Zassenhaus Lemma

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:

$\dfrac {N_1 \paren {H_1 \cap H_2} } {N_1 \paren {H_1 \cap N_2} } \cong \dfrac {H_1 \cap H_2} {\paren {H_1 \cap N_2} \paren {N_1 \cap H_2} } \cong \dfrac {N_2 \paren {H_1 \cap H_2} } {N_2 \paren {N_1 \cap H_2} }$

where:

$N_1 \paren {H_1 \cap H_2}$, and so on, denotes subset product

$\cong$ denotes group isomorphism.

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:

$(1): \quad N_1 \paren {H_1 \cap N_2} \lhd N_1 \paren {H_1 \cap H_2}$

$(2): \quad \paren {H_1 \cap N_2} \paren {N_1 \cap H_2} \lhd H_1 \cap H_2$

This needs considerable tedious hard slog to complete it. 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.

Proof of Isomorphism

Let:

$H = H_1 \cap H_2$

$N = N_1 \paren {H_1 \cap N_2}$

Then:

\(\ds \frac {N_1 \paren {H_1 \cap N_2} \paren {H_1 \cap H_2} } {N_1 \paren {H_1 \cap N_2} }\)

\(=\)

\(\ds \frac {N H} N\)

\(\ds \)

\(\cong\)

\(\ds \frac H {H \cap N}\)

Second Isomorphism Theorem

\(\ds \)

\(=\)

\(\ds \frac {H_1 \cap H_2} {H_1 \cap H_2 \cap N_1 \paren {H_1 \cap N_2} }\)

From Intersection with Subgroup Product of Superset:

If $X, Y, Z$ are subgroups of a group $\struct {G, \circ}$, and $Y \subseteq X$, then:

$X \cap \paren {Y \circ Z} = Y \circ \paren {X \cap Z}$

Thus:

\(\ds N_1 \paren {H_1 \cap N_2} \paren {H_1 \cap H_2}\)

\(=\)

\(\ds N_1 \paren {H_1 \cap H_2}\)

by taking $X = H_1, Y = H_1 \cap N_2, Z = H_2$

\(\ds H_1 \cap H_2 \cap N_1 \paren {H_1 \cap N_2}\)

\(=\)

\(\ds \paren {H_1 \cap N_2} \paren {N_1 \cap H_2}\)

by taking $X = H_1 \cap H_2, Y = H_1 \cap N_2, Z = N_1$

Hence the result.

$\blacksquare$

Also known as

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

either the third isomorphism theorem or the fourth isomorphism theorem, which names are not recommended because of the lack of any consistent naming convention for the Isomorphism Theorems in the literature.

Source of Name

This entry was named for Hans Julius Zassenhaus.

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Group Homomorphism and Isomorphism: $\S 70$. The Third Isomorphism Theorem