Show that if $N$ is a normal subgroup of $G$ which contains all commuters then $G/N$ is abelian.

Question

I am working on my proof for class and I was wondering if this look ok?

Let $N$ be a normal subgroup of $G$ we want to show that $G/N$ is abelian, or $(aN)(aN) = abN = baN = (bN)(aN)$. Since $N$ contains all commutors, then let $aba^{-1}b^{-1}N = N$ for some $a,b \in G$ then, \begin{align*} aba^{-1}b^{-1}N &= N && \text{ Given}\\ ab(ba)^{-1}N &= N && \text{ Definition of inverse} \\ abN(ba)^{-1} &= N && \text{ since N is normal}\\ abN &= N(ba) && \text{ right multiply by ba}\\ abN &= baN && \text{ since N is normal}\\ \end{align*} Which is what we wanted to show.

We are using Abstract Algebra by Judson and I tried to mimic one of the proofs in the book plus add my reasons behind doing so.

Your proof is good. Also, you might be interested to know that any subgroup that contains the commutators has to be normal. — Stephen Montgomery-Smith, Apr 15 '14 at 00:29

score 2 · Accepted Answer · answered Apr 15 '14 at 00:29

2

Commutators. No, not "for some $a,b\in G$, rather, "for all $a,b\in G$, $[a,b]N=N$. This says precisely that $G/N$ is abelian.

answered Apr 15 '14 at 00:29

Pedro

122,002

Good catch, the devil is in the details. – spitfiredd Apr 15 '14 at 00:31

Show that if $N$ is a normal subgroup of $G$ which contains all commuters then $G/N$ is abelian.

1 Answers1