# 4-subnormal not implies finite-conjugate-join-closed subnormal

From Groupprops

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., 4-subnormal subgroup) neednotsatisfy the second subgroup property (i.e., finite-conjugate-join-closed subnormal subgroup)

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Get more facts about 4-subnormal subgroup|Get more facts about finite-conjugate-join-closed subnormal subgroup

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## Statement

It is possible to have a group and a finite collection of conjugate 4-subnormal subgroups of whose join is not subnormal. In fact, it is possible to have just two conjugate 4-subnormal subgroups whose join is not subnormal.

## Facts used

- 3-subnormal not implies finite-automorph-join-closed subnormal (in fact, it is possible to have just two automorphic 3-subnormal subgroups whose join is not subnormal.
- Left residual of finite-conjugate-join-closed subnormal by normal is finite-automorph-join-closed subnormal

## Proof

### Hands-on proof

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### Proof using facts

Suppose that every 4-subnormal subgroup of a group is finite-conjugate-join-closed subnormal. In particular, this means that if is a 3-subnormal subgroup of , then whenever is a normal subgroup of a group , is finite-conjugate-join-closed subnormal in .

Fact (2) says that this implies that is finite-automorph-join-closed subnormal in . Thus, we have shown that every 3-subnormal subgroup is finite-automorph-join-closed subnormal. This, however, contradicts fact (1).