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Let $G$ be a group in which, for some integer $n>1$, $(ab)^{n}=a^{n}b^{n}$ for all $a,b \in G$. Show that $G^{(n)}=\{x^{n} \mid x \in G\}$ is a normal subgroup of $G$.

$G$ could be easily shown to be abelian and therefore proved, if $(ab)^{n}=a^{n}b^{n}$ is true for all $n$, but the question says it to be so for some $n$. Or does that make any difference?

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I added an alternative proof to my answer. You may want to check it out. – M. Vinay Jul 13 '14 at 12:12
up vote 6 down vote accepted

We need the given property only to prove that $G^{(n)}$ is a subgroup.
Let $x^n, y^n \in G^{(n)}$.
Then $x^n(y^n)^{-1} = x^n(y^{-1})^n = (xy^{-1})^n$ (by the given property).
Thus, for every $x^n, y^n \in G^{(n)}$, it is true that $x^n(y^n)^{-1} \in G^{(n)}$ too, which implies that $G^{(n)}$ is a subgroup of $G$ (as $e = e^n \in G^{(n)}$ ensures that $G^{(n)}$ is a non-empty subset of $G$).

Now, to show that it is normal, let $x^n \in G^{(n)}$. Consider any $g \in G$.
Then $(gxg^{-1})^n = gx^ng^{-1} \Rightarrow gx^ng^{-1} \in G^{(n)}$ (as it can be written in the form $y^n$ for $y = gxg^{-1}$).
Therefore, $G^{(n)}$ is normal in $G$.

Alternative Proof
Let $f:G \to G,\ x \mapsto x^n$. Then $f(ab) = (ab)^n = a^nb^n = f(a)f(b) \Rightarrow$ $f$ is a group homomorphism $\Rightarrow \text{image}(f)$ is a subgroup of $G$. But $\text{image}(f) = \{x^n \mid x \in G \} = G^{(n)} \Rightarrow$ $G^{(n)}$ is a subgroup.

$\forall g, x \in G, (gxg^{-1})^n = gx^ng^{-1} \Rightarrow \forall g \in G, x^n \in G^{(n)},\ gx^ng^{-1} \in G^{(n)} \Rightarrow G^{(n)}$ is a normal subgroup of $G$.

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It is not true, in general, that $G$ must be abelian if $(ab)^n = a^nb^n$, for some $n$. (Unless that $n$ happens to be equal to $2$.)

But, the set $G^{(n)}$ is still a normal subgroup. That it is normal is actually the easier part, since $(g^{-1}xg)^n = g^{-1}x^ng$, for any $x,g\in G$.

Clearly, the identity element $1\in G$ is of the form $1^n$, so $1\in G^{(n)}$ and $G^{(n)}$ is non-empty.

If $a,b\in G^{(n)}$, write $a = x^n$ and $b=y^n$, for $x,y\in G$. Then $$ab^{-1} = x^n(y^n)^{-1} = x^n(y^{-1})^n = (xy^{-1})^n,$$ so $ab^{-1}\in G^{(n)}$. Therefore, $G^{(n)}$ is a subgroup, and also normal in $G$.

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