Permutability is not finite-intersection-closed

From Groupprops
Jump to: navigation, search
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).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about permutable subgroup|Get more facts about finite-intersection-closed subgroup property|

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.

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