Symmetric group of degree six or higher is not weak subset-conjugacy-closed in general linear group over rationals

Statement

Suppose $n \ge 6$. Consider the symmetric group $S_n$ of degree $n$ as a subgroup of the general linear group of degree $n$ over the field of rational numbers, denoted $GL(n,\mathbb{Q})$, via the embedding as permutation matrices. Then, $S_n$ is not a weak subset-conjugacy-closed subgroup of $GL(n,\mathbb{Q})$. In other words, we can find subsets $A$ and $B$ of $S_n$ that are conjugate as subsets in $GL(n,\mathbb{Q})$ but not in $S_n$.

We can take $A$ and $B$ to be subgroups.

Related facts

Similar facts

• Analogue of Brauer's permutation lemma fails over rationals for every non-cyclic finite group: In fact, the statement generates a very large collection of examples, though for our purposes we need only one example.

Opposite facts

• Brauer's permutation lemma guarantees that any two subgroups of $S_n$ that are conjugate in $GL(n,\mathbb{Q})$ have an isomorphism between them that preserves cycle types of elements.

Proof

Example for $n = 6$

Consider the following two subsets of the symmetric group $S_6$ acting on the set $\{ 1,2,3,4,5,6 \}$:

$A = \{ (), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \}$

and

$B = \{ (), (1,2)(3,4), (1,2)(5,6), (3,4)(5,6) \}$

These are not conjugate inside $S_6$, because $A$ has two global fixed points while $B$ has none. However, they are conjugate inside $GL(6,\mathbb{Q})$. We can see this either by finding explicit matrices that perform this conjugation, or by noting that the two representations have the same character, and hence, since the Klein four-group is a rational representation group, they must be conjugate over the rational numbers.

Example for higher $n$

We can use the same example as for $n = 6$, fixing all the $n - 6$ elements.