Zassenhaus isomorphism theorem

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This article is about an isomorphism theorem in group theory.
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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 $L\leq M$, 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 $H\triangleleft K\leq M$ 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 $A(B\cap D)$, 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.

Proof by application of the correspondence theorem

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