# Permutability is not finite-intersection-closed

From Groupprops

This article gives the statement, and possibly proof, of a subgroup property (i.e., permutable subgroup)notsatisfying a subgroup metaproperty (i.e., finite-intersection-closed subgroup property).

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## Contents

## Statement

### Verbal statement

The intersection of two permutable subgroups of a group need not be permutable.

### Symbolic statement

It is possible to find a group and subgroups and of such that and are both permutable subgroups (viz quasinormal subgroups) but is not.

## Related facts

### Related facts that don't hold for permutable subgroups

### Related facts that do hold for permutable subgroups

- Permutability is strongly join-closed
- Permutability satisfies image condition
- Permutability satisfies inverse image condition
- Permutability satisfies intermediate subgroup condition
- Permutability satisfies transfer condition

## Proof

### Construction of the counterexample

Setup: Let be an odd prime.

- is a semidirect product of cyclic group of prime-square order and cyclic group of prime order. More specifically it is a group generated by two elements subject to the relations and . Alternatively is the semidirect product of the additive group modulo by the multiplicative group of order in the multiplicative group of automorphisms. Note that is a non-abelian group of order .
- is a cyclic group of prime-square order: It is a cyclic group of order , generated by an element .
- .
- .
- , and .
- .

We claim that and are both permutable in , but their intersection is not permutable.

- is permutable: is a direct factor of so it is clearly a normal subgroup and hence a permutable subgroup.
- is permutable: Since permutability satisfies the inverse image condition, we see that if is permutable in , then is permutable in . Thus, it suffices to show that is permutable as a subgroup of . This can easily be checked by verifying that commutes with all the cyclic subgroups of . (a proof of this is provided in an example for permutable not implies normal).
- is not permutable in : Consider the cyclic subgroup generated by . The claim is that . To prove this notice that . This is clearly not in .

## Further fact shown by the example

This example shows some further facts:

- The intersection of a permutable subgroup with a direct factor need not be a permutable subgroup. In this example, for instance, is a direct factor, but its intersection with is still not a permutable subgroup.
- A permutable subgroup of a direct factor need not be a permutable subgroup. In this case is a permutable subgroup inside , which itself is a direct factor.
- Permutability is not a direct product-closed subgroup property