Subnormality of fixed depth satisfies intermediate subgroup 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., intermediate subgroup condition)
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Statement

Verbal statement

A subnormal subgroup of a group is also subnormal in every intermediate subgroup. In fact, its subnormal depth in any intermediate subgroup is bounded from above by the subnormal depth in the whole group.

Property-theoretic statement

The subgroup property of being a subnormal subgroup satisfies the subgroup metaproperty called the intermediate subgroup condition -- any subnormal subgroup of the whole group is also subnormal in every intermediate subgroup.

Statement with symbols

Suppose H is a subnormal subgroup of a group G. Then, for any intermediate subgroup K (i.e., H \le K \le G), H is subnormal in K. Moreover, if H is k-subnormal in G, H is also k-subnormal in K. (Here, when we say k-subnormal, we mean the subnormal depth is at most k).

Related facts

Generalizations

Related facts about normality and subnormality

Facts used

  1. Normality satisfies transfer condition: If H, K \le G are subgroups such that H is normal in G, then H \cap K is normal in K.

Proof

Hands-on proof

Given: A group G, a k-subnormal subgroup H, a subgroup K \le G such that H \le K.

To prove: H is k-subnormal in K.

Proof: Consider a subnormal series for H of length k:

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

where H_i is normal in H_{i+1} for each i. We claim that the series:

H = H_0 \le H_1 \cap K \le H_2 \cap K \le \dots \le H_k \cap K = K

is a subnormal series for H in K. For this, observe that:

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

We know that H_i is normal in H_{i+1}, so by fact (1), H_i \cap (H_{i+1} \cap K) is normal in H_{i+1} \cap K, yielding that H_i \cap K is normal in H_{i+1} \cap K, as desired.