Third Isomorphism Theorem/Groups/Proof 1

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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 $H$.

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.

$\blacksquare$


Sources

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