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?

  • 4
    $\begingroup$ If $H$ is not normal then $G/H$ cannot be recognized as an (abelian) group. $\endgroup$ – drhab Mar 15 '17 at 13:12
  • 2
    $\begingroup$ You use the fact that $H$ is normal when you start computing in the quotient group $G/H$. $\endgroup$ – Andreas Caranti Mar 15 '17 at 13:12
  • $\begingroup$ So where have I gone wrong? $\endgroup$ – Shaun Mar 15 '17 at 13:14
  • $\begingroup$ @drhab Oh, yeah, I remember now. Thank you. $\endgroup$ – Shaun Mar 15 '17 at 13:16
  • $\begingroup$ @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. $\endgroup$ – 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.

| cite | improve this answer | |

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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