Show that $T$ is isomorphic to $f(T)$

Question

I'm having trouble figuring out why $T$ is isomorphic to $f(T)$.
$G$ and $H$ are groups and $f:G\to H$ a group isomorphism. $T$ a proper subgroup of $G$ and $f(T)$ a proper subgroup of $H$.
Now, I think because $T$ is a proper subgroup of $G$, and $f$ being group isomorphism, then $T$ is automatically also isomorphic to $f(T)$, but I don't feel confident saying that.

Answer 1

Since $f$ is one-to-one, the restriction of $f$ to $T$ is also ono-to-one. And by definition of $f(T)$, the restriction of $f$ to $T$ is a surjective map from $T$ onto $f(T)$.

Since, furthermore, $f$ is a group homomorphism and $T$ s a subgroup, the restriction of $f$ to $T$ is a group homomorphism. Therefore, the restriction of $f$ to $T$ is an isomorphism from $T$ onto $f(T)$.