# Zassenhaus isomorphism theorem

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## 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 $A(B \cap C)$ is normal in $A(B \cap D)$, and we have an isomorphism: $\frac{C(D \cap B)}{C(D \cap A)} \cong \frac{A(B \cap D)}{A(B \cap C)}$.

## Facts used

1. 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. 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. 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. 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 $P/(P \cap Q)$.

## Proof

### 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 $A(C \cap D) is normal in [itex]A(B \cap D)$, and we have an isomorphism: $\frac{C(D \cap B)}{C(D \cap A)} \cong \frac{A(B \cap D)}{A(B \cap C)}$.

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:
• $\frac{C(D \cap B)}{C(D \cap A)} \cong \frac{D \cap B}{D \cap B \cap C(D \cap A)}$.
6. (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:
• $\frac{C(D \cap B)}{C(D \cap A)} \cong \frac{D \cap B}{D \cap B \cap CA} = \frac{D \cap B}{D \cap B \cap AC}$.
7. $A (B \cap C)$ is normal in $A(B \cap D)$, both are subgroups, and we have (by analogous reasoning to steps (1) - (6)):
• $\frac{A(B \cap D)}{A(B \cap C)} \cong \frac{B \cap D}{B \cap D \cap AC} = \frac{B \cap D}{B \cap D \cap CA}$.
8. The right sides for steps (6) and (7) are equal, hence the left sides are isomorphic, completing the proof.