Permutability is not finite-intersection-closed

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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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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

Related facts that do hold for permutable subgroups

Proof

Construction of the counterexample

Setup: Let $p$ be an odd prime.

$A$ is a group generated by two elements $a, b$ subject to the relations $a^{p^2} = 1, b^p = 1$ and $a b = b a p + 1$ . 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 $p 3$ .
$C$ 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 $B 0 D$ .

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

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Permutability is not finite-intersection-closed

From Groupprops

Contents

Statement

Verbal statement

Symbolic statement

Related facts

Related facts that don't hold for permutable subgroups

Related facts that do hold for permutable subgroups

Proof

Construction of the counterexample

Further fact shown by the example

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