Group Epimorphism Induces Bijection between Subgroups

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Theorem

Let $G_1$ and $G_2$ be groups whose identities are $e_{G_1}$ and $e_{G_2}$ respectively.

Let $\phi: G_1 \to G_2$ be a group epimorphism.

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


Let $\mathbb H_1 = \set {H \subseteq G_1: H \le G_1, K \subseteq H}$ be the set of subgroups of $G_1$ which contain $K$.

Let $\mathbb H_2 = \set {H \subseteq G_2: H \le G_2}$ be the set of subgroups of $G_2$.


Then there exists a bijection $Q: \mathbb H_1 \leftrightarrow \mathbb H_2$ such that:

$\forall N \lhd G_1: \map Q N \lhd G_2$
$\forall N \lhd G_2: \map {Q^{-1} } N \lhd G_1$

where $N \lhd G_1$ denotes that $N$ is a normal subgroup of $G_1$.


That is, normal subgroups map bijectively to normal subgroups under $Q$.


Corollary

Let $H \le G$ denote that $H$ is a subgroup of $G$.


Then:

$\forall H \le G, K \subseteq H: \phi \sqbrk H \cong H / K$

where $H / K$ denotes the quotient group of $H$ by $K$.


Proof

Let $Q$ be the mapping defined as:

$\forall H \le \mathbb H_1: \map Q H = \set {\map \phi h: h \in H}$

Let $H$ be a subgroup of $G_1$ such that $K \subseteq H$.

From Group Homomorphism Preserves Subgroups, $\phi \sqbrk H$ is a subgroup of $G_2$.

This establishes that $Q$ is actually a mapping.


Let $N \lhd G_1$.

From Group Epimorphism Preserves Normal Subgroups, $\phi \sqbrk N$ is a normal subgroup of $G_2$.

This establishes that:

$\forall N \lhd G_1: \map Q N \lhd G_2$


Next it is shown that $Q$ is a bijection.


Injective Nature of $Q$

Let $H, J \in \mathbb H_1$.

Let $\map Q H = \map Q J$.

Let $h \in H$.

\(\ds \map \phi h\) \(\in\) \(\ds \map Q H\)
\(\ds \leadsto \ \ \) \(\ds \map \phi h\) \(\in\) \(\ds \map Q J\)
\(\ds \leadsto \ \ \) \(\ds \exists j \in J: \, \) \(\ds \map \phi j\) \(=\) \(\ds \map \phi h\)
\(\ds \leadsto \ \ \) \(\ds e_{G_2}\) \(=\) \(\ds \paren {\map \phi j}^{-1} \paren \phi h\) Definition of Inverse Element
\(\ds \) \(=\) \(\ds \map \phi {j^{-1} } \map \phi h\) Group Homomorphism Preserves Inverses
\(\ds \) \(=\) \(\ds \map \phi {j^{-1} h}\) morphism property of $\phi$
\(\ds \leadsto \ \ \) \(\ds j^{-1} h\) \(\in\) \(\ds K\) Definition of Kernel of Group Homomorphism
\(\ds \leadsto \ \ \) \(\ds \exists k \in K: \, \) \(\ds j^{-1} h\) \(=\) \(\ds k\)
\(\ds \leadsto \ \ \) \(\ds h\) \(=\) \(\ds j k\)
\(\ds \) \(\in\) \(\ds J\) as $K \subseteq J$ and so $k \in j$
\(\ds \leadsto \ \ \) \(\ds H\) \(\subseteq\) \(\ds J\)

A similar argument shows that $J \subseteq H$.

So by definition of set equality:

$H = J$

Thus:

$\map Q H = \map Q J \implies H = J$

So by definition, $Q$ is injective.

$\Box$


Surjective Nature of $Q$

Now let $N' \in \mathbb H_2$.

By definition of $\mathbb H_2$, $N'$ is a subgroup of $G_2$.

Let $N = \set {x: \map \phi x = N'}$.

We have from Identity of Subgroup that $e_{G_2} \in N'$.

Thus by definition of kernel, $K \subseteq N$.

Now suppose $\map \phi x, \map \phi y \in N'$.

Then:

\(\ds \map \phi {x y^{-1} }\) \(=\) \(\ds \map \phi x \map \phi {y^{-1} }\) Definition of Group Homomorphism
\(\ds \) \(=\) \(\ds \map \phi x \map \phi {y^{-1} }\) Group Homomorphism Preserves Inverses
\(\ds \) \(\in\) \(\ds N'\) One-Step Subgroup Test
\(\ds \leadsto \ \ \) \(\ds x y^{-1}\) \(\in\) \(\ds N\)

So by the One-Step Subgroup Test, $N$ is a subgroup of $G_1$.

It has been established that $K \subseteq N$, and so $N \in \mathbb H_1$.

Thus it follows that for all $N' \in \mathbb H_2$ there exists $N \in H_1$ such that $\map Q N = N'$.

So $Q$ is a surjection.

$\Box$


So $Q$ has been shown to be both an injection and a surjection, and so by definition is a bijection.


Finally, it can then be shown that if $N'$ is normal in $G_2$, it follows that $N = \map {Q^{-1} } {N'}$ is normal in $G_1$.

This establishes that:

$\forall N \lhd G_2: \map {Q^{-1} } N \lhd G_1$


This needs considerable tedious hard slog to complete it.
In particular: if $N'$ is normal in $G_2$, it follows that $N = \map {Q^{-1} } {N'}$ is normal in $G_1$.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Finish}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.


$\blacksquare$


Also see


Sources

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