Permutability is not finite-intersection-closed: Difference between revisions

Latest revision as of 20:35, 11 August 2010

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 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 $a b = b a^{p + 1}$ . 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 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 $B_{0} D \neq D B_{0}$ . To prove this notice that $D B_{0} ∋ (a, c) (b, e) = (a b, c) = (b a^{p + 1}, c)$ . 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

@@ Line 1: / Line 1: @@
 {{subgroup metaproperty dissatisfaction|
 property = permutable subgroup|
-metaproperty = intersection-closed subgroup property}}
+metaproperty = finite-intersection-closed subgroup property|
+corollary1 = intersection-closed subgroup property|
+corollary2 = strongly intersection-closed subgroup property|
+corollary3 = strongly finite-intersection-closed subgroup property}}
 ==Statement==
@@ Line 13: / Line 16: @@
 It is possible to find a group <math>G</math> and subgroups <math>H</math> and <math>K</math> of <math>G</math> such that <math>H</math> and <math>K</math> are both [[permutable subgroup]]s (viz quasinormal subgroups) but <math>H \cap K</math> is not.
-==Proof==
+==Related facts==
-===Construction of the counterexample===
+===Related facts that don't hold for permutable subgroups===
-Let <span class="texhtml">''A''</span> be a group generated by two elements <math>a,b</math> subject to the relations <math>a^{p^2} = 1, b^p = 1</math> and <math>ab = ba^{p+1}</math>. Alternatively <math>A</math> is the semidirect product of the additive group modulo <math>p^2</math> by the multiplicative group of order <math>p</math> in the multiplicative group of automorphisms.
+* [[Permutability is not upper join-closed]]
+* [[Permutable not implies normal]]
-Note that <math>A</math> is a non-Abelian group of order <math>p^3</math>.
+===Related facts that do hold for permutable subgroups===
-Let <math>C</math> be a cyclic group of order <math>p^2</math>, generated by an element <math>c</math>.
+* [[Permutability is strongly join-closed]]
+* [[Permutability satisfies image condition]]
+* [[Permutability satisfies inverse image condition]]
+* [[Permutability satisfies intermediate subgroup condition]]
+* [[Permutability satisfies transfer condition]]
-Set <math>G = A \times C</math>.
+==Proof==
-We claim that: <math>A</math> and <math>\{ b,c \}</math> are both permutable subgroups of <math>G</math>, but their intersection (which is just the cyclic subgroup <math>B</math> generated by <math>b</math>) is not.
+===Construction of the counterexample===
-===Proof of the claim===
+Setup: Let <math>p</math> be an odd prime.
-<math>A</math> is a [[direct factor]] of <math>G</math> so it is clearly a [[normal subgroup]] and hence a permutable subgroup.
+* <math>A</math> 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 <math>a,b</math> subject to the relations <math>a^{p^2} = 1, b^p = 1</math> and <math>ab = ba^{p+1}</math>. Alternatively <math>A</math> is the semidirect product of the additive group modulo <math>p^2</math> by the multiplicative group of order <math>p</math> in the multiplicative group of automorphisms. Note that <math>A</math> is a non-abelian group of order <math>p^3</math>.
+* <math>C</math> is a [[particular example::cyclic group of prime-square order]]: It is a cyclic group of order <math>p^2</math>, generated by an element <math>c</math>.
+* <math>G = A \times C</math>.
+* <math>\! B = \{ b \}</math>.
+* <math>H = A \times \{ e \}</math>, and <math>K = B \times C = \{b , c\}</math>.
+* <math>B_0 = H \cap K = B \times \{ e \}</math>.
-Since permutability satisfies the inverse image condition, it suffices to show that <math>B</math> is permutable as a subgroup of <math>A</math>. This can easily be checked by verifying that <math>B</math> commutes with all the cyclic subgroups of <math>A</math>.
+We claim that <math>H</math> and <math>K</math> are both permutable in <math>G</math>, but their intersection <math>B_0 = H \cap K</math> is not permutable.
-To prove that <math>B</math> is not permutable, consider the cyclic subgroup <math>D</math> generated by <math>(a,c)</math>. The claim is that <math>BD \ne DB</math>. To prove this notice that <math>(a,c)(b,1) = (ab,c) = (ba^{p+1},c)</math>. This is clearly not in <math>BD</math>.
+* <math>H = A \times \{ e \}</math> is permutable: <math>H</math> is a [[direct factor]] of <math>G</math> so it is clearly a [[normal subgroup]] and hence a permutable subgroup.
+* <math>K = B \times C = \{ b,c \}</math> is permutable: Since permutability satisfies the inverse image condition, we see that if <math>B</math> is permutable in <math>A</math>, then <math>B \times C = \{ b, c\}</math> is permutable in <math>G</math>. Thus, it suffices to show that <math>B</math> is permutable as a subgroup of <math>A</math>. This can easily be checked by verifying that <math>B</math> commutes with all the cyclic subgroups of <math>A</math>. (a proof of this is provided in an example for [[permutable not implies normal]]).
+* <math>B_0 = H \cap K = B \times \{ e \}</math> is not permutable in <math>G</math>: Consider the cyclic subgroup <math>D</math> generated by <math>(a,c)</math>. The claim is that <math>B_0D \ne DB_0</math>. To prove this notice that <math>DB_0 \ni (a,c)(b,e) = (ab,c) = (ba^{p+1},c)</math>. This is clearly not in <math>B_0D</math>.
 ==Further fact shown by the example==