# Is the commutator subgroup of a subgroup the same as the commutator subgroup of the group intersected with that subgroup?

I might be overthinking this, but anyway:

Let $$G$$ be a group and $$H$$ a subgroup. Let $$K'$$ be the commutator subgroup of $$H$$, i.e. $$K' = \langle [x, y] \mid x, y, \in H \rangle$$. Is it true that $$K' = G' \cap H$$?

Attempt: I believe that $$K' \subset G' \cap H$$, because if $$k \in K'$$ then $$k$$ is a product of commutators in $$K$$, so $$k \in G'$$. By closure, $$k$$ is in $$H$$, so $$k \in G' \cap H$$. I'm uncertain about the $$\supset$$ direction.

This is false. Take $$G$$ a non-abelian simple group. Then its commutator subgroup is all of $$G$$. So letting $$H$$ be any abelian (edit: oops! thanks to the commentors) subgroup of $$G$$ gives a counterexample.
• What about $G = A_6, H = A_5$? Then $K' = A_5$, and $G' \cap H = A_6 \cap A_5 = A_5$. Does this not hold? Commented Jul 13, 2020 at 14:57
• There's a typo in the answer. $H$ is meant to be abelian. Commented Jul 13, 2020 at 15:10
Another small counterexample: Let $$G$$ be the group of permutations of $$3$$ objects (the smallest nonabelian group) and let $$H$$ be its cyclic subgroup of order $$3$$. The commutator subgroup of $$G$$ is $$H$$ but the commutator subgroup of $$H$$ is trivial.