math-inline"><math>a, b, c \in G</math></span> which contradicts the associativity of <span class="math-inline"><math>\cdot</math></span> on the group <span class="math-inline"><math>G</math></span>. Hence <span class="math-inline"><math>\cdot</math></span> must actually be associative on <span class="math-inline"><math>G</math></span>.</li>
 +
</ul>
 +
<ul>
 +
<li>Now since <span class="math-inline"><math>S</math></span> is closed under <span class="math-inline"><math>\cdot</math></span> and for <span class="math-inline"><math>a \in S</math></span> there exists an <span class="math-inline"><math>a^{-1} \in S</math></span> such that <span class="math-inline"><math>a \cdot a^{-1} = e</math></span> and <span class="math-inline"><math>a^{-1} \cdot a = e</math></span> we must have that <span class="math-inline"><math>e \in S</math></span> and furthermore, <span class="math-inline"><math>a \cdot e = a</math></span> and <span class="math-inline"><math>e \cdot a = a</math></span>.</li>
 +
</ul>
 +
<ul>
 +
<li>Therefore <span class="math-inline"><math>(S, \cdot)</math></span> is a group, and in particular since <span class="math-inline"><math>S \subseteq G</math></span> we have that <span class="math-inline"><math>(S, \cdot)</math></span> is a subgroup of <span class="math-inline"><math>(G, \cdot)</math></span>. <span class="math-inline"><math>\blacksquare</math></span></li>
 +
</ul>
 +
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 +
<td><strong>Theorem 2:</strong> If <span class="math-inline"><math>(G, \cdot)</math></span> is a group with the identity <span class="math-inline"><math>e \in G</math></span> of <span class="math-inline"><math>\cdot</math></span> and <span class="math-inline"><math>S \subseteq G</math></span> then <span class="math-inline"><math>(S, \cdot)</math></span> is a subgroup of <span class="math-inline"><math>(G, \cdot)</math></span> if and only if <span class="math-inline"><math>H \neq \emptyset</math></span> and for all <span class="math-inline"><math>a, b \in H</math></span> we have that <span class="math-inline"><math>a \cdot b^{-1} \in H</math></span>.</td>
 +
</blockquote>
 +
<ul>
 +
<li><strong>Proof:</strong> <span class="math-inline"><math>\Rightarrow</math></span> If <span class="math-inline"><math>(S, \cdot) </math> is a subgroup of <math> (G, \cdot)</math></span> then this direction is trivial.</li>
 +
</ul>
 +
<ul>
 +
<li><span class="math-inline"><math>\Leftarrow</math></span> Suppose that <span class="math-inline"><math>S \neq \emptyset</math></span> and for all <span class="math-inline"><math>a, b \in S</math></span> we have that <span class="math-inline"><math>a \cdot b^{-1} \in S</math></span>. Since <span class="math-inline"><math>S \neq \emptyset</math></span> there exists an <span class="math-inline"><math>a \in S</math></span>. So <span class="math-inline"><math>a \cdot a^{-1} = e \in S</math></span>.</li>
 +
</ul>
 +
<ul>
 +
<li>Now if <span class="math-inline"><math>a \in S</math></span>, then since <span class="math-inline"><math>e \in S</math></span> we have that <span class="math-inline"><math>e \cdot a^{-1} = a^{-1} \in S</math></span>. So if <span class="math-inline"><math>a \in S</math></span> then <span class="math-inline"><math>a^{-1} \in H</math></span>.</li>
 +
</ul>
 +
<ul>
 +
<li>Lastly, if <span class="math-inline"><math>a, b \in S</math></span> then <span class="math-inline"><math>a, b^{-1} \in S</math></span>. Thus <span class="math-inline"><math>a \cdot (b^{-1})^{-1} = a \cdot b \in S</math></span>. So <span class="math-inline"><math>S</math></span> is closed under the operation <span class="math-inline"><math>\cdot</math></span>. Thus <span class="math-inline"><math>(S, \cdot)</math></span> is a subgroup of <span class="math-inline"><math>(G, \cdot)</math></span>. <span class="math-inline"><math>\blacksquare</math></span></li>
 +
</ul>
 +
 +
==Order of an element in a group==
 +
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 +
<td><strong>Definition:</strong> Let <span class="math-inline"><math>(G, *)</math></span> be a group and let <span class="math-inline"><math>a \in G</math></span>. The <strong>Order</strong> of <span class="math-inline"><math>a</math></span> denoted by <span class="math-inline"><math>|a|</math></span> or <span class="math-inline"><math>\text{ord}(a)</math></span> is the smallest positive integer <span class="math-inline"><math>m</math></span> such that <span class="math-inline"><math>a^m = \underbrace{a \cdot a \cdot ... \cdot a}_{m : \text{many factors}} = e</math></span> (where <span class="math-inline"><math>e</math></span> is the identity element of <span class="math-inline"><math>G</math></span>). If no such <span class="math-inline"><math>m</math></span> exists then <span class="math-inline"><math>a</math></span> is said to have order <span class="math-inline"><math>\infty</math></span>.</td>
