Showing a particular subgroup is normal

$G$ is a finite group with identity $e$. $G$ has the property that, for some fixed integer $n > 1$, $$(xy)^n = x^ny^n\qquad \text{for all x,y in G.}\tag{1}$$

Let $G_1 = \{g \in G : g^n = e\}$, $G_2 = \{x^n : x \in G\}$. We are asked to show that $G_1$ and $G_2$ are normal subgroups.

I succeeded in showing $G_1$ is normal by showing it is the kernel of the homomorphism that takes $y\in G$ to $y^n$, which is a homomorphism because $G$ satisfies property (1). We also know from this that $G_2$ is isomorphic to $G/G_1$.

I am having trouble showing $G_2$ is normal. I tried showing it was closed under conjugation but didn't manage to. And I couldn't come up with a homomorphism whose kernel is $G_2$.

• All you need is $$yx^ny^{-1} = (yxy^{-1})^n$$ – Crostul Feb 25 '18 at 22:48

From the given identity, it's easily verified that $G_2$ is a subgroup of $G$.
To prove it's a normal subgroup, let $x\in G_2$, and let $g\in G$.
Since $x\in G_2$, we have $x=y^n$, for some $y\in G$.
Note that \begin{align*} (gyg^{-1})^2 &= (gyg^{-1})(gyg^{-1})=gy^2g^{-1}\\[4pt] (gyg^{-1})^3 &= (gyg^{-1})(gyg^{-1})(gyg^{-1})=gy^3g^{-1}\\[4pt] &\;\;\vdots\\[4pt] (gyg^{-1})^n &= (gyg^{-1})\cdots(gyg^{-1})=gy^ng^{-1}\\[4pt] \end{align*} hence $$gxg^{-1} = gy^ng^{-1} = (gyg^{-1})\cdots (gyg^{-1})=(gyg^{-1})^n$$ so $gxg^{-1}\in G_2$.
Hence, $G_2$ is normal in $G$.
Since $G_2$ is the image of a homomorphism, it is a subgroup. And much more is true: $G_2$ is a fully invariant subgroup. Indeed, if $\varphi$ is an endomorphism of $G$, then $$\varphi(x^n)=(\varphi(x))^n$$