Let $\phi : G \rightarrow G'$ be a homomorphism. We want to show that $H \leq G$ implies that $\phi(H) \leq G'$.
This is one of those problems that seems obviously true to the intuition, yet I'm having a devil of a time trying to prove it formally. A sketch of my efforts so far:
Let us define the codomain of $H$ under $\phi$ to be $H'$, and $h'$ to be an arbitrary element of $H'$ such that $\phi(h)=h'$ (for $h \in H$).
To achieve the desired result, we have to show that $h' \in G'$.
What I want to say is that, since $H \leq G$, then every $h$ in $H$ is also in $G$, and thus clearly $h' \in G'$. Is it that simple, or am I begging the question?
Hint: You have to show that, for $h_1,h_2$ we have $\phi(h_1)\phi(h_2)\in G'$ and $\phi(h_1)^{-1}\in G'$. Then $\phi(H)$ is a subgroup of $G'$.
Reference: This duplicate:
Image of subgroup and Kernel of homomorphism form subgroups