Prove that if $x\mapsto x^{2n}$ and $x\mapsto x^{3n}$ are surjective homomorphisms on a group, the group is abelian

Question

Let $(G, \cdot)$ be a group. If the functions $f, g : G \rightarrow G$, with $f(x) = x^{2n}$ and $g(x) = x^{3n}$ ($n \in \mathbb{N}^*$) are surjective homomorphisms, then prove that $G$ is abelian.

Here is what I did so far:

Since $f$ and $g$ are homomorphisms, we know that: $$f(xy) = f(x)f(y), \, \forall \, x, y \in G$$ $$g(xy) = g(x)g(y), \, \forall \, x, y \in G$$

Using the above results, we get $(xy)^{3n} = (xy)^n(xy)^{2n} = (xy)^nx^{2n}y^{2n}$ and $(xy)^{3n} = (xy)^{2n}(xy)^n = x^{2n}y^{2n}(xy)^n$. But $(xy)^{3n} = x^{3n}y^{3n}$. Using these three results, we can easily get:

$$(xy)^nx^{2n} = x^{3n}y^n$$ $$y^{2n}(xy)^n = x^ny^{3n}$$

From the first conditions, we have $y^{2n} = x^{-2n}(xy)^{2n}$ and $x^{2n} = (xy)^{2n}y^{-2n}$.

$$(xy)^nx^{2n} = x^{3n}y^n = x^{2n}x^ny^n = (xy)^{2n}y^{-2n}x^ny^n$$ $$y^{2n}(xy)^n = x^ny^{3n} = x^ny^ny^{2n} = x^ny^nx^{-2n}(xy)^{2n}$$

By making simplifications, we get: $$x^{2n} = (xy)^ny^{-2n}x^ny^n$$ $$y^{2n} = x^ny^nx^{-2n}(xy)^n$$

Using these into the first conditions, we have: $$(xy)^{2n} = (xy)^ny^{-2n}x^ny^nx^ny^nx^{-2n}(xy)^n$$

Again, by making simplifications, we get: $$y^{2n}x^{2n} = x^ny^nx^ny^n$$

That's all I could find so far.

Thank you in advance!

Answer 1

Hint: Try to use the equation $y^{2n}x^{2n}=x^ny^nx^ny^n$ you have found to prove that $x^{2n}y^{3n}=y^{3n}x^{2n}$. Can you see why this implies $G$ is abelian?

A full proof (given $y^{2n}x^{2n}=x^ny^nx^ny^n$) is hidden below:

From your equation $y^{2n}x^{2n}=x^ny^nx^ny^n$, we get $$y^{3n}x^{2n}=y^n(y^{2n}x^{2n})=y^nx^ny^nx^ny^n.$$ Swapping the roles of $x$ and $y$, we also have $x^{2n}y^{2n}=y^nx^ny^nx^n$, so $$x^{2n}y^{3n}=(x^{2n}y^{2n})y^n=y^nx^ny^nx^ny^n.$$ Comparing these two equations we see that $$y^{3n}x^{2n}=x^{2n}y^{3n}.$$ But since $f$ and $g$ are surjective, any element of $G$ can be written in the form $x^{2n}$ or $y^{3n}$, so this says any two elements of $G$ commute.