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I tried it using proof by contradiction:

Suppose that there are even number of elements of order $2$. Call them $x_1, ..., x_{2m}$, where $x_1,...,x_{2m} \neq e$. Then, consider the set $G=G \setminus \{e,x_1,...,x_n\}$. The size of $G'$ is odd since the size of $G$ is even.

Elements in G' does not have itself as an inverse, since $x^2=e \iff x=x^{-1}$.

Also, note that each elements in G' has a different inverses. (Suppose $a\neq b$, $\ a,b \in G'$, and $\exists c \in G'$ such that $ac=bc=e$. Then, $a=b=c^{-1}$ which is a contradiction.)

This shows that $|G'| =$ even which is a contradiction.

This seems complicated for no reason. Is this correct? and how do I make my proof neater?

axiom
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1 Answers1

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Pick all those elements which have different inverses. Pair each such element with its inverse. Basically they will be even in number. (Taking into account all of them). Call this set $ S $. Now $ G-S $ has even no of elements. But $ e \in G\S $. And except $ e $ all other elements in $ G\S $ have order $2$ and they are odd in number.

tattwamasi amrutam
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