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Chapter 16: Problem 4

If \(G\) is a group of even order, prove that there is an element \(a \in G\) with \(a \neq e\) and \(a=a^{-1}\).

Short Answer

Expert verified
If a group \(G\) is of even order, there are an even number of elements. Subtracting the identity element leaves an odd number of non-identity elements. By the cancellation law, these remaining elements can be paired with their inverses. As we cannot pair an odd number of elements without one being left over, there must be at least one element in the group which is its own inverse.

Step by step solution

01

Understand the Group's Order

The order of a group is the number of its elements. If a group \(G\) is of even order, that means the number of elements in \(G\) is an even number, say \(2n\), where \(n\) is a whole number.
02

Extract the Identity Element

Every group \(G\) has an identity element, often denoted by \(e\). This element has the property that for any element \(a\) in \(G\), \(ea=ae=a\). This leaves \(2n-1\) elements in \(G\) that are not the identity element.
03

Pair Non-Identity Elements With Their Inverses

The remaining \(2n-1\) non-identity elements can be paired with their inverses. The group's closure property ensures that for every element \(a\), there is an associated element \(a^{-1}\) such that \(aa^{-1}=a^{-1}a=e\). Because \(2n-1\) is an odd number, there must be at least one element that is not paired, i.e., it is its own inverse.
04

Conclude the Proof

Since for every non-identity element \(a \in G\), if \(a \neq a^{-1}\), \(a\) and \(a^{-1}\) are considered as a pair. As the total number of non-identity elements is an odd number, after pairing there should be at least one element left, which can only be its own inverse. Thus, in a group of even order, there is always an element \(a\) such that \(a \neq e\) and \(a=a^{-1}\). This completes the proof.

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Most popular questions from this chapter

For \(k, n \in \mathbf{Z}^{+}\)where \(n \geq 2\) and \(1 \leq k \leq n\), let \(P(n, k)\) denote the number of permutations \(\pi \in S_{n}\) that have \(k\) cycles. (For example, \((1)(23)\) is counted in \(P(3,2)\), (12) \((34)\) is counted in \(P(4,2)\), and \((1)(23)(4)\) is counted in \(P(4,3) .)\) a) Verify that \(P(n+1, k)=P(n, k-1)+n P(n, k)\). b) Determine \(\sum_{k=1}^{n} P(n, k)\).

Let \(f: G \rightarrow H\) be a group homomorphism onto \(H\). If \(G\) is a cyclic group, prove that \(H\) is also cyclic.

a) Find all the elements of order 10 in \(\left(\mathrm{Z}_{40},+\right)\). b) Let \(G=\langle a\rangle\) be a cyclic group of order 40 . Which elements of \(G\) have order 10?

If \(G\) is a group of order \(n\) and \(a \in G\), prove that \(a^{n}=e\).

The following provides an alternative way to establish Lagrange's Theorem. Let \(G\) be a group of order \(n\), and let \(H\) be a subgroup of \(G\) of order \(m\). a) Define the relation \(\mathscr{\text { on }} G\) as follows: If \(a, b \in G\), then \(a \mathscr{F} b\) if \(a^{-1} b \in H\). Prove that \(\mathscr{\text { is an equivalence relation on }} G\). b) For \(a, b \in G\), prove that \(a \mathscr{b}\) if and only if \(a H=b H .\) c) If \(a \in G\), prove that \([a]\), the equivalence class of \(a\) under \(\mathscr{A}\), satisfies \([a]=a H .\) d) For any \(a \in G\), prove that \(|a H|=|H|\). e) Now establish the conclusion of Lagrange's Theorem, namely that \(|H|\) divides \(|G|\).
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