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:

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


Corollary 1

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:

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


Corollary 2

Let $G$ and $H$ be groups whose identities are $e_G$ and $e_H$ respectively.

Let $\phi: G \to H$ be a group homomorphism.

Let $K$ be the kernel of $\phi$.

Let $N$ be a normal subgroup of $G$ such that $N \subseteq K$.

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


Then $N \subseteq K$ if and only if there exists a group homomorphism $\psi: \dfrac G N \to H$ such that:

$\phi = \psi \circ q$

and:

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


Proof

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 $x N = y N \implies y^{-1} x \in N$.

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:

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


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:

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


The result follows from the First Isomorphism Theorem.

$\blacksquare$


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