# Conjugate-permutable implies subnormal in finite

From Groupprops

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties, when the big group is a finite group. That is, it states that in a Finite group (?), every subgroup satisfying the first subgroup property (i.e., Conjugate-permutable subgroup (?)) must also satisfy the second subgroup property (i.e., Subnormal subgroup (?)). In other words, every conjugate-permutable subgroup of finite group is a subnormal subgroup of finite group.

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## Contents

## Statement

Any conjugate-permutable subgroup of a finite group is subnormal.

## Related facts

- Permutable implies subnormal in finite
- 2-subnormal implies conjugate-permutable
- Subnormal not implies conjugate-permutable
- Conjugate-permutable implies descendant in slender

## Facts used

- Conjugate-permutability satisfies intermediate subgroup condition
- Maximal conjugate-permutable implies normal

## Proof

**Given**: A finite group , a conjugate-permutable subgroup of .

**To prove**: is a subnormal subgroup of .

**Proof**:

- (
**Fact used**: fact (1)): Define a descending chain as follows: , and if properly contains , is a maximal element among the proper conjugate-permutable subgroups of that contain .- (Well-definedness): Note that by fact (1), is conjugate-permutable in , so the collection of proper conjugate-permutable subgroups of containing is nonempty. Since is finite, it has a maximal element.
- (Terminates at in finitely many steps): The chain is a strictly descending chain of subgroups until it reaches . Since is finite, it terminates in finitely many steps at . Thus, there exists such that .
- (
**Fact used**: fact (2)): By definition, is a maximal conjugate-permutable subgroup of , so fact (2) tells us that is normal in . - (
**Conclusion**): The s thus form a subnormal series for in , making a subnormal subgroup of .