4
$\begingroup$

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.

  • 0
    @Asaf Karagila: I understood. Thanks for taking the care.2011-02-02

2 Answers 2

5

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......

  • 1
    @Marla, just$a$LaTeX tip: If you want to write $a^{-1}$, you need to write$a^{-1}$with { and } around the -1.2011-02-02
5

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.

  • 0
    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.2011-02-03