Let $G$ be a group of even order. Define $t(G)$ to be the set $\{ g \in G\mid g\neq g^{-1}\}$. Show that $t(G)$ has an even number of elements.

Question

Let $G$ be a group of even order. Define $t(G)$ to be the set $\{g\in G\mid g\neq g^{-1}\}$. Show that $t(G)$ has an even number of elements.

I saw there are many solutions related to this question but I want to verify whether my solution is correct or not because I think it might be different from Prove that any finite group $G$ of even order contains an element of order $2.$

Here is my work,

take any $x \in t(G)$ now I consider $2$ cases where the order of $x$ is odd and even.

Case 1: When the order of $x$ is odd, then it is clear that for any non-identity element $a \in\langle x\rangle$, $a \in t(G)$ by the Lagrange theorem. so it immediately follows $\langle x\rangle\setminus \{ e\} $ have even number of element.

Case 2: When the order of $x$ is even ($|x|=2k$), we have only the $x^k$ has order $2$ in $\langle x\rangle$ and other elements also in $t(G)$ so it immediately follows $\langle x\rangle\setminus \{ x^k,e\} $ have even number of the element.

So by case $1$ and case $2$, we can conclude that $t(G)$ has an even number of elements.

Can anyone verify my answer?

Answer 1

Note this statement is true, no matter $|G|$ is even or odd. Whenever $g\in t(G)\Rightarrow g\neq g^{-1}$, then we have $g^{-1}\in t(G)$, because $(g^{-1})^{-1}\neq g^{-1}$. This implies $|t(G)|$ is even.

Answer 2

Well, if $g\in t(G)$ so does $g^{-1}$. And, from definition they are different. So, one can pair elements of $t(G)$.