Third Isomorphism Theorem/Groups

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.