# Group Epimorphism Induces Bijection between Subgroups

## Contents

## 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 $K := \ker \left({\phi}\right)$ be the kernel of $\phi$.

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

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

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

- $\forall N \lhd G_1: Q \left({N}\right) \lhd G_2$
- $\forall N \lhd G_2: Q^{-1} \left({N}\right) \lhd G_1$

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:

- $\forall H \le \mathbb H_1: Q \left({H}\right) = \left\{{\phi \left({h}\right): h \in H}\right\}$

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

From Group Homomorphism Preserves Subgroups, $\phi \left({H}\right)$ 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, $\phi \left({N}\right)$ is a normal subgroup of $G_2$.

This establishes that:

- $\forall N \lhd G_1: Q \left({N}\right) \lhd G_2$

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

### Injective Nature of $Q$

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

Let $Q \left({H}\right) = Q \left({J}\right)$.

Let $h \in H$.

\(\displaystyle \phi \left({h}\right)\) | \(\in\) | \(\displaystyle Q \left({H}\right)\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \phi \left({h}\right)\) | \(\in\) | \(\displaystyle Q \left({J}\right)\) | ||||||||||

\(\displaystyle \implies \ \ \) | \(\, \displaystyle \exists j \in J: \, \) | \(\displaystyle \phi \left({j}\right)\) | \(=\) | \(\displaystyle \phi \left({h}\right)\) | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle e_{G_2}\) | \(=\) | \(\displaystyle \left({\phi \left({j}\right)}\right)^{-1} \phi \left({h}\right)\) | definition of inverse element | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \phi \left({j^{-1} }\right) \phi \left({h}\right)\) | Group Homomorphism Preserves Inverses | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \phi \left({j^{-1} h}\right)\) | morphism property of $\phi$ | ||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle j^{-1} h\) | \(\in\) | \(\displaystyle K\) | definition of kernel | |||||||||

\(\displaystyle \implies \ \ \) | \(\, \displaystyle \exists k \in K: \, \) | \(\displaystyle j^{-1} h\) | \(=\) | \(\displaystyle k\) | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle h\) | \(=\) | \(\displaystyle j k\) | ||||||||||

\(\displaystyle \) | \(\in\) | \(\displaystyle J\) | as $K \subseteq J$ and so $k \in j$ | ||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle H\) | \(\subseteq\) | \(\displaystyle J\) |

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

So by definition of set equality:

- $H = J$

Thus:

- $Q \left({H}\right) = Q \left({J}\right) \implies H = J$

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 $N = \left\{{x: \phi \left({x}\right) = N'}\right\}$.

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

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

Now suppose $\phi \left({x}\right), \phi \left({y}\right) \in N'$.

Then:

\(\displaystyle \phi \left({x y^{-1} }\right)\) | \(=\) | \(\displaystyle \phi \left({x}\right) \phi \left({y^{-1} }\right)\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \phi \left({x}\right) \phi \left({y}\right)^{-1}\) | |||||||||||

\(\displaystyle \) | \(\in\) | \(\displaystyle N'\) | One-Step Subgroup Test | ||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle x y^{-1}\) | \(\in\) | \(\displaystyle N\) |

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 $Q \left({N}\right) = N'$.

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 $N = Q^{-1} \left({N'}\right)$ is normal in $G_1$.

This establishes that:

- $\forall N \lhd G_2: Q^{-1} \left({N}\right) \lhd G_1$

$\blacksquare$

## Also see

## Sources

- 1966: Richard A. Dean:
*Elements of Abstract Algebra*... (previous) ... (next): $\S 1.10$: Theorem $29$