Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
I added an alternative proof to my answer. You may want to check it out. –  M. Vinay Jul 13 at 12:12

2 Answers 2

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 $G$.

share|improve this answer

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 $G$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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