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., finite-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.

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 $a^{p^{2}}=1,b^{p}=1$ 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 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 $B_{0}D\neq DB_{0}$. To prove this notice that $DB_{0}\ni (a,c)(b,e)=(ab,c)=(ba^{p+1},c)$. This is clearly not in $B_{0}D$.

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