9
$\begingroup$

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 $-4$.

$\endgroup$
2
  • $\begingroup$ To be clear, $G_1$, $G_2$, and $G_3$ are given subgroups of $S_n$? They are not varying in the sum itself? $\endgroup$ Jan 25, 2015 at 19:06
  • $\begingroup$ Exactly, the capital G's are fixed, the sum is over all triplets in $G_1 \times G_2 \times G_3$ with the extra condition that the product of the elements is the identity. $\endgroup$ Jan 25, 2015 at 19:11

2 Answers 2

16
$\begingroup$

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 $-1$.

$\endgroup$
1
  • $\begingroup$ Ah, nice! I had a feeling there was some counterexample. The case I was studying in particular, only concerns certain subgroups. Perhaps the statement is true for these, see my edit... $\endgroup$ Jan 25, 2015 at 19:38
19
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ That is a very nice observation! $\endgroup$ Jan 26, 2015 at 0:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.