Preimage of Image of Subgroup under Group Epimorphism

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Theorem

Let $\struct {G_1, \circ}$ and $\struct {G_2, *}$ be groups.

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

Let $K = \map \ker \phi$ denote the kernel of $\phi$.


Then:

$\phi^{-1} \sqbrk {\phi \sqbrk H} = H \circ K$

where:

$\phi \sqbrk H$ denotes the image of $H$ under $\phi$
$\phi^{-1} \sqbrk {\phi \sqbrk H}$ denotes the preimage of $\phi \sqbrk H$ under $\phi$
$H \circ K$ denotes the subset product of $H$ with $K$.


Proof

Let $e_2$ be the identity element of $\struct {G_2, *}$.

Let $x \in \phi^{-1} \sqbrk {\phi \sqbrk H}$.

Then:

\(\ds x\) \(\in\) \(\ds \phi^{-1} \sqbrk {\phi \sqbrk H}\)
\(\ds \leadsto \ \ \) \(\ds \map \phi x\) \(\in\) \(\ds \phi \sqbrk H\) Definition of Preimage of Element under Mapping
\(\ds \leadsto \ \ \) \(\ds \exists b \in H: \, \) \(\ds \map \phi x\) \(=\) \(\ds \map \phi b\) Definition of Image of Element under Mapping
\(\ds \leadsto \ \ \) \(\ds \map \phi x * \paren {\map \phi b}^{-1}\) \(=\) \(\ds e_2\) Definition of Inverse Element in $G_2$
\(\ds \leadsto \ \ \) \(\ds \map \phi {x \circ b^{-1} }\) \(=\) \(\ds e_2\) as $\phi$ is a homomorphism
\(\ds \leadsto \ \ \) \(\ds x \circ b^{-1}\) \(\in\) \(\ds K\) Definition of Kernel of Group Homomorphism
\(\ds \leadsto \ \ \) \(\ds b^{-1} \circ x\) \(\in\) \(\ds K\) Definition of Normal Subgroup: Kernel is Normal Subgroup of Domain
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds H \circ K\) as $x = b \circ \paren {b^{-1} \circ x}$


So we have shown that:

$\phi^{-1} \sqbrk {\phi \sqbrk H} \subseteq H \circ K$


Now suppose that $x \in H \circ K$.

Then:

\(\ds x\) \(\in\) \(\ds H \circ K\)
\(\ds \leadsto \ \ \) \(\ds \exists b \in H, a \in K: \, \) \(\ds x\) \(=\) \(\ds b \circ a\)
\(\ds \leadsto \ \ \) \(\ds \map \phi x\) \(=\) \(\ds \map \phi b * \map \phi a\) as $\phi$ is a homomorphism
\(\ds \) \(=\) \(\ds \map \phi b * e_2\) as $a \in K$
\(\ds \) \(=\) \(\ds \map \phi b\)
\(\ds \leadsto \ \ \) \(\ds \map \phi x\) \(\in\) \(\ds \phi \sqbrk H\) as $b \in H$
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds \phi^{-1} \sqbrk {\phi \sqbrk H}\) Definition of Preimage of Subset under Mapping


So we have shown that:

$H \circ K \subseteq \phi^{-1} \sqbrk {\phi \sqbrk H}$


Hence the result by definition of set equality.

$\blacksquare$


Sources

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