$|G|= 2^n, o(g)=2~\forall g \in G, g \neq e$ show that G is abelian

Question

Question as in title: I have attempted this question by saying that the homomorpism that maps G to $Z_2$ is onto, $Z_2$ is abelian so G must be abelian.

This question has been asked multiple times on this site (so go and look for the answer) and you do not have to assume that $|G|$ is a power of $2$. — Nicky Hekster, May 12 '18 at 13:33
Removing the question is not necessary. Just next time take a closer look on this site. — Nicky Hekster, Oct 15 '18 at 18:19

score 0 · Accepted Answer · answered May 12 '18 at 13:34

0

We have, for all $a,b\in G$, $$abba=e=baba$$ Hence $ab=ba$

answered May 12 '18 at 13:34

BAI

2,565

$|G|= 2^n, o(g)=2~\forall g \in G, g \neq e$ show that G is abelian

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