Group Epimorphism Induces Bijection between Subgroups/Corollary

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Theorem

Let $G_1$ and $G_2$ be groups whose identities are $e_{G_1}$ and $e_{G_2}$ respectively.

Let $\phi: G_1 \to G_2$ be a group epimorphism.

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

Let $H \le G$ denote that $H$ is a subgroup of $G$.


Then:

$\forall H \le G, K \subseteq H: \phi \sqbrk H \cong H / K$

where $H / K$ denotes the quotient group of $H$ by $K$.


Proof

Let $H$ be a subgroup of $G$ such that $K \subseteq H$.

Consider the restriction of $\phi$ to $H$.

By Group Homomorphism Preserves Subgroups, $\phi_{\restriction_H} \sqbrk H$ is a group.

From Group Epimorphism Induces Bijection between Subgroups it follows that the First Isomorphism Theorem can be applied.

Hence the result.

$\blacksquare$


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