Subgroup rank of symmetric group is about half the degree

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Statement

Let n be a natural number. Let G be the symmetric group of degree n, i.e., the symmetric group on a set of size n. Then, the subgroup rank of G is one of the numbers n/2, (n-1)/2, (n+1)/2.

In particular:

  • If n is even, the subgroup rank of G is n/2.
  • If n = 3, the subgroup rank of G is 2.
  • If n is any odd positive integer other than 3, the subgroup rank of G is (n - 1)/2.

The subgroup rank is defined as the maximum, over all subgroups, of the minimum size of generating set of that subgroup. In particular, this means that every subgroup of G has a generating set of size at most the subgroup rank, and there is at least one subgroup of G for which the minimum size of generating set equals the subgroup rank.

Particular cases

n n! (order of symmetric group) symmetric group of degree n subgroup rank
1 1 trivial group 0
2 2 cyclic group:Z2 1
3 6 symmetric group:S3 2
4 24 symmetric group:S4 2
5 120 symmetric group:S5 2
6 720 symmetric group:S6 3
7 5040 symmetric group:S7 3

Related facts

Proof

The easy part: finding subgroups where the maximum is attained

For simplicity, we take the symmetric group on the set \{ 1,2,3,\dots,n \}.

Case of even n: In this case, consider the subgroup:

\langle (1,2), (3,4), \dots, (n-1,n) \rangle

This is an elementary abelian group of rank n/2, so its minimum size of generating set is also n/2.

Case of n = 3: In this case, the whole group has minimum size of generating set equal to 2.

Case of odd n (n \ne 3): In this case, consider the subgroup:

\langle (1,2),(3,4), \dots,(n-2,n-1)\rangle

This is an elementary abelian group of rank (n-1)/2, so its minimum size of generating set is also n/2.

The hard part: showing that the subgroup rank is not higher