# Group Homomorphism Preserves Subgroups

## Theorem

Let $\struct {G_1, \circ}$ and $\struct {G_2, *}$ be groups.

Let $\phi: \struct {G_1, \circ} \to \struct {G_2, *}$ be a group homomorphism.

Then:

$H \le G_1 \implies \phi \sqbrk H \le G_2$

where:

$\phi \sqbrk H$ denotes the image of $H$ under $\phi$
$\le$ 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:

$H \ne \O \implies \phi \sqbrk H \ne \O$

and so $\phi \sqbrk H$ is not empty.

Next, let $x, y \in \phi \sqbrk H$.

Then:

$\exists h_1, h_2 \in H: x = \map \phi {h_1}, y = \map \phi {h_2}$

From the definition of Group Homomorphism, we have:

$\map \phi {h_1^{-1} \circ h_2} = x^{-1} * y$

Since $H$ is a subgroup:

$h_1^{-1} \circ h_2 \in H$

Hence:

$x^{-1} * y \in \phi \sqbrk H$

Thus from the One-Step Subgroup Test:

$\phi \sqbrk H \le G_2$

$\blacksquare$