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.