 +
</blockquote>
 +
<p><em>If the operation is multiplicative in nature then we usually define the order of <span class="math-inline"><math>a \in G</math></span> as above. If the operation is instead additive in nature then we define the order of <span class="math-inline"><math>a \in G</math></span> as the smallest positive integer <span class="math-inline"><math>m</math></span> such that <span class="math-inline"><math>ma = \underbrace{(a + a + ... + a)}_{m \; \text{many}} = e</math></span> or <span class="math-inline"><math>\infty</math></span> if no such positive integer <span class="math-inline"><math>m</math></span> exists.</em></p>
 +
<h3 id="toc1"><span>Example 1</span></h3>
 +
<p>If <span class="math-inline"><math>G</math></span> is any group with identity <span class="math-inline"><math>e</math></span> then the order of <span class="math-inline"><math>e</math></span> is <span class="math-inline"><math>1</math></span>.</p>
 +
<h3 id="toc2"><span>Example 2</span></h3>
 +
<p>Consider the group <span class="math-inline"><math>(\mathbb{Z}_5, +)</math></span> where <span class="math-inline"><math>+</math></span> is defined for all <span class="math-inline"><math>x, y \in \mathbb{Z}_5 = \{ 0, 1, 2, 3, 4 \}</math></span> to be:</p>
 +
<div style="text-align: center;"><math>\begin{align} \quad x + y = (x + y) \mod 5 \end{align}</math></div>
 +
<p>The order of <span class="math-inline"><math>e = 0</math></span> is trivially <span class="math-inline"><math>1</math></span>. The order of <span class="math-inline"><math>1</math></span> is <span class="math-inline"><math>5</math></span> since:</p>
 +
<div style="text-align: center;"><math>\begin{align} \quad 1^5 = (1 + 1 + 1 + 1 + 1) = 0 \mod 5 \end{align}</math></div>
 +
<p>The order of <span class="math-inline"><math>2</math></span> is also <span class="math-inline"><math>5</math></span>. In fact, the orders of <span class="math-inline"><math>3</math></span> and <span class="math-inline"><math>4</math></span> are also <span class="math-inline"><math>5</math></span>.</p>
 +
<h3 id="toc3"><span>Example 3</span></h3>
 +
<p>Consider the group <span class="math-inline"><math>(\mathbb{Z}_6, +)</math></span> where <span class="math-inline"><math>+</math></span> is defined for all <span class="math-inline"><math>x, y \in \mathbb{Z}_6 = \{ 0, 1, 2, 3, 4, 5 \}</math></span> to be:</p>
 +
<div style="text-align: center;"><math>\begin{align} \quad x + y = (x + y) \mod 6 \end{align}</math></div>
 +
<p>You should verify that the order of <span class="math-inline"><math>0</math></span> is <span class="math-inline"><math>1</math></span>, the order of <span class="math-inline"><math>1</math></span> is <span class="math-inline"><math>6</math></span>, the order of <span class="math-inline"><math>2</math></span> is <span class="math-inline"><math>3</math></span>, the order of <span class="math-inline"><math>3</math></span> is <span class="math-inline"><math>2</math></span>, the order of <span class="math-inline"><math>4</math></span> is <span class="math-inline"><math>3</math></span>, and the order of <span class="math-inline"><math>5</math></span> is <span class="math-inline"><math>6</math></span></p>
 +
<h3 id="toc4"><span>Example 4</span></h3>
 +
<p>Consider the group <span class="math-inline"><math>(\mathbb{R}, +)</math></span>. Then every nonzero <span class="math-inline"><math>x \in \mathbb{R}</math></span> has order infinity since the equation <span class="math-inline"><math>mx = 0</math></span>, <span class="math-inline"><math>m \geq 1</math></span>, has no solution in <span class="math-inline"><math>\mathbb{R} \setminus \{ 0 \}</math></span>.</p>
 +
<h3 id="toc5"><span>Example 5</span></h3>
 +
<p>Consider the group <span class="math-inline"><math>(\mathbb{R} \setminus \{ 0 \}, \cdot)</math></span>. The order of <span class="math-inline"><math>-1</math></span> is <span class="math-inline"><math>2</math></span> since <span class="math-inline"><math>(-1)^2 = 1</math></span>.</p>
 +
 +
==Basic Theorems Regarding the Order of Elements in a Group==
 +
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 +
<td><strong>Proposition 1:</strong> Let <span class="math-inline"><math>(G, \cdot)</math></span> be a group. Then:<br />
 +
<strong>a)</strong> <span class="math-inline"><math>\mathrm{order}(a) = 1</math></span> if and only if <span class="math-inline"><math>a = e</math></span> where <span class="math-inline"><math>e</math></span> is the identity element of <span class="math-inline"><math>G</math></span>.<br />
 +
