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

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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

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.