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 $-4$.



Certain signed sum over $S_n$
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 $-1$.


