1

Let $H$ be a normal subgroup of some group $G$, then I aim to prove that $[H,H]=\langle h^{-1}k^{-1}hk \mid h,k\in H \rangle$ is a normal subgroup of $G$.

I’ve tried to use the same strategy of showing the commutator is normal, but when I take $g^{-1}cg=c[c,g],c\in [H,H]$ the conclusion doesn’t follow as $[c,g]$ might not be in $[H,H].$

Mattan Feldman
  • 324

1 Answers1

6

The commutator subgroup of $H$ satisfies a property stronger than normality: it is a characteristic subgroup of $H$, meaning that for every automorphism $\phi : H \to H$ we have $\phi[H,H]=[H,H]$. This is pretty easy to prove, using that $\phi(hkh^{-1}k^{-1}) = \phi(h) \phi(k) \phi(h)^{-1} \phi(k)^{-1}$.

So now take any $g \in G$, and consider the inner automorphism $i_g :G \to G$ defined by $i_g(c)=gcg^{-1}$ for any $c \in G$.

Using that $H$ is a normal subgroup of $G$, it follows that the restriction $\phi = i_g \mid H$ is an automorphism $\phi : H \to H$.

And then it follows, from the first paragraph, that $i_g[H,H]=[H,H]$.

Lee Mosher
  • 135,265
  • this seems like an overkill argument. we haven’t discussed automorphisms much in class, so I don’t have a strong grasp on the properties and consequences. Is there a more direct argument that looks at the normalizer? – Mattan Feldman Oct 02 '24 at 20:09
  • 3
    You could show normality directly: take $g\in G$, then for any $h,k\in H$ you have $g(hkh^{-1}k^{-1})g^{-1}=(ghg^{-1})(gkg^{-1})(ghg^{-1})^{-1}(gkg^{-1})^{-1}$. and this is in $[H, H]$ because $H$ is normal. This shows the condition of normality for the generators of $[H, H]$, and I think you can argue yourself why it holds for the rest of the elements. This is @Lee Mosher 's argument but without the "automorphism" talk. – Mor A. Oct 02 '24 at 20:20
  • 1
    I wouldn't say it's overkill, although is a bit more abstract. On the other hand this solution motivates a broader understanding of what's going on. The key thing to take away from this solution, besides just solving the problem, is to learn about automorphisms and characteristic subgroups. – Lee Mosher Oct 02 '24 at 22:10
  • You may want to say "characteristic in $H$". (It is of course also fully invariant and verbal in $H$, but that is even more overkill....) – Arturo Magidin Oct 02 '24 at 22:21