Zassenhaus isomorphism theorem

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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
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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 \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

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