If $P$ and $Q$ are groups and $f: P \to Q$ is a surjective homomorphism, is it possible that $f(P) = Q$ as well as $f(P_{\mathrm{sub}}) = Q$ where $P_{\mathrm{sub}}$ is a proper subgroup of $P$? This should be an easy question but for some reason I'm just blanking out.
You can suppose $P$ and $Q$ are abelian groups, but I guess that doesn't affect the question. I'm thinking of this in the context of https://math.stackexchange.com/a/525836.
Yes this can happen. You could take $Q$ to be the trivial group, and $f: P \rightarrow Q$ the trivial map, which is surjective. The image of any subgroup of $P$ under $f$ is clearly also $Q$.