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

From ProofWiki
Jump to navigation Jump to search

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 $\struct {K, \odot}$

if and only if

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:

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


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.

$\Box$


Necessary Condition


This theorem requires a proof.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.


$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Quotient_Epimorphism_Condition_for_Normal_Subgroup_Product_to_be_Internal_Group_Direct_Product&oldid=565624"