Let $(G, \cdot)$ be a finite group with $n^2 - n + 1$ elements. We know that $f : G \rightarrow G$, $f(x) = x^n$ is an endomorphism of $(G, \cdot)$. Prove that the given group is abelian.

By knowing that $f$ is an endomorphism, we get: $f(xy) = f(x)f(y)$, $\forall x, y \in G$.

So, $(xy)^n = x^ny^n$, and this can be rewritten as $(yx)^{n - 1} = x^{n - 1}y^{n - 1}$.

That's all I did so far.

Thank you!


2 Answers 2


If $f$ is an endomorphism, then so is $f\circ f\circ f$. But $$ f(f(f(x)))=x^{n^3}. $$ Remember that $$ n^3+1=(n+1)(n^2-n+1). $$ Then your remaining tasks are to conclude that

  1. The mapping $g(x)=x^{-1}$ is an endomorphism.
  2. This implies that the group is abelian.
  • $\begingroup$ Can you, please, elaborate a bit more on those two conclusions? $\endgroup$
    – George R.
    Dec 13, 2016 at 17:12
  • $\begingroup$ @GeorgeR. Those were written as extended hints to cater for the possibility that this is a homework :-) Anyway. You should first show that $x^{n^3}=x^{-1}$ for all $x\in G$. Then conclude that $f(f(f(x)))=x^{-1}$ for all $x\in G$. It is an oft recurring exercise that if $x\mapsto x^{-1}$ is an endomorphism, then the group is abelian. Recall (or redo) that result. Does this help? $\endgroup$ Dec 13, 2016 at 19:54

You have $f^{n-1}(x)=(x^n)^{n-1}=x^{n(n-1)}=x^{n^2-n}=x^{-1}$ since $n^2-n+1$ is the order of $G$. This implies that the inversion is a morphism of group and $G$ is commutative.


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