Third Isomorphism Theorem/Groups/Proof 3

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 $H / N$.
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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

1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.19 \ \text {(b)}$