Subgroup rank of symmetric group is about half the degree

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

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 $n/2$.