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At the very beginning, I will refer to this post:How to prove the cyclic property of the trace? which is a starting point for my question:

If cyclic permutations don't change $tr(A_1\cdot A_2\ldots A_n),$ prove that $n\;\text{factors}$ give at most $(n-1)!$ different traces.

My attempt: $$\text{cyclic permutation}\iff \text{initial trace}$$ $$\implies\text{number of non-cyclic permutations}=?$$ By the assumption and the proof I've seen on the link above: $$tr(A_1\cdot\underbrace{ A_2\ldots A_{n-1}}_{X} A_n)=tr(A_n\cdot A_1\cdot X)$$ $X$ consists of $(n-2)$ factors $\implies$ $(n-2)!-1$ non-cyclic permutations 'in' $X$, but I haven't achieved anything this way.

Then I considered derangements, but it wasn' t a very clear idea because I didn't know what to do with the order of factors with consecutive indexes and then I thought of a proof by induction, but after proving the base it didn't seem sufficient either.

Matcha Latte
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    Write out the 24 permutations of $A_1\cdot A_2\cdot A_3\cdot A_4$. Group together the ones that the cyclic property tells you must be equal. How many groups are there? What is the general principle here? – Will Orrick Dec 15 '19 at 16:45
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    @WillOrrick, so, I should get something like 'classes'? – Matcha Latte Dec 15 '19 at 17:58
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    Yes. Define two products to be equivalent when their factors differ by cyclic permutation. Equivalence classes then have size $n$. The largest number of different traces occurs when each equivalence class gives a different trace. – Will Orrick Dec 16 '19 at 12:06

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