# Zassenhaus Lemma

From ProofWiki

## Lemma

Let $G$ be a group.

Let $H_1$ and $H_2$ be subgroups of $G$.

Let:

- $N_1 \lhd H_1$
- $N_2 \lhd H_2$

where $\lhd$ denotes the relation of being a normal subgroup.

then:

- $\displaystyle \frac {N_1 \left({H_1 \cap H_2}\right)} {N_1 \left({H_1 \cap N_2}\right)} \cong \frac {H_1 \cap H_2} {\left({H_1 \cap N_2}\right) \left({N_1 \cap H_2}\right)} \cong \frac {N_2 \left({H_1 \cap H_2}\right)} {N_2 \left({N_1 \cap H_2}\right)}$

where:

- $N_1 \left({H_1 \cap H_2}\right)$ etc. denotes subset product
- $\cong$ denotes group isomorphism.

This lemma is also known as:

- the
**butterfly lemma**, so named because the Hasse diagram of the various subgroups involved can be drawn to resemble a butterfly - the
**fourth isomorphism theorem**, which name is not recommended because of the lack of any consistent naming convention for the Isomorphism Theorems in the literature.

## Proof

Because of symmetry, only the first of the isomorphisms needs to be proved.

### Proof of Normality

In order for the expressions to make sense, the expressions on the bottom of the fractions need to be normal subgroups.

That is, we need to show that:

- $(1): \quad N_1 \left({H_1 \cap N_2}\right) \lhd N_1 \left({H_1 \cap H_2}\right)$
- $(2): \quad \left({H_1 \cap N_2}\right) \left({N_1 \cap H_2}\right) \lhd H_1 \cap H_2$

### Proof of Isomorphism

Let:

- $H = H_1 \cap H_2$
- $N_1 \left({H_1 \cap N_2}\right)$

Then:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \frac {N_1 \left({H_1 \cap N_2}\right) \left({H_1 \cap H_2}\right)} {N_1 \left({H_1 \cap N_2}\right)}\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \frac {N H} H\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\cong\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \frac H {H \cap N}\) | \(\displaystyle \) | \(\displaystyle \) | Third Isomorphism Theorem | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \frac {H_1 \cap H_2} {H_1 \cap H_2 \cap N_1 \left({H_1 \cap N_2}\right)}\) | \(\displaystyle \) | \(\displaystyle \) |

From Intersection with Subgroup Product of Superset we have:

- If $X, Y, Z$ are subgroups of a group $\left({G, \circ}\right)$, and $Y \subseteq X$, then:
- $X \cap \left({Y \circ Z}\right) = Y \circ \left({X \cap Z}\right)$

Thus:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle N_1 \left({H_1 \cap N_2}\right) \left({H_1 \cap H_2}\right)\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle N_1 \left({H_1 \cap N_2}\right)\) | \(\displaystyle \) | \(\displaystyle \) | by taking $X = H_1, Y = H_1 \cap N_2, Z = H_2$ | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle H_1 \cap H_2 \cap N_1 \left({H_1 \cap N_2}\right)\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \left({H_1 \cap N_2}\right) \left({N_1 \cap H_2}\right)\) | \(\displaystyle \) | \(\displaystyle \) | by taking $X = H_1 \cap H_2, Y = H_1 \cap N_2, Z = N_1$ |

Hence the result.

$\blacksquare$

## Source of Name

This entry was named for Hans Julius Zassenhaus.

## Sources

- Allan Clark:
*Elements of Abstract Algebra*(1971)... (previous)... (next): $\S 70$