Subnormality satisfies transfer condition

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This article gives the statement, and possibly proof, of a subgroup property (i.e., subnormal subgroup) satisfying a subgroup metaproperty (i.e., transfer condition)
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Statement

Verbal statement

The intersection of a subnormal subgroup with any subgroup is subnormal in that subgroup. Moreover, the subnormal depth of the intersection is bounded from above by the subnormal depth of the original subgroup.

Statement with symbols

Suppose H is a subnormal subgroup of a group G, and K \le G is a subgroup. Then, H \cap K is a subnormal subgroup of K. Moreover, if H is k-subnormal in G, H \cap K is k-subnormal in K (i.e., its subnormal depth is at most k.

Related facts

Related facts about subnormality

Related facts about normality

Facts used

  1. Normality satisfies transfer condition: If H is a normal subgroup of a group G, and K \le G, then H \cap K is a normal subgroup of K.
  2. Transfer condition is subordination-closed: If p is a subgroup property satisfying the transfer condition, the subordination of p.

Proof

Hands-on proof

Given: A group G, a k-subnormal subgroup H of G, and a subgroup K of G.

To prove: H \cap K is a k-subnormal subgroup of K.

Proof: Since H is k-subnormal subgroup of G, we have a subnormal series:

H = H_0 \le H_1 \le \dots \le H_k = G.

We claim that the following is a subnormal series for H \cap K in K:

H \cap K = H_0 \cap K \le H_1 \cap K \le \dots H_k \cap K = K.

For this, we need to show that each H_i \cap K is normal in H_{i+1} \cap K. For this, note that:

H_i \cap K = H_i \cap (H_{i+1} \cap K).

Since H_i is normal in H_{i+1}, fact (1) tells us that H_i \cap (H_{i+1} \cap K) = H_i \cap K is normal in H_{i+1} \cap K, completing the proof.

Property-theoretic proof

This follows directly from facts (1) and (2).