# If $G$ is a non-trivial group in which every non-identity element has order $2$, then $|G|$ is even

I'm doing some exercises on group-theory but got stuck in the following question:

Let $G$ be a group such that for all $a \in G\setminus\{e\}$, $a^2 = e$. Prove that $|G|$ is even.

I tried using Lagrange's theorem but perhaps I still don't understand it fully for I was not capable of finishing it.

Could you shed some enlightenment on the matter?

Thanks in advance.

• what about the group just containing one element, the identity? – Chris Card Feb 2 '11 at 15:56
• You're right. I forget to write that a can't be e. Thanks for noticing it. – Marla Feb 2 '11 at 16:03
• It doesn't matter. Ofc, $e^2 = e$. – Calle Feb 2 '11 at 16:04
• it does not matter if you assume $a\not= e$, but you need to say that $G$ is nontrivial, otherwise as Chris noted the claim is false, even in its revised form (as for the trivial group it is vacuously true.) – Asaf Karagila Feb 2 '11 at 16:52
• @Asaf Karagila: I understood. Thanks for taking the care. – Marla Feb 2 '11 at 17:08

## 2 Answers

HINT: Let $a\not=e$. If $a^2=e$, how many elements you got in the subgroup generated by $a$? Apply Lagrange......

NOTE: written at the same time as Myself's answer......

• <a> = {a, a^2 = e, a^-1}, right? – Marla Feb 2 '11 at 16:01
• Sure, and what is $a^{-1}$ in this case? – Calle Feb 2 '11 at 16:05
• a-Ha, it is 'a' itself! I got it. Thanks for not telling me that right away! – Marla Feb 2 '11 at 16:09
• @Marla: Ok, so you now see that it's enough to have just ONE element $a\not=e$ such that $a^2=e$ for $|G|$ to be even. Now you can try the following EXERCISE: Suppose $|G|$ even. Then there exists $a\not=e$ such that $a^2=e$. – Andrea Mori Feb 2 '11 at 16:17
• @Marla, just a LaTeX tip: If you want to write $a^{-1}$, you need to write a^{-1} with { and } around the -1. – Calle Feb 2 '11 at 17:32

One must obviously assume $G$ is a nontrivial finite group.

In this case $G$ has a subgroup $H = \{e,a\}$ of order 2, for any $a\in G$ that is not the unit. Therefore by Lagranges theorem the order of H divides the order of G, in other words: $|G|$ is even.

• That's the straightest answer I can think of. But no need to look at H. If for all $a \in G \text{ such that } a \neq e$, $a^2 = e$, then they all are of order two. However, you don't need all of them to satisfy this. If one element $x$, not being $e$, satisfies $x^2 = e$, then its order divides the order of $G$, that's as simple as this. Thus the result. – Alp Mestanogullari Feb 2 '11 at 16:33
• Actually, proving that the order of an element divides the order of the group is done via the Lagrange theorem, resoning with the cyclic group generated by that element. So one way or another, you're using that group. However, on an even more basic level you could state that $x\mapsto x\cdot a$ (where a satisfies $a^2=e$) is a involutory permutation on $G$ with no fixed points, hence the result. (Note that this is in fact equivalent to proving Lagranges thorem in this special case.) – Myself Feb 2 '11 at 17:38
• Yeah sure. My remark was about my point of view on the clarity of exposure of the argument, not about the argument itself, since as I said that's exactly what I would have answered. I just thought adding H in there wasn't really necessary when just invoking the theorem, but this is equivalently correct as using it, of course. – Alp Mestanogullari Feb 3 '11 at 0:32