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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 May 29 '14 at 8:05

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  • $\begingroup$ Hint If $G$ is non -Abelian, then $G$ has a non-identity inner automorphism. $\endgroup$ – 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.


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