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:
 * $(1): \quad N_1 \paren {H_1 \cap N_2} \lhd N_1 \paren {H_1 \cap H_2}$
 * $(2): \quad \paren {H_1 \cap N_2} \paren {N_1 \cap H_2} \lhd H_1 \cap H_2$

Proof of Isomorphism
Let:
 * $H = H_1 \cap H_2$
 * $N = N_1 \paren {H_1 \cap N_2}$

Then:

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:

Hence the result.

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.