I think I have somewhat of a roundabout way to show that the commutator subgroup $[G,G]$ is normal in $G$. Please check my proof for errors/improvements.

Let $G$ be a group and define $S := \{xyx^{-1}y^{-1}: x,y \in G\}$, then $$[G,G] = \langle S\rangle = \bigcap_{S \subseteq H < G} H, $$ so it is clear that $[G,G]$ is at least a subgroup. For normality consider the group action homomorphism $$ \psi \colon G \to \text{Perm}(\langle S \rangle), \quad g \mapsto \phi_g $$ where $\phi_g \colon \langle S \rangle \to \langle S \rangle $ is defined by $ s \mapsto g^{-1} s g$. The conjugation map is a homomorphism (i.e. $\phi_g$ is a homomorphism for every $g \in G$), and in this case is an automorphism so $\langle S \rangle$ is normal in $G$.

This is a bit different than the "standard" way of showing normality (i.e. via the definition), so I just wanted to make sure my reasoning is correct.


This is a good effort, but there's an issue to address. By considering the homomorphism

$$\psi\colon G \to \operatorname{Perm}(\langle S \rangle),$$ you are tacitly assuming that $\langle S \rangle = [G,G]$ is normal. More specifically: the statement that for all $g$ in $G$, $\phi_g$ takes elements of $\langle S \rangle$ to elements of $\langle S \rangle$ is equivalent to the statement that $\langle S \rangle$ is normal.

However, you've come close to one nice way of showing that $[G,G]$ is normal. Some general set-up: suppose $H = \langle a_i \rangle_{i\in I} \subset G$ is a subgroup. To show that $H$ is normal, it's sufficient to show that $g a_i g^{-1} \in H$ for all $i \in I$ and $g \in G$.

We can apply this to the generating set $S$ that you define: given $xyx^{-1}y^{-1} \in S$,

$$g(xyx^{-1}y^{-1})g^{-1}=(gxg^{-1})(gyg^{-1})(gx^{-1}g^{-1})(gy^{-1}g^{-1})$$ $$= (gxg^{-1})(gyg^{-1})(gxg^{-1})^{-1}(gyg^{-1})^{-1} \in \langle S \rangle.$$

This is essentially the same argument that you need to give to show that map $\psi \colon G\to\operatorname{Perm}([G,G])$ is well-defined.

  • 1
    $\begingroup$ Thank you for the correction. Do you know of a way to express $[G,G]$ as the kernel of some homomorphism? This was my original idea, but I couldn't think of a map that didn't require the co-domain to be abelian. $\endgroup$ Oct 1 '15 at 0:59

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.