# 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.

## 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 $B_0D \ne DB_0$. 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