<strong>b)</strong> If <span class="math-inline"><math>a \in G</math></span> then <span class="math-inline"><math>\mathrm{ord}(a^{-1}) = \mathrm{ord}(a)</math></span>.<br />
 +
<strong>c)</strong> If <span class="math-inline"><math>a, b \in G</math></span> then <span class="math-inline"><math>\mathrm{ord}(a \cdot b) = \mathrm{ord}(b \cdot a)</math></span>. _</td>
 +
</blockquote>
 +
<ul>
 +
<li><strong>Proof of a)</strong> <span class="math-inline"><math>\Rightarrow</math></span> Suppose that <span class="math-inline"><math>\mathrm{ord}(a)</math></span>. Then <span class="math-inline"><math>a^1 = e</math></span>. So <span class="math-inline"><math>a = e</math></span>.</li>
 +
</ul>
 +
<ul>
 +
<li><span class="math-inline"><math>\Leftarrow</math></span> The smallest positive integer such that <span class="math-inline"><math>e^n = e</math></span> is <span class="math-inline"><math>n = 1</math></span>. So <span class="math-inline"><math>\mathrm{ord}(e) = 1</math></span>. <span class="math-inline"><math>\blacksquare</math></span></li>
 +
</ul>
 +
<ul>
 +
<li><strong>Proof of b)</strong> Let <span class="math-inline"><math>a \in G</math></span>. Suppose that <span class="math-inline"><math>a</math></span> has finite order, say <span class="math-inline"><math>\mathrm{ord}(a) = n</math></span>. Then <span class="math-inline"><math>n</math></span> is the smallest positive integer such that <span class="math-inline"><math>a^n = e</math></span>. So <span class="math-inline"><math>a^{-n} = e</math></span>. So <span class="math-inline"><math>\mathrm{ord}(a^{-1}) \leq n</math></span>. If <span class="math-inline"><math>\mathrm{ord}(a^{-1}) = m < n</math></span> then <span class="math-inline"><math>a^{-m} = e</math></span> implies that <span class="math-inline"><math>a^m = e</math></span>, and since <span class="math-inline"><math>m < n = \mathrm{ord}(a)</math></span> we have arrived at a contradiction. Thus <span class="math-inline"><math>\mathrm{ord}(a^{-1}) = \mathrm{ord}(a)</math></span>.</li>
 +
</ul>
 +
<ul>
 +
<li>Now suppose that <span class="math-inline"><math>a</math></span> has infinite order. If <span class="math-inline"><math>a^{-1}</math></span> has finite order, say <span class="math-inline"><math>\mathrm{ord}(a^{-1}) = m < \infty</math></span> then <span class="math-inline"><math>a^{-m} = e</math></span> implies that <span class="math-inline"><math>a^m = e</math></span> - contradicting <span class="math-inline"><math>a</math></span> having infinite order. Thus <span class="math-inline"><math>a^{-1}</math></span> must also have infinite order.</li>
 +
</ul>
 +
<ul>
 +
<li><strong>Proof of c)</strong> Let <span class="math-inline"><math>a, b \in G</math></span>. First, suppose that <span class="math-inline"><math>\mathrm{ord}(a \cdot b) = n < \infty</math></span>. Then <span class="math-inline"><math>n</math></span> is the smallest positive integer such that <span class="math-inline"><math>(a \cdot b)^n = e</math></span>. So <span class="math-inline"><math>(a \cdot b)^n \cdot a = a</math></span>. Now observe that:</li>
 +
</ul>
 +
<div style="text-align: center;"><math>\begin{align} \quad (a \cdot b)^n \cdot a = \underbrace{(a \cdot b) \cdot (a \cdot b) \cdot ... \cdot (a \cdot b)}_{n \; \text{factors}} \cdot a = a \cdot \underbrace{(b \cdot a) \cdot (b \cdot a) \cdot ... \cdot (b \cdot a)}_{n \; \text{factors}} = a \cdot (b \cdot a)^n \end{align}</math></div>
 +
<ul>
 +
<li>Therefore <span class="math-inline"><math>(a \cdot b)^n \cdot a = a \cdot (b \cdot a)^n</math></span>. But <span class="math-inline"><math>(a \cdot b)^n = e</math></span>, so <span class="math-inline"><math>a = a \cdot (b \cdot a)^n</math></span>. Thus <span class="math-inline"><math>(b \cdot a)^n = e</math></span>. So <span class="math-inline"><math>\mathrm{ord}(b \cdot a) \leq n = \mathrm{ord}(a \cdot b)</math></span>.</li>
 +
</ul>
 +
<ul>
 +
<li>By symmetry, we see that <span class="math-inline"><math>\mathrm{ord}(a \cdot b) \leq \mathrm{ord}(b \cdot a)</math></span>. Thus <span class="math-inline"><math>\mathrm{ord}(a \cdot b) = \mathrm{ord}(b \cdot a)</math></span>.</li>
 +
</ul>
 +
<ul>
 +
<li>Now suppose that <span class="math-inline"><math>\mathrm{ord}(a \cdot b)</math></span> is infinite. If <span class="math-inline"><math>\mathrm{ord}(b \cdot a) = n < \infty</math></span> then <span class="math-inline"><math>(b \cdot a)^n = e</math></span>. By the same argument above, we see that <span class="math-inline"><math>(a \cdot b)^n = e</math></span> - contradicting <span class="math-inline"><math>a \cdot b</math></span> having infinite order. Thus <span class="math-inline"><math>\mathrm{ord}(b \cdot a)</math></span> is infinite. <span class="math-inline"><math>\blacksquare</math></span></li>
 +
</ul>
  
