Max-equivalent to normal and intermediate subgroup condition implies subnormal in finite

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Statement

Suppose p is a subgroup property satisfying the following two conditions:

Then, every subgroup of a finite group satisfying property p is a subnormal subgroup.

Related facts

Generalizations

Applications

Proof

These two proofs are essentially the same; one is presented in an inductive style, and the other is presented in terms of construction of the subnormal series.

Inductive proof

Given: A subgroup H of a group G satisfying property p. p is max-equivalent to normality and satisfies the intermediate subgroup condition.

To prove: H is subnormal in G.

Proof:

  1. Consider the collection \mathcal{C} of proper subgroups of G containing H and satisfying property p. Note that \mathcal{C} is nonempty (it contains H) and finite, so it has maximal elements under inclusion. Let K be such an element.
  2. Since p is max-equivalent to normality, K is normal in G.
  3. Since p satisfies the intermediate subgroup condition, H satisfies p in K. Thus, by induction on the order, H is subnormal in K.
  4. Combining the previous two steps, we see that H is subnormal in G.

Proof by construction of a subnormal series

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