Does $A_n$ split a complement of the stabilizer of a partition of $[n]$?

Question

Let $G=S_n$ be the symmetric group on $[n]$, and consider its natural action on the set of partitions of $[n]$. (I mean set partitions, not like in number theory.) Let $\pi$ be a partition and let $G_\pi$ be its stabilizer. Define a set-complement of $G_\pi$ in $G$ to be a set $S\subset G$ that intersects every coset of $G_\pi$ in exactly one element. Let $s=|G|/|G_\pi|$ be the cardinality of the orbit of $\pi$. Then for any set-complement $S$, we have $|S|=s$.

I'd like conditions on $\pi$ or $s$, such that there exist set-complements that are either contained or "exactly half-contained" in the alternating group $A_n$. I conjecture:

If $s$ is odd, then there exists a set-complement $S$ having $S\subset A_n$.

If $s$ is even, then there exists a set-complement $S$ having $S\cap A_n = s/2$.

Is this true? Is there a better or simpler way to guarantee the existence of an $S$ of one of these two types?

A couple simple examples:

1. If $G=S_3$ and $\pi=\{\{1,2\},\{3\}\}$, then $G_\pi=S_2$ and $s=3$. A "good" set-complement is $S=\{e,(123),(132)\}$, which is $A_3$ itself. On the other hand, a "bad" set-complement would be the set $\{e,(13),(23)\}$, which is $1/3$ contained in $A_3$.
2. If $G=S_4$ and $\pi=\{1,2,3\},\{4\}\}$, then $G_\pi=S_3$ and $s=4$. A "good" set-complement is $S=\langle(1234)\rangle$, which is exactly half-contained in $A_4$. A "bad" set-complement would be $\{e,(14),(24),(34)\}$, which is $1/4$ contained in $A_4$.

In the above examples, the "good" set-complements were in fact subgroups of $G$, but I can't count on those:

• A set-complement that is also a subgroup is simply called a complement, or a factor in an internal Zappa–Szép product.
• There are many possible set-complements. By contrast, a complementary subgroup might not exist. For example, let $n=6$ and $\pi=\{\{1,2,3,4\},\{5\},\{6\}\}$, so that $G_\pi=S_4$. There are $24^{60}$ possible set-complements of $S_4$ in $S_6$, but none of them is a subgroup. See David Joyner, "Complements in the symmetric group".
• That's too bad, because for complementary subgroups, the first part of the conjecture is easy.

The motivation is to use something like the orbit-stabilizer theorem to count the number of $n\times n$ matrices having some condition on their determinants, and letting $S_n$ act by exchanging rows. It's more convenient to enumerate representatives of orbits of $S_n$ than $A_n$. Then for each orbit of $S_n$, one wants to know if all of the matrices are related to the chosen representative via $A_n$, or only half of them, so that the other half have a determinant of the opposite sign. (That's only the gist; I'm leaving out details that make the real problem harder.)

Answer 1

By the time I typed up the question, I realized the answer. We typically have so much freedom in choosing $S$ that it's no problem to meet either condition. Note that the following are equivalent:

• $\pi$ is the discrete partition
• $G_\pi$ is trivial
• $s=n!$

and conversely, the following are equivalent:

• $\pi$ is non discrete; it has an equivalence class containing at least two members, $x\neq y$.
• $G_\pi$ is not trivial
• $G_\pi$ contains an odd permutation $(xy)$
• $s<n!$

Now we easily conclude:

If $s$ is even, then there exists a set-complement $S$ having $S\cap A_n = s/2$.

If $G_\pi$ is trivial and $s=n!$ is even, then $n\geq2$, so take the only choice $S=G$; it suffices because $|S_n/A_n|=2$. If $G_\pi$ is nontrivial, then it has an odd permutation. So every coset of $G_\pi$ has both even and odd elements. Therefore we can freely pick even elements from $m/2$ cosets and odd elements from the other $m/2$ cosets.

If $s$ is odd, then there exists a set-complement $S$ having $S\subset A_n$.

This is clear for $n\leq 2$. For $n>2$, we can do even better: If $s<n!$, then there exists a set-complement $S$ having $S\subset A_n$. The same logic as before, but this time, freely pick even elements from all cosets.