Third Isomorphism Theorem/Groups

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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.


Corollary

Let $G$ be a group.

Let $N$ be a normal subgroup of $G$.

Let $q: G \to \dfrac G N$ be the quotient epimorphism from $G$ to the quotient group $\dfrac G N$.

Let $K$ be the kernel of $q$.


Then:

$\dfrac G N \cong \dfrac {G / K} {N / K}$


Proof 1

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

We define a mapping:

$\phi: G / N \to G / H$ by $\map \phi {g N} = g H$

Since $\phi$ is defined on cosets, we need to check that $\phi$ is well-defined.

Suppose $x N = y N$.

Then:

$y^{-1} x \in N$

Then:

$N \le H \implies y^{-1} x \in H$

and so:

$x H = y H$

So:

$\map \phi {x N} = \map \phi {y N}$

and $\phi$ is indeed well-defined.


Now $\phi$ is a homomorphism, from:

\(\ds \map \phi {x N} \map \phi {y N}\) \(=\) \(\ds \paren {x H} \paren {y H}\)
\(\ds \) \(=\) \(\ds x y H\)
\(\ds \) \(=\) \(\ds \map \phi {x y N}\)
\(\ds \) \(=\) \(\ds \map \phi {x N y N}\)


Also, since $N \subseteq H$, it follows that:

$\order N \le \order H$

So:

$\order {G / N} \ge \order {G / H}$, indicating $\phi$ is surjective.


So:

\(\ds \map \ker \phi\) \(=\) \(\ds \set {g N \in G / N: \map \phi {g N} = e_{G / H} }\)
\(\ds \) \(=\) \(\ds \set {g N \in G / N: g H = H}\)
\(\ds \) \(=\) \(\ds \set {g N \in G / N: g \in H}\)
\(\ds \) \(=\) \(\ds H / N\)


The result follows from the First Isomorphism Theorem.

$\blacksquare$


Proof 2

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

Let $q_H$ and $q_N$ be the quotient mappings from $G$ to $\dfrac G H$ and $G$ to $\dfrac G N$ respectively.

Hence:

$N \subseteq \map \ker {q_H}$


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From Quotient Theorem for Group Homomorphisms: Corollary 2, it therefore follows that:

there exists a group epimorphism $\psi: \dfrac G N \to \dfrac G H$ such that $\psi \circ q_N = q_H$

Then we have that :

there exists a group epimorphism $\phi: \dfrac {G / N} {H / N} \to \dfrac G N$ such that $\phi \circ q_{H / N} = \psi$

if and only if:

$H / N \subseteq \map \ker \psi$


This needs considerable tedious hard slog to complete it.
In particular: Just need to prove that $H / N \subseteq \map \ker \psi$ and the job is done.
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Thus we form the composite:

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

and it remains to be shown that $\phi$ is an isomorphism.


This needs considerable tedious hard slog to complete it.
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Proof 3

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.
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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.


Also known as

This result is also referred to by some sources as the first isomorphism theorem, and by others as the second isomorphism theorem.


Also see


Sources

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