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

## Theorem

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

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

Let $\struct {G, \odot}$ be the internal group direct product of $\struct {H, \odot}$ and $\struct {K, \odot}$.

Then:

- 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

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

- $\forall x \in G: \map {q_H} x = x \odot H$

From Restriction of Homomorphism is Homomorphism:

- the restriction of $q_H$ to $K$ is a homomorphism.

From Lagrange's Theorem (Group Theory):

- $\card {G / H} = \card K$

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 $G$

From Restriction of Homomorphism is Homomorphism:

- the restriction of $q_K$ to $H$ is a homomorphism.

From Lagrange's Theorem (Group Theory):

- $\card {G / K} = \card H$

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.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.9 \ \text{(a)}$