# When the automorphism group is trivial… [duplicate]

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If $G$ is a trivial group , obviously $Aut(G)$ is trivial . Does the converse hold $?$ .

When we are given that a group $G$ has trivial automorphism group , can we conclude that the group will be trivial $?$ .

I guess not . For one thing , we know that $$G/Z(G)\cong Inn(G)$$. So , $Aut(G)$ being trivial would imply that $$G\cong Z(G)$$ i.e. $G$ is commutative . But I see nothing about $G'$s being trivial .

The problem is that while I cannot prove it , I could not find a counter example either , from among the groups I am familiar with .

What is the non-trivial commutative group that has trivial automorphism group $?$

## marked as duplicate by Dietrich Burde, Derek Holt 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(); } ); }); }); Dec 11 '15 at 9:31

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• this question provides an answer: $\mathrm{Aut}(G)=1$ iff $G=1$ or $G\cong\mathbf C_2$. – Myself Dec 11 '15 at 9:19

## 1 Answer

If $G$ is abelian, $x \mapsto x^{-1}$ is an automorphism. If it is trivial, all elements have order at most $2$, hence $G$ is a $\mathbb F_2$-vectorspace. A vectorspace has non-trivial automorphism if the dimension is at least $2$ (any permutation of the basis will do). Hence the dimension must be $\leq 1$ and we end up with $G=\mathbb F_2$ or $G$ trivial.