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 $G$ 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

Necessary Condition

Suppose $\psi$ exists as defined.

Then:

$K = \phi^{-1} \left({e_H}\right) = q^{-1} \left({\psi^{-1} \left({e_H}\right)}\right)$

That is: $\psi^{-1} \left({e_H}\right)$ is the kernel of $\psi$.

So by Kernel is Normal Subgroup of Domain, $\psi^{-1} \left({e_H}\right)$ 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:

$\forall g N \in G / N: \psi \left({g N}\right) = \phi \left({g}\right) \in H$

We have that:

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

So:

$\phi \left({n^{-1} g}\right) = \phi \left({n}\right)^{-1} \phi \left({g}\right) = e_H$

and so $\phi \left({n}\right) = \phi \left({g}\right)$.

Thus $\psi$ is well-defined.


As $N \subseteq K$ , the Quotient Theorem for Group Homomorphisms has that $\operatorname{Im} \left({\phi}\right)$ is isomorphic to $G / K$.

We have that:

$\operatorname{Im} \left({\phi}\right) = \operatorname{Im} \left({\psi \circ q}\right) = \operatorname{Im} \left({\psi}\right)$

and:

$\ker \left({\psi}\right) = K / N = \left\{{k n: k \in K}\right\}$


Thus:

$\operatorname{Im} \left({\psi}\right)$ is isomorphic to $G / K$

and also:

$\operatorname{Im} \left({\psi}\right)$ is isomorphic to the quotient of the domain $G / N$ of $\psi$ modulo the kernel $K / N$ of $\psi$.

$\blacksquare$


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