Certain signed sum over $S_n$

Question

The following question appeared in my research:

Let $G_1,G_2,G_3$ all be subgroups of $S_n$, and consider the sum

$$ \sum_{g_i \in G_i, g_1g_2g_3 = id} \epsilon(g_1) $$ that is, we only consider triplets whose product is the identity permutation, and we sum all the signs of $g_1$.

The question is: is this sum always non-negative?

EDIT:

What if we restrict $G_1$ to be the entire $S_n$? What if we restrict $G_2$ and $G_3$ to be groups of type $G_T=\{\sigma \in S_n : c(\sigma) \leq T \}$ where $c$ indicates the cycle type, and $T$ is a fixed set partition, and the inequality indicates refinement.

Update

Even with the updated version of subgroups, there is a counter-example. The three set partitions $$S=((1, 2, 3, 4), (5, 6, 7, 8)) \quad T=((1, 5),(2, 8),(3, 6),(4, 7)) \quad U=((1, 6),(2, 4, 5),(3, 7, 8))$$

give rise to a sum with value $-4$.

Answer 1

There is an example in $S_5$ where the sum is negative. Take $G_1 = \langle (12345), (1325) \rangle$, $G_2 = \langle (135) \rangle$, $G_3 = \langle (35), (235) \rangle$. Then

$ \bigl\{(g_2,g_3) : g_2 \in G_2, g_3 \in G_3, g_2g_3 \in G_1 \bigr\} = \bigl\{ (\mathrm{id}, \mathrm{id}), ((135), (25)), ((153), (23))\bigr\} $

so the sum of the signs is $-1$.

Answer 2

If $G_1 = S_n$ as in your edit, then every pair $(g_2,g_3) \in G_2 \times G_3$ occurs exactly once in the sum, and for each such summand we have $\epsilon(g_1) = \epsilon(g_2g_3)$. So the sum simplifies to $$\sum_{g_2 \in G_2}\sum_{g_3 \in G_3} \epsilon(g_2g_3) = \left( \sum_{g_2 \in G_2} \epsilon(g_2)\right)\left(\sum_{g_3 \in G_3} \epsilon(g_3)\right)$$ which is nonnegative by the following observation: if $G \subset S_n$ is a subgroup, then the sum $\sum_{g \in G} \epsilon(g)$ is either $0$ or $\vert G \vert$, depending on whether the map $\epsilon \colon G \to \{\pm 1\}$ is trivial or not.