Quotient Theorem for Group Homomorphisms/Corollary 2

Theorem

Let $\struct {G, \odot}$ and $\struct {H, *}$ be groups whose identities are $e_G$ and $e_H$ respectively.

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

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

Let $N$ be a normal subgroup of $G$.

Let $q_N: G \to G / N$ denote the quotient epimorphism from $G$ to the quotient group $G / N$.

Then:

$N \subseteq K$

if and only if:

there exists a group epimorphism $\psi: G / N \to H$ such that $\phi = \psi \circ q_N$

Proof 1

From Quotient Theorem for Group Homomorphisms: Corollary 1:

$N \subseteq K$

if and only if:

there exists a group homomorphism $\psi: G / N \to H$ such that $\phi = \psi \circ q_N$

From Surjection if Composite is Surjection, it follows that the group homomorphism $\psi$ is a surjection.

Hence by definition, $\psi$ is an epimorphism.

$\blacksquare$

Proof 2

Let $e$ be the identity element of $G$.

Let $\RR$ be the congruence relation defined by $N$ in $G$.

Let $\RR_\phi$ be the equivalence relation induced by $\phi$.

From Condition for Existence of Epimorphism from Quotient Structure to Epimorphic Image:

there exists an epimorphism $\psi$ from $G / N$ to $H$ which satisfies $\psi \circ q_N = \phi$

if and only if:

$\RR \subseteq \RR_\phi$

It remains to be shown that:

$\RR \subseteq \RR_\phi$

if and only if:

$N \subseteq K$

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Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.14$