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This article gives the statement, and possibly proof, of a subgroup property (i.e., permutable subgroup) not satisfying a subgroup metaproperty (i.e., finite-intersection-closed subgroup property).
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Contents 1 Statement 1.1 Verbal statement 1.2 Symbolic statement 2 Related facts 2.1 Related facts that don't hold for permutable subgroups 2.2 Related facts that do hold for permutable subgroups 3 Proof 3.1 Construction of the counterexample 4 Further fact shown by the example

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 G and subgroups H and K of G such that H and K are both permutable subgroups (viz quasinormal subgroups) but is not.

Related facts

Related facts that don't hold for permutable subgroups

• Permutability is not upper join-closed
• Permutable not implies normal

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 p be an odd prime.

• A 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 a,b subject to the relations and ab = ba^{p + 1}. Alternatively A is the semidirect product of the additive group modulo p² by the multiplicative group of order p in the multiplicative group of automorphisms. Note that A is a non-abelian group of order p³.
• C is a cyclic group of prime-square order: It is a cyclic group of order p², generated by an element c.
• .
• .
• , and .
• .

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

• is permutable: H is a direct factor of G 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 B is permutable in A, then is permutable in G. Thus, it suffices to show that B is permutable as a subgroup of A. This can easily be checked by verifying that B commutes with all the cyclic subgroups of A. (a proof of this is provided in an example for permutable not implies normal).
• is not permutable in G: Consider the cyclic subgroup D generated by (a,c). The claim is that . To prove this notice that . This is clearly not in B₀D.

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, A is a direct factor, but its intersection with C 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 A, which itself is a direct factor.
• Permutability is not a direct product-closed subgroup property