Permutability is not finite-intersection-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., permutable subgroup) not satisfying a subgroup metaproperty (i.e., intersection-closed subgroup property).
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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 $H\cap K$ is not.

Proof

Construction of the counterexample

Let A be a group generated by two elements $a,b$ subject to the relations $a^{p^2}=1,b^p=1$ and $ab=b{a}^{p+1}$. 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 $p^3$.

Let $C$ be a cyclic group of order $p^2$, generated by an element $c$.

Set $G=A\times C$.

We claim that: $A$ and $\{b,c\}$ are both permutable subgroups of $G$, but their intersection (which is just the cyclic subgroup $B$ generated by $b$) is not.

Proof of the claim

$A$ is permutable: $A$ is a direct factor of $G$ so it is clearly a normal subgroup and hence a permutable subgroup.

$B\times C=\{b,c\}$ 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 $A$.

$B$ is not permutable: Consider the cyclic subgroup $D$ generated by $(a,c)$. The claim is that $BD\ne DB$. To prove this notice that $(a,c)(b,1)=(ab,c)=(b{a}^{p+1},c)$. This is clearly not in $BD$.

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