Third Isomorphism Theorem/Groups/Proof 2

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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 $q_H$ 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.
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.



Sources

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