TITLE: Is the abelianization of a subgroup $H$ a subgroup of the abelianization of a group $G$?
QUESTION [5 upvotes]: Let $G$ be a finite group and $H<G$. Then, is it true that $H^{ab} < G^{ab}$, that is, the abelianization of $H$ is a subgroup of the abelianization of $G$?
To me, it would make sense if is was indeed true. However, I do not know exactly how to prove it. For any group $G$, we know that $G^{ab} = G/G' < G$, where $G'$ is the commutator subgroup of $G$. Thus, $H^{ab} < H < G$ and so by transitivity, $H^{ab} < G$ and so we obtain $H^{ab} < G$ and $G^{ab} < G$. I am stuck here...
Thanks for your help.

REPLY [2 votes]: It does not always hold. Consider the following hint.
Hint: For an element $a$ of the derived subgroup, $a\in G^{\prime}$, consider the cyclic subgroup $H=\langle a\rangle$.