Preimage of Image of Subgroup under Group Epimorphism

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 epimorphism.

Let $K = \map \ker \phi$ denote the kernel of $\phi$.

Then:
 * $\phi^{-1} \sqbrk {\phi \sqbrk H} = H \circ K$

where:
 * $\phi \sqbrk H$ denotes the image of $H$ under $\phi$
 * $\phi^{-1} \sqbrk {\phi \sqbrk H}$ denotes the preimage of $\phi \sqbrk H$ under $\phi$
 * $H \circ K$ denotes the subset product of $H$ with $K$.

Proof
Let $e_2$ be the identity element of $\struct {G_2, *}$.

Let $x \in \phi^{-1} \sqbrk {\phi \sqbrk H}$.

Then:

So we have shown that:
 * $\phi^{-1} \sqbrk {\phi \sqbrk H} \subseteq H \circ K$

Now suppose that $x \in H \circ K$.

Then:

So we have shown that:
 * $H \circ K \subseteq \phi^{-1} \sqbrk {\phi \sqbrk H}$

Hence the result by definition of set equality.