Prove that if G is a cyclic group with more than 2 elements, then there always exists an isomorphism ϕ:G→G that is not the identity mapping

Question

The full question is

Prove that if G is a cyclic group with more than 2 elements, then there always exists an isomorphism ϕ:G→G that is not the identity mapping.I have no idea where to start?

Answer 1

$G$ is cyclic, say of order $n>2$. Then $\varphi(n)>1$. So...?

Spoiler

Then we can find $k$ with $(k,n)=1$, $1<k<n$, and if $G=\langle x\rangle$; then $x\mapsto x^k$ is a nonidentity isomorphism.

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Just for the sake of addressing Gina's comment: if $G$ is abelian and not $2$-elementary, $x\mapsto x^{-1}$ is a group automorphism that is not the identity. If $G$ is non abelian and $g$ is a noncentral element, then $x\mapsto gxg^{-1}$ is a nonidentity group automorphism. If $G$ is $2$-elementary i.e of the form $\Bbb Z_2^n$ for some $n$, you can argue ${\rm Aut}G\simeq {\rm GL}(n,2)$ which has size $(2^n-2^{n-1})\cdots (2^n-1)$.

Answer 2

By exhibiting a nonidentity mapping. There is a certain mapping, defined on any group, that you already know of (it's inherent in one of the axioms of a group). Another exercise in group theory asks a student to show this mapping is a group homomorphism iff the group is abelian. Can you guess which map I'm talking about?