Does $x\mapsto x^n$ is an automorphism on $G$ imply $G$ is abelian?

Question

I am seeking to generalize this result (which is a particular case: $n=3$).

Let $G$ be a finite group and $n\in\mathbb Z^+$. Find the values of $n$ for which the following implication holds. $$ x\mapsto x^n \text{ is an automorphism on $G$} \implies G \text{ is abelian}.$$

I believe that it's true for all $n\neq 1$.

Observations:

1. $(xy)^n=x^n y^n\implies (yx)^{n-1}=x^{n-1}y^{n-1}$.
2. If $x\mapsto x^n$ is an automorphism then $x^{n-1}\in Z(G)$ for all $x\in G$. (Proof).

For $n=2$: observation $1$ suffices.
For $n=3$: We have $(yx)^2 = x^2y^2$ (from obs. 1). And by obs. 2, $x^2y^2= y^2x^2$. Thus, $(yx)^2 = y^2x^2$. (We're done using $n=2$ case.)
For $n=4$: $(yx)^3 \overset{\text{obs. }1}= x^3y^3 \overset{\text{obs. }2}= y^3x^3$.

This gives an idea for an inductive proof.

Am I right? I'm posting this because I have seen homework problems with $n=2$ and $n=3$ on this site but no generalization. So I'm suspicious that there may be something wrong in my proof.

Update: I spotted the mistake. In the $n=4$ case, I have proved that if $x\mapsto x^4$ is an automorphism then it holds that $(xy)^3=x^3y^3$. This result is correct. However, this result alone doesn't imply that $G$ is abelian. We also need that $x^3=e\iff x=e$. Thus, $n=3$ case doesn't contribute anything to $n=4$ case.

Answer 1

The flaw in your argument has been identified, so let us answer the more general question: for which values of $n$ does the implication hold?

A group for which the $n$th power map is an endomorphism is said to be $n$-abelian. Alperin determined the structure of $n$-abelian groups for $n\gt 0$. They are quotients of subgroups of groups of the form $A\times M\times N$, where $A$ is abelian, $M$ is a group of exponent $n-1$, and $N$ is a group of exponent $n$.

Take $n\gt 0$ first. If we also require the $n$th power map to be one-to-one, then the group will be a quotient of a subgroup of a group of the form $A\times M$, and $A$ will have order prime to $n$. For this to necessarily be an abelian group, we require groups of exponent $n-1$ to necessarily be abelian. This happens when $n=2$ and $n=3$ only, as for any $k\gt 2$ there is a group of exponent $k$ that is not abelian.

Conversely, $2$-abelian groups are necessarily abelian, so $2$-abelian groups with no elements of order $2$ are abelian. And $3$-abelian groups with no elements of order $3$ will be quotients of subgroups of a group of the form $A\times M$, with $A$ abelian and $M$ of exponent $2$... hence abelian.

In addition, if the map $x\mapsto x^{-1}$ is an automorphism then the group is abelian. What other negative values might work?

If a group satisfies $(xy)^n = x^ny^n$ for all $x$ and $y$, then it also satisfies $(ab)^{1-n} = a^{1-n}b^{1-n}$ for all $a$ and $b$.

Indeed, for positive $n$, we have $$b(ab)^{n-1}a = (ba)^n = b^na^n,$$ which gives $(ab)^{1-n}=b^{n-1}a^{n-1}$. Taking inverses, we obtain $(ab)^{1-n} = a^{1-n}b^{1-n}$.

If $m\lt 0$ and $(xy)^m = x^my^m$ for all $x$ and $y$, then $(xy)^{-m}=y^{-m}x^{-m}$, so $$(ab)^{1-m} = a(ba)^{-m}b = aa^{-m}b^{-m}b = a^{1-m}b^{1-m}.$$

So for $n\lt 0$, if $x\mapsto x^n$ is an automorphism, then $x\mapsto x^{1-n}$ is an endomorphism, so the group is $(1-n)$-abelian; so we want a group that is $(1-n)$-abelian with no elements of order dividing $n$ to be abelian. This will fail if $1-n\geq 3$, or equivalently if $n\leq -2$. Thus $n=-1$ is the only negative value for which this works.

Finally, for trivial reasons, if the map $x\mapsto x^0$ is an automorphism, then the group is abelian... because it must be the trivial map.

Thus, the only values of $n$ for which the implication $$x\mapsto x^n\text{ is an automorphism of }G\implies G\text{ is abelian}$$ holds are $n=-1$, $n=0$, $n=2$, and $n=3$.

Answer 2

The problem is that 2. uses the fact that $x\mapsto x^n$ is surjective (in general, you only get commuting with $n$-th powers).

Given that $x\mapsto x^n$ is a homomomorphism, this is true if and only if $n$ is coprime with the order of $G$.

In your inductive step, you only argue that $x\mapsto x^{n-1}$ is a homomorphism.

Thus, for the induction to actually work, you need to know that every $2\leq m\leq n$ is coprime with the order of $G$.

For a counterexample, consider $G=S_3\times C_2$. Then $x\mapsto x^7$ is identity on $G$, although $7\not\equiv 1\pmod {12}$.

Answer 3

Take an arbitrary finite group $G$ and let $e$ be the exponent of $G$, i.e. the lcm of all element orders. Set $n=e+1$.

Then for any $x\in G$ we have that $x^n=x$, so the map $x\mapsto x^n$ is an automorphism, regardless whether $G$ is abelian.

This shows that the statement is clearly false for arbitrary $n$.