Zassenhaus isomorphism theorem

Statement
Suppose $$A, B, C, D$$ are subgroups of a group $$G$$ such that $$A$$ is a normal subgroup of $$B$$ and $$C$$ is a normal subgroup of $$D$$. Then, $$C(D \cap A)$$ is normal in $$C(D \cap B)$$ and $$A(C \cap D)$$ is normal in $$A(B \cap D)$$, and we have an isomorphism:

$$\frac{C(D \cap B)}{C(D \cap A)} \cong \frac{A(B \cap D)}{A(C \cap D)}$$.

Facts used

 * 1) uses::Normality satisfies transfer condition: If $$H$$ is normal in $$L$$ and $$K$$ is a subgroup of $$L$$, $$H \cap K$$ is normal in $$K$$.
 * 2) uses::Modular property of groups: If $$L \le M$$, then $$L(M \cap N) = M \cap LN$$ and $$(M \cap N)L = M \cap NL$$.
 * 3) uses::Join lemma for normal subgroup of subgroup with normal subgroup of whole group: If $$H \triangleleft K \le M$$ and $$L$$ is normal in $$M$$, then the subgroup $$\langle H, L \rangle = HL$$ is normal in the subgroup $$\langle K, L \rangle = KL$$.
 * 4) uses::Second isomorphism theorem: If $$P, Q$$ are subgroups of a group $$G$$ such that $$Q$$ is normal in $$PQ$$, then $$PQ/Q$$ is isomorphic to $$P/(P \cap Q)$$.

Hands-on proof
Given: $$A, B, C, D$$ are subgroups of a group $$G$$ such that $$A$$ is a normal subgroup of $$B$$ and $$C$$ is a normal subgroup of $$D$$.

To prove: $$C(D \cap A)$$ is normal in $$C(D \cap B)$$ and A(C \cap D) is normal in $$A(B \cap D)$$, and we have an isomorphism:

$$\frac{C(D \cap B)}{C(D \cap A)} \cong \frac{A(B \cap D)}{A(C \cap D)}$$.

Proof:


 * 1) (Given data used: $$A \triangleleft B$$; Fact used: fact (1)): $$D \cap A$$ is normal in $$D \cap B$$: This follows from fact (1), setting $$H = A, L = B, K = D \cap B$$.
 * 2) (Given data used: $$C \triangleleft D$$): $$D \cap B$$ and $$D \cap A$$ normalize $$C$$: Since $$C$$ is normal in $$D$$, the normalizer of $$C$$ contains $$D$$, hence it also contains $$D \cap B$$.
 * 3) $$C(D \cap B)$$ and $$C(D \cap A)$$ are subgroups: This follows directly from step (2).
 * 4) $$C(D \cap A)$$ is normal in $$C(D \cap B)$$: By step (2), $$C$$ is normal in $$C(D \cap B)$$, and by step (1), $$D \cap A$$ is normal in $$D \cap B$$. Thus, by fact (3), we obtain that $$C(D \cap A)$$ is normal in $$C(D \cap B)$$. (Here, we set $$H = D \cap A, K = D \cap B, L = C, M = C(D \cap B)$$).
 * 5) (Facts used: fact (4)): Setting $$P = D \cap B$$ and $$Q = C(D \cap A)$$, we observe that by step (4), $$Q$$ is normal in $$PQ$$, so applying fact (4) yields:
 * 6) * $$\frac{C(D \cap B)}{C(D \cap A)} \cong \frac{D \cap B}{D \cap B \cap C(D \cap A)}$$.
 * 7) (Facts used: fact (2)): Applying fact (2) with $$L = C, M = D, N = A$$ yields $$C(D \cap A) = D \cap CA = D \cap C$$. (Note that both are equal because of the fact that $$C$$, being normal in $$D$$, permutes with $$D \cap A$$. We thus have:
 * 8) * $$\frac{C(D \cap B)}{C(D \cap A)} \cong \frac{D \cap B}{D \cap B \cap CA} = \frac{D \cap B}{D \cap B \cap AC}$$.
 * 9) $$A (C \cap D)$$ is normal in $$A(B \cap D)$$, both are subgroups, and we have (by analogous reasoning to steps (1) - (6)):
 * 10) * $$\frac{A(B \cap D)}{A(C \cap D)} \cong \frac{B \cap D}{B \cap D \cap AC} = \frac{B \cap D}{B \cap D \cap CA}$$.
 * 11) The right sides for steps (6) and (7) are equal, hence the left sides are isomorphic, completing the proof.