# If $f(g) = g^k$ is a homomorphism on a finite group $G$ and $k < |G|$ does not divide $|G|$, must $G$ be abelian?

There was a question that asked to prove that if $f(g) = g^3$ is a homomorphism on a finite group $G$ and $3$ does not divide $|G|$, then $G$ is abelian. Does this extend to any $k < |G|$ where $k$ does not divide $|G|$, instead of just the case $k = 3$? I.e. if $f(g) = g^k$ is a homomorphism and $k < |G|$ does not divide $|G|$, is $G$ abelian? If not, what is a counterexample?

No, the Quaternion group $Q$ of order $8$ is non-abelian, but the fifth power map is a homomorphism $Q\to Q$ (the identity).

Consider a non-abelian $p$-group of exponent $p$, with $p$ an odd prime. For instance the group of order $p^{3}$ and exponent $p$ will do.

Then $$(ab)^{p+1} = (ab)^{p} a b = a b = a^{p} a b^{p} b = a^{p+1} b^{p+1}.$$

In general, if $G$ is any finite group of exponent $e < \lvert G \rvert$, then $k = e+1$ should do. See for instance the example by James.

• Obvious follow-up question: if $x\mapsto x^k$ is an automorphism of $G$ then does $k\equiv1\bmod(\exp G)$. – whacka Apr 10 '15 at 17:25
• @whacka, good question, I don't have an answer right now. You have any ideas? – Andreas Caranti Apr 11 '15 at 10:06
• @whacka, no, wait, I do have an example in which $k \not\equiv 1 \pmod{\exp(G)}$. Take an odd prime $p$, and let $G$ be the nonabelian group of order $p^{3}$ and exponent $p^{2}$. Then $x \mapsto x^{p+1}$ is an automorphism, as $(xy)^{p+1} = x^{p+1} y^{p+1} [y, x]^{\binom{p+1}{2}} = x^{p+1} y^{p+1}$. We have used the fact that $G'$ has order $p$, and that $p$ divides $\binom{p+1}{2}$. But I think your conjecture can be refined, at least for the case of finite $p$-groups, for instance to something like $k \equiv 1 \pmod{\exp(G')}$. – Andreas Caranti Apr 11 '15 at 10:13

We have $(xyx^{-1})^3 = x^3y^3x^{-3}$ by that homomorphism. By writing out the product we get $xy^3x^{-1} = x^3y^3x^{-3}$. Therefore, $y^3x^2 = x^2y^3$. Since the $3$ does the divide the order of $|G|$, every element in $G$ is a cube. Therefore, the squares commute with $G$. Now write $(ab)^3 = a^3b^3$, expand, and commute the squares, and you get $ab=ba$.

Now if we try to do is more generally, if $k$ does not divide $|G|$, then the claim that every element of $G$ is a $k$-th power does not generalize. Unless we know that $k$ is a prime. But even if $k$ is a prime and does not divide $|G|$, then I am not sure how to conclude that it is abelian. For example, if $k=5$, then we reach $(b^3a^3) = (ab)^3$, and not sure if this is enough anymore.