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

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.

## marked as duplicate by Rob Arthan, N. F. Taussig, Henno Brandsma, Nicky Hekster group-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 12 '18 at 13:45

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