This is an exercise from "Introduction to Abstract Algebra" by Timothy J. Ford. I really don't know where to start proving this, so if you could give me a start on how to do this, but not reveal the proof, that'd be great.
The questions is in the title, but I'll re-state it.
For a group homomorphism $f: G \to G'$, we wish to show that $H < G$ implies $f(H) \leq G'$.
I came up with this solution, if someone is able to verify it:
Let $a,b \in f(H)$, so $a = f(h_1), b = f(h_2)$ for some $h_1,h_2 \in H$
We wish to show $f(h_1)(f(h_2))^{-1} \in H$
$f(h_1)(f(h_2))^{-1} = f(h_1)f(h_2^{-1})$ because homomorphisms send inverses to inverses $= f(h_1h_2^{-1})$ by the definition of a homomorphism
$f(h_1h_2^{-1}) \in f(H)$ because $h_1h_2^{-1} \in H$ by $H \leq G$.
Therefore, $f(H) \leq G'$.
There is a list of requirements for a subset to be a subgroup. You know that $H\subseteq G$ fulfills all of them. There is a list of requirements for a function to be called a homomorphism. You know that $f$ fulfills them.
There is a list of requirements for a subset to be a subgroup. You are asked whether $f(H)\subseteq G'$ fulfills this list. Use the facts in the above paragraph to show that the requirements are indeed fulfilled.