Group Homomorphism Preserves Subgroups

Theorem
Let $\left({G_1, \circ}\right)$ and $\left({G_2, *}\right)$ be groups.

Let $\phi: \left({G_1, \circ}\right) \to \left({G_2, *}\right)$ be a group homomorphism.

Then:
 * $H \le G_1 \implies \phi \left({H}\right) \le G_2$

where $\le$ denotes subgroup.

That is, group homomorphism preserves subgroups.

Proof
Let $H \le G_1$.

First note that from Null Relation is Mapping iff Domain is Empty Set:
 * $H \ne \varnothing \implies \phi \left({H}\right) \ne \varnothing$

and so $\phi \left({H}\right)$ is not empty.

Next, let $x, y \in \phi \left({H}\right)$.

Then:
 * $\exists h_1, h_2 \in H: x = \phi \left({h_1}\right), y = \phi \left({h_2}\right)$

From the definition of Group Homomorphism, we have:
 * $\phi \left({h_1^{-1} \circ h_2}\right) = x^{-1} * y$

Since $H$ is a subgroup, $h_1^{-1} \circ h_2 \in H$.

Hence $x^{-1} * y \in \phi \left({H}\right)$.

Thus from the One-Step Subgroup Test, $\phi \left({H}\right) \le G_2$.