Zassenhaus isomorphism theorem

This article describes a fact or result that is not basic but it still well-established and standard. The fact may involve terms that are themselves non-basic
View other semi-basic facts in group theory
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|
VIEW: Survey articles about this

This article is about an isomorphism theorem in group theory.
View a complete list of isomorphism theorems| Read a survey article about the isomorphism theorems

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 $A(C \cap D) is normal in <math>A(B \cap D)$, 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

PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.