# Proving commutator subgroup is normal

Part of Aluffi II.7.11 suggests proving this claim.

It is actually quite straightforward (especially after some earlier problems in the book): I've managed to prove it by noticing that for arbitrary $$a, b, g$$: $$g aba^{-1} b^{-1} g^{-1} = (gag^{-1})(gbg^{-1})(gag^{-1})^{-1}(gbg^{-1})^{-1}$$ which is in $$[G, G]$$, thus it's trivial to show that for arbitrary $$n \in [G, G]$$ the element $$gng^{-1}$$ is also in $$[G, G]$$ by interspersing it with some number of $$g^{-1}g$$'s.

My question, though, is a bit different (and that's what's different from all other questions about normality of commutators that MSE suggests as similar). Aluffi gives a hint that $$g \cdot aba^{-1}b^{-1} \cdot g^{-1} = \gamma_g(aba^{-1}b^{-1})$$, where $$\gamma_g : a \mapsto gag^{-1}$$ is an inner automorphism on $$G$$ defined by the element $$g$$. I don't think I've followed this hint and thus I haven't built a "more categorical" proof that Aluffi might have expected here. So, is there a way to prove the claim using this hint?

• I think you have followed the hint. – arctic tern Jun 3 at 23:13
• Aluffi's hint is, as explained in Berci's answer, just a different way of organizing your calculation. But this reorganization has real value, because it shows that the commutator subgroup is invariant under all automorphisms of $G$, not just the inner ones. (In other words, it shows that the commutator subgroup is a characteristic subgroup, which is a stronger property than normality.) – Andreas Blass Jun 4 at 2:16

## 2 Answers

You implicitly did follow the hint:

The point is that $$\gamma_g$$ is an inner automorphism, implying it's a group homomorphism, so that $$\gamma_g(xy) =\gamma_g(x)\gamma_g(y)$$ which was the crucial part in your proof, as well.

I think @Andreas Blass in his comment has the key to the "more categorical" approach.

The commutator can be viewed as a functor on the category of groups. The Wikipedia page is good on this.

It follows that the commutator subgroup is a fully characteristic subgroup,(ie, it is stable under any endomorphism).

Thus it is automatically a normal subgroup. This is because the normality condition can be phrased as invariance under inner automorphisms.

• I believe the usual terminology is “fully invariant”, rather than “fully characteristic”; at least, that’s the one I normally encounter. (In fact, the commutator subgroup is even more than fully invariant: it is verbal, which is what yields the functoriality) – Arturo Magidin Jun 5 at 4:42