For a group $G,$ define $G_2=\{g\in G: |g|=2 \}.$

Prove that if $G_2$ is finite, then $|G_2|$ is odd.

Can you help me please?


Not true, just consider $C_3$, which has no element of order 2.

If you meant that $G$ is finite and even, then the statement is true.

Indeed, you've got that:

$$G = \{1\} \dot \cup G_2 \dot \cup \{ g \in G : o(g) > 2\}$$


$$|G_2| = |G|- 1 - |\{ g \in G : o(g) > 2\}|$$

Now, $| \{ g \in G : o(g) > 2\} |$ is even, since you can pair each element with its inverse and therefore there are a even number of them. Therefore, since $|G|$ is even, $|G_2| = |G|- 1 - |\{ g \in G : o(g) > 2\}|$ is odd.

  • $\begingroup$ I think that G is a group with at least one element g such that |g|=2. $\endgroup$ – user114952 Apr 6 '14 at 22:08
  • $\begingroup$ and you used G is finite.. but there is no restriction on G.. $\endgroup$ – user114952 Apr 6 '14 at 22:10
  • $\begingroup$ The argument would work if $G_2$ generates a subgroup of $G$ that is finite, but I don't see any reason why this must hold. $\endgroup$ – RghtHndSd Apr 6 '14 at 22:25

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