# If $H$ is a normal subgroup of $G$ with $G/H$ abelian, then the commutator subgroup of $G$ is in $H$.

This is part of Exercise 2.7.9 of F. M. Goodman's "Algebra: Abstract and Concrete".

Let $C$ be the commutator subgroup of a group $G$. Show that if $H$ is a normal subgroup of $G$ with $G/H$ abelian, then $C\subseteq H$.

The following seems to be wrong.

## My Attempt:

The commutator subgroup $C$ of $G$ is the subgroup generated by all elements of the form $xyx^{-1}y^{-1}$ for $x, y\in G$.

Since $G/H$ is abelian, we have for $x, y\in G$, \begin{align} xyx^{-1}y^{-1}H&=xyy^{-1}x^{-1}H \\ &=H, \end{align} so that all elements of the form $xyx^{-1}y^{-1}$ are in $H$. Thus $C\subseteq H$.

But I don't use the fact that $H$ is normal. What have I done wrong and what is the right proof?

• If $H$ is not normal then $G/H$ cannot be recognized as an (abelian) group. – drhab Mar 15 '17 at 13:12
• You use the fact that $H$ is normal when you start computing in the quotient group $G/H$. – Andreas Caranti Mar 15 '17 at 13:12
• So where have I gone wrong? – Shaun Mar 15 '17 at 13:14
• @drhab Oh, yeah, I remember now. Thank you. – Shaun Mar 15 '17 at 13:16
• @Shaun you just neglected to note that $xyx^{-1}y^{-1}H$ is only relevant because it is equal to $(xH)(yH)(x^{-1}H)(y^{-1}H)$ because $H$ is normal. – Adam Hughes Mar 15 '17 at 13:16

$G/H=\{gH: g\in G\}$ by definition. this is only a group under $(gH)(g'H) = (gg')H$ if $Hg' = g'H$. But this is just another way of stating the definition of $H$ being normal. In your proof you just neglected to note that $xyx^{-1}y^{-1}H$ is only relevant because it is equal to $(xH)(yH)(x^{-1}H)(y^{-1}H)$ because $H$ is normal.
I would call this "incomplete" rather than "wrong" if anything, as the problem is a few steps beyond reproving the basic fact that $G/H$ is only a group when $H$ is normal. I think you just forgot that that's what makes $G/H$'s group operation well-defined.
The fact that $G/H$ is abelian gives us the second equality of:$$xyH=(xH)(yH)=(yH)(xH)=yxH$$ Consequently we find: $$x^{-1}y^{-1}xy=(yx)^{-1}xy\in H$$
This for every $x,y\in G$ so we are allowed to conclude that $H$ contains the commutator subgroup.