If $\phi : G \rightarrow G'$ is a group isomorphism, and $H$ is a subgroup of $G$, prove that the image set of $H$ is a subgroup of $G'$

Question

Well the full question is this:

Suppose that $ \phi : G \rightarrow G'$ is a group isomorphism. Let $H$ be a subgroup of $G$. Prove that $\phi [H] = \left\{ \phi (h) | h \in H \right\}$, i.e the image set of $H$, is a subgroup of $G'$

How will I prove this? Should I use the definition of a subgroup to prove this? Or should I use the Compact Criterion for subgroup (A nonempty subset $H$ of a group $G$ is a subgroup iff $ab^{-1} \in H$ for all $a,b\in H$). Any help will be greatly appreciated. Thank You.

Answer 1

Let $y,z\in \phi(H)$ then $y=\phi(h)$ and $z=\phi(k)$ so easily we have $z^{-1}=\phi(k^{-1}$) and then

$$yz^{-1}=\phi(hk^{-1})\in\phi(H)$$