Permutability is not finite-intersection-closed

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 $$H \cap K$$ is not.

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

Construction of the counterexample
Setup: Let $$p$$ be an odd prime.


 * $$A$$ is a particular example::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 $$a^{p^2} = 1, b^p = 1$$ and $$ab = ba^{p+1}$$. Alternatively $$A$$ is the semidirect product of the additive group modulo $$p^2$$ by the multiplicative group of order $$p$$ in the multiplicative group of automorphisms. Note that $$A$$ is a non-abelian group of order $$p^3$$.
 * $$C$$ is a particular example::cyclic group of prime-square order: It is a cyclic group of order $$p^2$$, generated by an element $$c$$.
 * $$G = A \times C$$.
 * $$\! B = \{ b \}$$.
 * $$H = A \times \{ e \}$$, and $$K = B \times C = \{b, c\}$$.
 * $$B_0 = H \cap K = B \times \{ e \}$$.

We claim that $$H$$ and $$K$$ are both permutable in $$G$$, but their intersection $$B_0 = H \cap K$$ is not permutable.


 * $$H = A \times \{ e \}$$ is permutable: $$H$$ is a direct factor of $$G$$ so it is clearly a normal subgroup and hence a permutable subgroup.
 * $$K = B \times C = \{ b,c \}$$ is permutable: Since permutability satisfies the inverse image condition, we see that if $$B$$ is permutable in $$A$$, then $$B \times C = \{ b, c\}$$ 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).
 * $$B_0 = H \cap K = B \times \{ e \}$$ is not permutable in $$G$$: Consider the cyclic subgroup $$D$$ generated by $$(a,c)$$. The claim is that $$B_0D \ne DB_0$$. To prove this notice that $$DB_0 \ni (a,c)(b,e) = (ab,c) = (ba^{p+1},c)$$. This is clearly not in $$B_0D$$.

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 $$B = A \cap C$$ is a permutable subgroup inside $$A$$, which itself is a direct factor.
 * Permutability is not a direct product-closed subgroup property