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.

Related facts

 * Jerrum's filter gives an algorithmic method for finding a Jerrum-reduced generating set for any subgroup, and such a generating set must have size at most $$n - 1$$.
 * Sims filter gives an algorithmic method for finding a Sims-reduced generating set for any subgroup, and such a generating set must have size at most $$n(n-1)/2$$.

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 $$n/2$$.