 
==Licensing==
 
==Licensing==
 
Content obtained and/or adapted from:
 
Content obtained and/or adapted from:
* [] under a CC BY-SA license
+
* [http://mathonline.wikidot.com/groups Groups, mathonline.wikidot.com] under a CC BY-SA license
 +
* [http://mathonline.wikidot.com/basic-theorems-regarding-groups Basic Theorems Regarding Groups, mathonline.wikidot.com] under a CC BY-SA license
 +
* [http://mathonline.wikidot.com/subgroups-and-group-extensions Subgroups and Group Extensions, mathonline.wikidot.com] under a CC BY-SA license

Latest revision as of 17:12, 16 November 2021

Groups

Recall that an operation on is said to be associative if for all we have that and is said to be commutative if for all we have that .

An element is the identity element of under if for all we have that and .

We can now begin to describe our first type of algebraic structures known as groups, which are a set equipped with a binary operation that is associative, contains an identity element, and contains inverse elements under for each element in .

Definition: A Group is a pair where is a set and is a binary operation on with the following properties:

1. For all , (Associativity of ).
2. There exists an such that for all , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e \cdot a = a} (The existence of an Identity Element).
3. For all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \in G} there exists an such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \cdot a^{-1} = e} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^{-1} \cdot a = e} (The existence of inverses).

