Maximal subgroup of permutations for which trace is invariant

Question

Recall that trace is invariant under cyclic permutations of a product of matrices.

Are cyclic permutations the maximal subgroup of permutations for which trace doesn't change?

Answer 1

Suppose $\sigma$ is any permutation of $\{ 1, 2, \dots, n \}$ other than a power of $(1\ 2\ 3\ \dots\ n)$. We want to show that there exist matrices $A_1, \dots A_n$ such that

$$\text{tr}(A_1 \cdots A_n) \neq \text{tr}(A_{\sigma(1)} \cdots A_{\sigma(n)}).$$

By permuting cyclically, we can assume WLOG that $\sigma(1) = 1$. Then $\sigma$ isn't a power of $(1\ 2\ 3\ \dots\ n)$ iff it isn't the identity. So if it isn't the identity, there is some minimal $i$ such that $\sigma(i) \neq i$, which means that $\sigma(i) > i$.

Set all of the $A_i$ to be the identity except $A_1, A_i$, and $A_{\sigma(i)}$. Then we want to set these three matrices such that

$$\text{tr}(A_1 A_i A_{\sigma(i)}) \neq \text{tr}(A_1 A_{\sigma(i)} A_i).$$

This isn't hard to do; random $2 \times 2$ matrices should work, but I'm out of time for now.