Exercise 1.1.31 (Any finite group $G$ of even order contains an element of order $2$)

Question

Prove that any finite group $G$ of even order contains an element of order $2$. [Let $t(G)$ be the set $\{g \in G \mid g
e g^{-1}\}$. Shows that $t(G)$ has an even number of elements and every non identity element of $G - t(G)$ has order $2$.]

Richard Ganaye · Accepted Answer

\begin{proof} Let $G$ be a finite group of even order $2n$. We define the set
$$t(G) = \{g \in G \mid g
e g^{-1}\}.$$
The map $\varphi : G 	o G$ defined by $\varphi(g) = g^{-1}$ is an involution (i.e., satisfies $\varphi \circ \varphi = \mathrm{id}_G$). Therefore 
$$G = \bigcup_{g \in G} \{g, g^{-1}\},$$ 
where $\{g, g^{-1}\}$ has two distinct elements if $g \in t(G)$, and one element if $g \in G - t(G)$.

So we can group the elements of $t(G)$ by pairs $\{g,g^{-1}\}$, where $g 
e g^{-1}$. Therefore $t(G)$ has an even number of elements.
Hence $G - t(G)$ has also an even number of elements, and
\begin{align*}
G - t(G) &= \{g \in G \mid g =  g^{-1}\}\
&= \{g \in G \mid g^2 = 1\}.
\end{align*}
Since $1 \in G - t(G)$, there at least another element  $a \in G$, which satisfies $a 
e 1$ and $a^2 = 1$, so $|a| = 2$.

So any finite group $G$ of even order contains an element of order $2$.
\end{proof}

Exercise 1.1.31 (Any finite group $G$ of even order contains an element of order $2$)

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Exercise 1.1.31 (Any finite group $G$ of even order contains an element of order $2$)

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