Furthermore, if is a finite set then the group Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (G, \cdot)} is said to be a Finite Group and if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G} is an infinite set then the group is said to be an Infinite Group. More generally, the Order of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (G, \cdot)} (or **Size of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (G, \cdot)} ) is the size of and is denoted Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle | G |} .

When we use the multiplication symbol Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cdot} to denote the operation on , we often call Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G} a “multiplicative group”. When the operation of the group is instead denoted by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle +} (instead of ) then we often call Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G} an “additive group”, and we write the inverse of each Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \in G} as (instead of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^{-1}} ).

Some of the sets and binary operations we have already seen can be considered groups. For example, is a group under standard addition since the sum of any two real numbers is a real number, , is associative, an additive identity exists and inverse elements exist for every (namely ).

Furthermore, is also a group under the operation of standard addition since the sum of any two integers is an integer, addition is associative, the additivity identity is , and for all we have as additive inverses.

We will examine many other (more interesting) groups later on, but for now, let's look at an example of a set and a binary operation that does NOT form a group.

Example 1

Consider the set of integers and define for all by:

(Where the on the righthand side is usual addition of numbers). We will show that is NOT a group by showing that is not associative. Let . Then is not associative since:

Clearly so does not form a group under the operation .

Basic Theorems Regarding Groups

A group is a set with a binary operation such that:

  • 1) is associative, i.e., for all , .
  • 2) There exists an identity element such that for all .
  • 3) For each there exists an such that .

We will now look at some rather basic results regarding groups which we can derive from the group axioms above.

Proposition 1: Let be a group and let be the identity for this group. Then:

a) The identity element is unique.
b) For each , the corresponding inverse is unique.
c) For each , .
d) For all , .
e) For all , if then and .

f) If then .

  • Proof of a) Suppose that and are both identities for . Then:
  • Therefore so the identity for is unique.
  • Proof of b) Suppose that and are both inverses for under . Then:
  • Therefore so the inverse for is unique.
  • Proof of c) Let . Then is the inverse to . However, the inverse to is and by (b) we have shown that the inverse of each element in is unique. Therefore .
  • Proof of d) If we apply the operation between and we get:
  • Therefore the inverse of is . We also have that the invere of is . By (b), the inverse of is unique and so:
  • Proof of e) Suppose that . Then:
  • Similarly:
  • Proof of f) Suppose that . Then:
  • Hence . Alternatively we see that if then the inverse of with respect to is , that is . Multiplying both sides of this equation by gives us that .


Subgroups and Group Extensions

Definition: Let be a group. If and forms a group under the same operation Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cdot} then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (S, \cdot)} is said to be a Subgroup of . If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S} is a subgroup of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G} then we write Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S \leq G} .

Definition: Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (G, \cdot)} be a group. If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (H, \cdot)} is a group such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G \subseteq H} then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (H, \cdot)} is said to be a Group Extension of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (G, \cdot)} .

For example, consider the group of complex numbers under the operation of standard addition, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\mathbb{C}, +)} . We know that the set of real numbers is a subset of the set of complex numbers, that is, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{R} \subset \mathbb{C}} and so the group of real numbers under the operation of standard addition, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\mathbb{R}, +)} is a subgroup of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\mathbb{C}, +)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\mathbb{C}, +)} is a group extension of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\mathbb{R}, +)} .

We will now look at a nice theorem which tells us that to determine if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (S, \cdot)} is a subgroup of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (G, \cdot)} where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S \subseteq G} , that then we only need to check two of the four group axioms for verification

Theorem 1: If is a group with the identity Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e \in G} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S \subseteq G} then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (S, \cdot)} is a subgroup of if and only if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S} is closed under Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cdot} and for all there exists an Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^{-1} \in S} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \cdot a^{-1} = e} and .

  • Proof: Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (G, \cdot)} be a group with the identity Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e \in G} of and let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S \subseteq G} .
  • Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow} Suppose that is a subgroup of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (G, \cdot)} . Then by definition, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (S, \cdot)} is a group itself and satisfies all of the group axioms - namely that is closed under the operation Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cdot} and that for all there exists an Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^{-1} \in S} such that and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^{-1} \cdot a = e} .
  • Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Leftarrow} Now suppose that is closed under Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cdot} and that for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \in S} there exists an such that and . These are precisely two of the group axioms we have looked at, and to show that is a subgroup of we only need to show that the other two axioms hold.
  • First suppose that and that , that is, suppose that is not associative on . Since we must have that for this particular which contradicts the associativity of on the group . Hence must actually be associative on .
  • Now since is closed under and for there exists an such that and we must have that and furthermore, and .
  • Therefore is a group, and in particular since we have that is a subgroup of .

