Unique Epimorphism from Quotient Group to Quotient Group

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Theorem

Let $\struct {G_1, \circledcirc}$ and $\struct {G_2, \circledast}$ be groups.

Let $\phi: \struct {G_1, \circledcirc} \to \struct {G_2, \circledast}$ be a group epimorphism.

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


Let $H_1$ be a normal subgroup of $\struct {G_1, \circledcirc}$.

Let $H_2 := \phi \sqbrk {H_1}$ be the image of $H_1$ under $\phi$.


Then there exists a unique epimorphism $\psi$ from $G_1 / H_1$ to $G_2 / H_2$ such that:

$\psi \circ q_{H_1} = q_{H_2} \circ \phi$

where:

$G_1 / H_1$ and $G_2 / H_2$ are quotient groups of $G_1$ by $H_1$ and of $G_2$ by $H_2$ respectively
$q_{H_1}: G_1 \to G_1 / H_1$ and $q_{H_2}: G_2 \to G_2 / H_2$ are their respective quotient mappings.


The kernel of $\psi$ is given by:

$\map \ker \psi = \dfrac {H_1 \circledcirc K} {H_1}$


Proof

First we note that by Group Epimorphism Preserves Normal Subgroups, $H_2$ is indeed a normal subgroup of $\struct {G_2, \circledast}$.

Hence $G_2 / H_2$ and $q_{H_2}: G_2 \to G_2 / H_2$ are appropriately defined.

Let $\RR$ be the congruence relation induced on $\struct {G_1, \circledcirc}$ by $H$.

Let $q_\RR$ be the quotient mapping induced by $\RR$.

Then by definition $q_\RR = q_{H_1}$.


We are given that $\phi$ is a group epimorphism.

We also have from Quotient Group Epimorphism is Epimorphism that $q_{H_2}$ is also a group epimorphism.

Hence from Composite of Group Epimorphisms is Epimorphism, $q_{H_2} \circ \phi$ is also a group epimorphism.

So from Quotient Theorem for Epimorphisms, there exists a unique epimorphism $\psi$ from $G_1 / H_1$ to $G_2 / H_2$ such that:

$\psi \circ q_{H_1} = q_{H_2} \circ \phi$


It remains to be shown that:

$\map \ker \psi = \dfrac {H_1 \circledcirc K} {H_1}$





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