Let $H\le G$ and $\pi:G\rightarrow G'$ be a homomorphism. If this is wrong or a duplicate let me know, but I haven't yet found a proof detailing $H\le G \implies \pi(H) \le \pi(G)$. I have found proofs of $H\le G \implies \pi(H) \le G'$.
Since $\pi$ could fail to be surjective, this is important to know, right? I feel uncertain about whether this line of reasoning is correct, or perhaps I am wrong about the whole thing. At the very least, it would be helpful to get verification that (4) is indeed a true statement. Thank you!
Assuming (1) $H\le G \implies \pi(H) \le G'$,
$$we\;have$$
$$\,\quad (2)\quad G\le G \implies \pi(G) \le G'\quad .$$
We also know since $H\le G$, $H\subseteq G.$ This means:
$$(3)\quad \pi(H)\subseteq \pi(G)\quad.$$
But since $H\le G \implies \pi(H) \le G'$, we now can conclude that:
$$(4)\quad H\le G \implies \pi(H) \le \pi(G)$$