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 Mapping:
 * $$H \ne \varnothing \implies \phi \left({H}\right) \ne \varnothing$$

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, $$\phi \left({H}\right) \le G_2$$.