The product of the commutator subgroup with any subgroup

Yes. This follows from the Correspondence Theorem, because $G'H/G'$ is a subgroup of the abelian group $G/G'$, so $G'H/G'$ is a normal subgroup of $G/G'$. Thus $G'H$, which is the full preimage of $G'H/G'$, is normal in $G$, which is the full preimage of $G/G'$.

You can see this explicitly, without the correspondence theorem, as follows:

• Note that if $N\unlhd G$ and $H\leq G$ then $NH\leq G$ (this is an easy exercise). Hence, $G^{\prime}H\leq G$.

• If $g\in G$, $c\in G^{\prime}$ and $h\in H$ then $g^{-1}chg=(g^{-1}cg)(g^{-1}hgh^{-1})h$. As both $g^{-1}cg, g^{-1}hgh^{-1}\in G^{\prime}$, the result follows.