Automorph-conjugacy is not finite-intersection-closed

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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Statement

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 $G$.

Proof

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.