Gallian: Ch. 3 #35

Problem Statement: Prove that a group of even order must have an element of order 2.

Proof: Let G be a group such that \left|G\right|=2k and consider the set S=\{g\in G:g\neq g^{-1}\}. S is the set of all elements of G that are not of order 2.

I claim that \left|S\right| is even. Let s\in S, then s\neq s^{-1} and so s^{-1}\in S as well. So no element may be in S unless it’s inverse is also an element of S, thus, \left|S\right| is even. Let \left|S\right|=2j.

Note that any element of G is either in S or in G-S, by construction. Furthermore, S\cap(G-S)=\varnothing and S\cup(G-S)=G and so it follows that \left|G\right|=\left|S\right|+\left|G-S\right|. This implies that 2k=2j+n and so it must be that n=\left|G-S\right| is even. Let \left|G-S\right|=2m.

G-S=\{e, b_2, b_3, \dots, b_{2m} \} where m\geq 1. Since m must be \geq 1 it follows that there is always some b\in G-S that is not the identity element. Since b\in G-S it follows that b=b^{-1} which implies that b^2=e and so \left|b\right|=2.

Thus, any group of even order must have at least one element of order 2.

\Box

Reflection: The key to this proof was breaking the group into two subsets, elements of order 2 and elements that were not of order 2. The other major idea is that every element is the inverse of it’s inverse. i.e. g=(g^{-1})^{-1}. This forces our set S to have even order which in turn forces the set we are interested in to also have even order. So, even though we know the identity is always in G-S, since the order is even there must always be at least one other, non-identity element in G-S as well.

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