Third Isomorphism Theorem/Groups/Corollary 2

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Theorem

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 $G$.

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:

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


This can be illustrated by means of the following commutative diagram:


$\begin{xy}\[email protected][email protected]+1em{ G \ar[dr]^*{\phi} \ar[d]_*{q} & \\ G / N \[email protected]{-->}[r]_*{\psi} & H }\end{xy}$


Proof

Necessary Condition

Suppose $\psi$ exists as defined.

Then:

$K = \map {\phi^{-1} } {e_H} = \map {q^{-1} } {\map {\psi^{-1} } {e_H} }$

That is: $\map {\psi^{-1} } {e_H}$ is the kernel of $\psi$.

So by Kernel is Normal Subgroup of Domain, $\map {\psi^{-1} } {e_H}$ is a normal subgroup of $G / N$.

This corresponds via $q^{-1}$ to $K$, which must then be a normal subgroup of $G$ which contains $N$.

$\Box$


Sufficient Condition

Suppose $N \subseteq K$.

Let $\psi: G / N \to H$ be defined as:

$\forall g N \in G / N: \map \psi {g N} = \map \phi g \in H$

We have that:

$n \in g N \implies n^{-1} g \in N \subseteq K$

So:

$\map \phi {n^{-1} g} = \map \phi n^{-1} \map \phi g = e_H$

and so $\map \phi n = \map \phi g$.

Thus $\psi$ is well-defined.


As $N \subseteq K$ , the Quotient Theorem for Group Homomorphisms has that $\Img \phi$ is isomorphic to $G / K$.

We have that:

$\Img \phi = \Img {\psi \circ q} = \Img \psi$

and:

$\map \ker \psi = K / N = \set {k n: k \in K}$


Thus:

$\Img \psi$ is isomorphic to $G / K$

and also:

$\Img \psi$ is isomorphic to the quotient of the domain $G / N$ of $\psi$ modulo the kernel $K / N$ of $\psi$.

$\blacksquare$


Sources