# Particular nontrivial group has a nonidentity automorphism [duplicate]

If $G$ is a nontrivial group that is not cyclic of order 2, then $G$ has a nonidentity automorphism.

This is the exercise of hungerford algebra in the chapter $IV$ MODULES. Can you help me please?

## marked as duplicate by 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(); } ); }); }); May 29 '14 at 8:05

• Hint If $G$ is non -Abelian, then $G$ has a non-identity inner automorphism. – Geoff Robinson May 29 '14 at 3:45
If $G$ is not abelian, take $a,b\in G$ such that $ab\ne ba$. Then $x \mapsto axa^{-1}$ is a nonidentity automorphism.
If $G$ is abelian, then $x \mapsto x^{-1}$ is a nonidentity automorphism, unless $G$ is a product of $C_2$'s. In this case, write $G=C_2 \times C_2 \times H$. Then $(x,y,z)\mapsto (y,x,z)$ is a nonidentity automorphism.
If $G$ is not abelian, then there exists $g\in G$ that does not lie in the center. Define $\varphi:G\rightarrow G$, $\varphi(x)=gxg^{-1}$. This is a non-trivial automorphism.
If $G$ is abelian an has an element of order $>2$, then $\varphi(x)=x^{-1}$ defines a non-trivial automorphism. Finally, if $G$ is abelian and every element has order $\leq 2$, then pick an automorphism that non-trivially permutes a generating set for $G$. This will create a non-trivial automorphism.