# If every nonidentity element in a group is of order $2$, the group is abelian [duplicate]

Let $$G$$ be a group. Show that if every non-identity element in $$G$$ has order $$2$$ then $$G$$ is abelian.

Proof:

Let $$a,b$$ be non-identity elements in $$G$$. Since $$|a|=|b|=2$$ , that means $$ab=babaab$$ $$=$$ $$ba$$.

Is the proof correct? How can I improve it?

## marked as duplicate by rschwieb abstract-algebra 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 6 at 13:46

• The proof is correct as written. – rubikscube09 May 4 at 21:04
• I must be tired by I don't see how to justfify the first equality. – elidiot May 4 at 21:08
• @elidiot (baba)ab=(e)ab=ab. bab(aa)b=babb=ba – topologicalmagician May 4 at 21:09
• It uses $|ba|=2$ on the left. – Berci May 4 at 21:10
• @rschwieb thanks, i'll keep that in mind next time. – topologicalmagician May 7 at 14:07

Alternative proof : $$a^2=1$$ so every element is its one inverse.
So, $$(ab)^{-1}=ab$$, but $$(ab)^{-1}=b^{-1}a^{-1}=ba$$ by using twice again the remark.