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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.

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2 Answers 2

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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.

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    $\begingroup$ 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? $\endgroup$
    – Junglemath
    Commented Jul 13, 2020 at 14:57
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    $\begingroup$ There's a typo in the answer. $H$ is meant to be abelian. $\endgroup$
    – David A. Craven
    Commented Jul 13, 2020 at 15:10
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    $\begingroup$ In fact you need to relace "non-abelian" by "non-trivial abelian". $\endgroup$
    – Derek Holt
    Commented Jul 13, 2020 at 15:21
  • $\begingroup$ thanks! edited in. $\endgroup$
    – hunter
    Commented Jul 13, 2020 at 15:24
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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.

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