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If $G$ is a finite group and $\phi(x) = x^n$ is an automorphism of $G$ does this imply $G$ is abelian?

I've been reading this page.

Def: A group $G$ is said to be $n$-abelian if $(ab)^n=a^nb^n$ for every $a,b\in G$.

which convinces me my question must be true....but I just want to make sure I'm not misinterpreting anything.

If yes, how would I go about proving this. i.e: showing $xy=yx$ -Thank u

I've seen the proof for $\phi(x) =x^2$. For arbitrary $n$ is this the same?

user26857
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abe
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4 Answers4

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It is not true for arbitrary $n$,

Being homomorphism only give us that $(xy)^n=x^ny^n$ and it is not enough for arbitrary $n$.

here is required condition.(if it is true for three consecutive $n$, you can conclude $G$ is abelian)

Note: But you can show that it is true for these numbers $n=-1,1,2$

Community
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mesel
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Not necessarily true for endomorphisms. For example, there are nonabelian groups where every element has order $3$; so $x\mapsto x^3=1$ is the trivial map, so trivially a group endomorphism.

Pedro
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Let $G$ be any finite non-abelian group and consider its order $|G|$. Then the map $x\mapsto x^{|G|+1}$ is simply the identity, so it is trivially an automorphism.

For the result you want, there must be some condition between $n$ and $|G|$, specifically $\operatorname{mcd}(|G|,n(n-1))=1$. See groupprops.

Luiz Cordeiro
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  • Presumably one means to take $2 < n < |G|-1$, at which point the question is nontrivial. –  Jun 10 '14 at 01:48
  • In general it works for $n = {\rm exponent}(G)+1$. So, for example, for $A_4$, it works for $n=7$. – Derek Holt Jun 10 '14 at 08:03
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Take any non-abelian group $G$ of order $n$. Then $G$ is $(n+1)$-abelian, since the map $g\mapsto g^{n+1}$ is the identity (hence, an automorphism). For a specific example, the symmetric group of degree $3$ is $7$-abelian.

James
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