Theorem 2: If is a group with the identity of and then is a subgroup of if and only if and for all we have that .

  • Proof: If is a subgroup of then this direction is trivial.
  • Suppose that and for all we have that . Since there exists an . So .
  • Now if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \in S} , then since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e \in S} we have that . So if then .
  • Lastly, if then . Thus . So is closed under the operation . Thus is a subgroup of .

Order of an element in a group

Definition: Let be a group and let . The Order of denoted by or Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{ord}(a)} is the smallest positive integer Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m} such that (where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e} is the identity element of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G} ). If no such Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m} exists then is said to have order Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \infty} .

If the operation is multiplicative in nature then we usually define the order of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \in G} as above. If the operation is instead additive in nature then we define the order of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \in G} as the smallest positive integer Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ma = \underbrace{(a + a + ... + a)}_{m \; \text{many}} = e} or if no such positive integer Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m} exists.

Example 1

If is any group with identity Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e} then the order of is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1} .

Example 2

Consider the group Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\mathbb{Z}_5, +)} where is defined for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x, y \in \mathbb{Z}_5 = \{ 0, 1, 2, 3, 4 \}} to be:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \quad x + y = (x + y) \mod 5 \end{align}}

The order of is trivially Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1} . The order of is since:

The order of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2} is also . In fact, the orders of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 4} are also .

Example 3

Consider the group Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\mathbb{Z}_6, +)} where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle +} is defined for all to be:

You should verify that the order of is , the order of is , the order of is , the order of is , the order of is , and the order of is

Example 4

Consider the group . Then every nonzero has order infinity since the equation , , has no solution in .

Example 5

Consider the group . The order of is since .

Basic Theorems Regarding the Order of Elements in a Group

Proposition 1: Let be a group. Then:

a) if and only if where is the identity element of .
b) If then .

c) If then . _

  • Proof of a) Suppose that . Then . So .
  • The smallest positive integer such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^n = e} is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n = 1} . So Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{ord}(e) = 1} . Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \blacksquare}
  • Proof of b) Let . Suppose that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} has finite order, say Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{ord}(a) = n} . Then is the smallest positive integer such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^n = e} . So Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^{-n} = e} . So . If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{ord}(a^{-1}) = m < n} then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^{-m} = e} implies that , and since we have arrived at a contradiction. Thus .
  • Now suppose that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} has infinite order. If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^{-1}} has finite order, say Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{ord}(a^{-1}) = m < \infty} then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^{-m} = e} implies that - contradicting Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} having infinite order. Thus Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^{-1}} must also have infinite order.
  • Proof of c) Let . First, suppose that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{ord}(a \cdot b) = n < \infty} . Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} is the smallest positive integer such that . So Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a \cdot b)^n \cdot a = a} . Now observe that:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \quad (a \cdot b)^n \cdot a = \underbrace{(a \cdot b) \cdot (a \cdot b) \cdot ... \cdot (a \cdot b)}_{n \; \text{factors}} \cdot a = a \cdot \underbrace{(b \cdot a) \cdot (b \cdot a) \cdot ... \cdot (b \cdot a)}_{n \; \text{factors}} = a \cdot (b \cdot a)^n \end{align}}
  • Therefore . But Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a \cdot b)^n = e} , so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a = a \cdot (b \cdot a)^n} . Thus . So Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{ord}(b \cdot a) \leq n = \mathrm{ord}(a \cdot b)} .
  • By symmetry, we see that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{ord}(a \cdot b) \leq \mathrm{ord}(b \cdot a)} . Thus .
  • Now suppose that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{ord}(a \cdot b)} is infinite. If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{ord}(b \cdot a) = n < \infty} then . By the same argument above, we see that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a \cdot b)^n = e} - contradicting having infinite order. Thus Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{ord}(b \cdot a)} is infinite. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \blacksquare}

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