# Quotient Epimorphism Condition for Normal Subgroup Product to be Internal Group Direct Product

## Theorem

Let $\struct {G, \odot}$ be a group

Let $\struct {H, \odot}$ and $\struct {K, \odot}$ be normal subgroups of $\struct {G, \odot}$.

Then:

$\struct {G, \odot}$ is the internal group direct product of $\struct {H, \odot}$ and $\struct {K, \odot}$
the restriction of the quotient epimorphism $q_H$ to $K$ is an isomorphism from $K$ onto the quotient group $G / H$

and:

the restriction of the quotient epimorphism $q_K$ to $H$ is an isomorphism from $H$ onto the quotient group $G / K$

## Proof

### Sufficient Condition

Recall that the quotient epimorphism $q_H: G \to G / H$ is defined as:

$\forall x \in G: \map {q_H} x = x \odot H$
the restriction of $q_H$ to $K$ is a homomorphism.
$\card {G / H} = \card K$

Then we have:

 $\ds \forall a, b \in K: \,$ $\ds \map {q_H} a$ $=$ $\ds \map {q_H} b$ by assumption $\ds \leadsto \ \$ $\ds a \odot H$ $=$ $\ds b \odot H$ Definition of Quotient Group Epimorphism $\ds \leadsto \ \$ $\ds \exists h, h' \in H: \,$ $\ds a \odot \paren {a^{-1} \odot h \odot a}$ $=$ $\ds b \odot \paren {b^{-1} \odot h' \odot b}$ as $H$ is a normal subgroup of $G$ $\ds \leadsto \ \$ $\ds h \odot a$ $=$ $\ds h' \odot b$ where $h, h' \in H$ and $a, b \in K$

But by definition of internal group direct product, a representation of $g \in G$ in the form $h \odot k$ where $h \in H$ and $k \in K$ is unique.

That is:

 $\ds h$ $=$ $\ds h'$ $\ds a$ $=$ $\ds b$

Hence the restriction of $q_H$ to $K$ is an injection.

Hence by Injection to Equivalent Finite Set is Bijection, it follows that:

$q_H$ is a bijection.

A bijective homomorphism is an isomorphism by definition.

$\Box$

Similarly, the quotient epimorphism $q_K: G \to G / K$ is defined as:

$\forall x \in G: \map {q_K} x = x \odot K = K \odot x$

as $K$ is a normal subgroup of $G$

the restriction of $q_K$ to $H$ is a homomorphism.
$\card {G / K} = \card H$

Then we have:

 $\ds \forall a, b \in K: \,$ $\ds \map {q_K} a$ $=$ $\ds \map {q_K} b$ by assumption $\ds \leadsto \ \$ $\ds K \odot a$ $=$ $\ds K \odot b$ Definition of Quotient Group Epimorphism $\ds \leadsto \ \$ $\ds \exists k, k' \in K: \,$ $\ds \paren {a \odot k \odot a^{-1} } \odot a$ $=$ $\ds \paren {b \odot k' \odot b^{-1} } \odot b$ as $K$ is a normal subgroup of $G$ $\ds \leadsto \ \$ $\ds a \odot k$ $=$ $\ds b \odot k'$ where $a, b \in H$ and $k, k' \in K$

But by definition of internal group direct product, a representation of $g \in G$ in the form $h \odot k$ where $h \in H$ and $k \in K$ is unique.

That is:

 $\ds k$ $=$ $\ds k'$ $\ds a$ $=$ $\ds b$

Hence the restriction of $q_K$ to $H$ is an injection.

Hence by Injection to Equivalent Finite Set is Bijection, it follows that:

$q_K$ is a bijection.

A bijective homomorphism is an isomorphism by definition.

Hence the result.

$\Box$

### Necessary Condition

$\blacksquare$