-1

Let $H$ be a commutator subgroup of $G$. I try to prove that $H$ is a normal subgroup.

I saw some proof using the definition. But I try to first consider the automorphism on $G$ by conjugation and then prove that $H$ is characteristic in $G$.

Let $\pi$ be an automorphism on $G$: $\pi: G\to G$ defined by $x\to gxg^{-1}$ for $g\in G$. This map is well-defined. For every element $xyx^{-1}y^{-1}\in H$, $x, y\in G$, then $$ \pi(xyx^{-1}y^{-1})=\pi(x)\pi(y)\pi(x)^{-1}\pi(y)^{-1}=(gxg^{-1})(gyg^{-1})(gx^{-1}g^{-1})(gy^{-1}g^{-1})=gxyx^{-1}y^{-1}g^{-1} $$

Can we say $\pi(x)\pi(y)\pi(x)^{-1}\pi(y)^{-1}$ this is a commutator in $H$?

If so, then $\pi(H)\le H$. Similarly, since $\pi^{-1}$ is also automorphism, then $\pi^{-1}(H)\le H$. Hence, $H=\pi\pi^{-1}(H)\le \pi(H)$. Hence, $\pi(H)=H$. This means $gHg^{-1}=H$ for all $g\in G$.

Hermi
  • 1,480
  • There only is one commutator subgroup by definition, so you mean let $H$ be the commutator subgroup. – AlgTop1854 Oct 03 '23 at 19:26
  • And not to ignore your attempt, I got stuck on the same point, is the element you came up with as being a commutator, seems like this should work I just didn't see it off the bat. Also not every element of $H$ is a commutator, rather these generate $H$. – AlgTop1854 Oct 03 '23 at 19:28
  • This is silly. Of course $\pi(xyx^{-1}y^{-1})$ is a commutator: it is the commutator of $\pi(x)$ and $\pi(y)$, which are elements of $G$. – Arturo Magidin Oct 03 '23 at 19:56
  • "Can we this is a commutator in $H$?" does not parse. – Arturo Magidin Oct 03 '23 at 20:02
  • @ArturoMagidin thanks - clear now after taking a few more minutes on it. – AlgTop1854 Oct 03 '23 at 20:19
  • @AlgTop1854 That $[G,G]$ is normal follows now because $\langle S\rangle \triangleleft G$ if and only if for every $g\in G$ and $s\in S$, $gsg^{-1}\in\langle S\rangle$. Here $S$ is the set of all commutators. – Arturo Magidin Oct 03 '23 at 20:22

1 Answers1

1

Found in the comments of a prior post, a gem of a simple proof.

Let $c\in C$ be any element of the commutator subgroup and let $g\in G$.

Then $gcg^{-1}c^{-1}$ is a commutator so $(gcg^{-1}c^{-1})c=gcg^{-1}\in C$.

How to show that the commutator subgroup is a normal subgroup

AlgTop1854
  • 1,344