Third Isomorphism Theorem/Groups/Proof 3

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Theorem

Let $G$ be a group, and let:

$H, N$ be normal subgroups of $G$
$N$ be a subset of $H$.


Then:

$(1): \quad N$ is a normal subgroup of $H$
$(2): \quad H / N$ is a normal subgroup of $G / N$
where $H / N$ denotes the quotient group of $H$ by $N$
$(3): \quad \dfrac {G / N} {H / N} \cong \dfrac G H$
where $\cong$ denotes group isomorphism.


Proof

From Normal Subgroup which is Subset of Normal Subgroup is Normal in Subgroup, $N$ is a normal subgroup of $H$.

Let $q_H$ denote the quotient mapping from $G$ to $\dfrac G H$.

Let $q_N$ denote the quotient mapping from $G$ to $\dfrac G N$.


Let $\RR$ be the congruence relation defined by $N$ in $G$.

Let $\TT$ be the congruence relation defined by $H$ in $G$.


Thus from Congruence Relation induces Normal Subgroup:

$q_H = q_\TT$

and:

$q_N = q_\RR$

where $q_\RR$ and $q_\TT$ denote the quotient epimorphisms induced by $\RR$ and $\TT$ respectively.


We have that:

\(\ds \tuple {x, y}\) \(\in\) \(\ds \RR\)
\(\ds \leadsto \ \ \) \(\ds x N\) \(=\) \(\ds y N\) Definition of Congruence Modulo Subgroup
\(\ds \leadsto \ \ \) \(\ds x y^{-1}\) \(\in\) \(\ds N\) Definition of Normal Subgroup
\(\ds \leadsto \ \ \) \(\ds x y^{-1}\) \(\in\) \(\ds H\) as $N \subseteq H$
\(\ds \leadsto \ \ \) \(\ds x H\) \(=\) \(\ds y H\) Definition of Normal Subgroup
\(\ds \tuple {x, y}\) \(\in\) \(\ds \TT\) Definition of Congruence Modulo Subgroup

That is:

$\RR \subseteq \TT$


Let $\SS$ be the relation on the quotient group $G / N$ which satisfies:

$\forall X, Y \in G / N: X \mathrel \SS Y \iff \exists x \in X, y \in Y: x \mathrel \TT y$

That is, by definition of $\TT$:

$\forall X, Y \in G / N: X \mathrel \SS Y \iff \exists x \in X, y \in Y: x H = y H$


Then by Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence: Corollary:

$\SS$ is a congruence relation on $G / N$.

Hence by Congruence Relation on Group induces Normal Subgroup:

$\SS$ induces a normal subgroup of $G / N$.


This needs considerable tedious hard slog to complete it.
In particular: We need to identify $q_\SS$ with $q_{H / N}$, that is, show that the normal subgroup defined above is $H / N$.
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 {{Finish}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.


Again from Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence: Corollary:

there exists a unique isomorphism $\phi$ from $G / N$ to $G / H$ which satisfies:
$\phi \circ q_\SS \circ q_\RR = q_\TT$
where $q_\SS$, $q_\RR$ and $q_\TT$ denote the quotient epimorphisms as appropriate.

That is:

$\phi \circ q_{H / N} \circ q_N = q_H$

and the result follows.


Sources

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