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The following results are well-known:

(i) If $f(x)=x^2$ is a homomorphism from a (finite) group to itself, then the group must be abelian.

(ii) If $g(x)=x^3$ is a homomorphism from a finite group $G$ to itself, and if $(3,|G|)=1$, then $G$ is abelian.

Is there a generalization of $(ii)$ for some $n$ instead of $3$? Does it fail for some $n$ instead of $3$?

More precisely: is there other integer $n\neq 3$, such that following statement is true:

If $G$ is any finite group with $(|G|,n)=1$ and $f:G\rightarrow G$, $x\mapsto x^n$ is homomorphism, then $G$ is abelian

Maths Rahul
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1 Answers1

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No. It doesn't work for any integers other than $1$, $2$ and $3$.

Let $G$ be a group of exponent $n-1$, i.e., all elements have order dividing $n$. Then $x\mapsto x^n$ is the identity map, so certainly is a homomorphism. Thus if there is a non-abelian group of exponent $n-1$ then $n$ cannot work for your question.

There are non-abelian groups of exponent any number greater than $2$, as is proved in the link below.

(The counterexamples are similar to those in this question: If $G$ is a group and $3$ consecutive integers $i$ s.t. $(ab)^i=a^ib^i$ for all $a,b\in G$, then $G$ is abelian. (Finding examples))

David A. Craven
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  • Thanks for suggestion. I saw (ii) as a question in a book of Alperin-Bell; Alperin also gives a reference for more general thing, but I did not get anything related to my question in his reference, so I posted it here. – Maths Rahul Sep 17 '21 at 09:00
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    @MathsRahul It can be quite hard in general to find duplicates on the site, but some people manage to do it very quickly. – David A. Craven Sep 17 '21 at 09:01
  • While posting question, I could not see in related questions, hence I wrote. – Maths Rahul Sep 17 '21 at 09:02