Prove $N$ is a normal subgroup of $G$ is $\forall a,b\in G, (ab)^n=a^n b^n$, and $N=\{x^n|x\in G\}$.

Question

In a group $G$, we have $(ab)^n=a^n b^n$ for all $a,b\in G$ and one particular 'n'. Let $N=\{x\in G| x^n\}$. Prove that $N$ is a normal subgroup of $G$.

Proving $N$ is a subgroup is easy. I can't figure out how to prove why $gx^n g^{-1}\in N$ for any $x^n\in N$ and $g\in G$. If $gx^n g^{-1}\in N$, then $gx^n g^{-1}=a^n$ for some $a\in G$.

I tried converting $gx^n g^{-1}$ into an $n^{th}$ power by equating it with $g^{-(n-1)} (gxg^{-1})^n g^{n-1}$, but it didn't give me anything I could work with.

Thanks in advance!

Is $n$ a fixed value? – Michael Albanese Aug 27 '13 at 10:54 — Michael Albanese, Aug 27 '13 at 10:54
$gx^ng^{-1}=(gxg^{-1})^n$. – Derek Holt Aug 27 '13 at 10:56 — Derek Holt, Aug 27 '13 at 10:56

score 3 · Accepted Answer · answered Aug 27 '13 at 10:55

3

$gx^n g^{-1}=(gx g^{-1})^n\in N$.

answered Aug 27 '13 at 10:55

Boris Novikov

17,470

Wow that really is so trivial. Thanks! – Aug 27 '13 at 11:00
+1. May I ask you to see my post elsewhere? Thanks – Mikasa Aug 27 '13 at 12:23
http://math.stackexchange.com/a/477204/8581. – Mikasa Aug 27 '13 at 12:34
@ Babak S. I sow it. All right. – Boris Novikov Aug 27 '13 at 14:06

Prove $N$ is a normal subgroup of $G$ is $\forall a,b\in G, (ab)^n=a^n b^n$, and $N=\{x^n|x\in G\}$.

1 Answers1