Automorph-conjugacy is not finite-intersection-closed

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This article gives the statement, and possibly proof, of a subgroup property (i.e., automorph-conjugate subgroup) not satisfying a subgroup metaproperty (i.e., finite-intersection-closed subgroup property).
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We can have a group G with two automorph-conjugate subgroups H, K \le G, such that H \cap K is not automorph-conjugate in G.


Example in the symmetric group

Further information: symmetric group:S6

(This example demonstrates the stronger fact that automorph-conjugacy is not finite-conjugate-intersection-closed).

Let G be the symmetric group on six letters: \{ 1,2,3,4,5,6 \}. Let H, K be the following 2-Sylow subgroups of G:

H = \langle (1,3,2,4) , (1,2), (5,6) \rangle ;\qquad K = \langle (1,2), (3,5,4,6), (5,6) \rangle

In other words, H is the internal direct product of a 2-Sylow subgroup on \{ 1,2,3,4\} with the 2-Sylow subgroup on \{ 5,6\}, while K is the internal direct product of the 2-Sylow subgroup on \{ 1,2 \} with a 2-Sylow subgroup on \{ 3,4,5,6 \}.

The intersection is given by:

H \cap K = \langle (1,2), (3,4), (5,6) \rangle.

Now, note that:

  • Both H and K are automorph-conjugate, because they are both Sylow subgroups, and Sylow implies automorph-conjugate.
  • H \cap K is not automorph-conjugate. To see this, note that G has an outer automorphism that sends transpositions to triple transpositions. Under this automorphism, H \cap K goes to a subgroup of G that contains three commuting triple transpositions. If this is conjugate to H \cap K, then H \cap K should also contain three commuting triple transpositions. But it doesn't.