0
$\begingroup$

$G$ 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 $G_2$.

$\endgroup$
  • 3
    $\begingroup$ All you need is $$yx^ny^{-1} = (yxy^{-1})^n$$ $\endgroup$ – Crostul Feb 25 '18 at 22:48
1
$\begingroup$

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 $G$.

$\endgroup$
0
$\begingroup$

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 $$ \varphi(x^n)=(\varphi(x))^n $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.