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Take a deck of cards, cut it in three, and rearrange the three packs in any order (for instance, the middle goes on top, the top in the middle, and the bottom stays). Repeat as many times as you want with different (and possibly uneven) cuts. What permutations can you generate?

Motivation: given a set of symmetric matrices, the trace of their product does not change if they are subject to a permutation generated in the way above.

Arthur B.
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    "Cut in three..(possibly uneven cuts)" So, would the top 50 cards in the first "cut", then the 51'st card in its own cut and the 52'nd card in its own cut count? So we have access to transpositions involving the last two positions. Using commutators, do we not then have access to every possible transposition involving the last position? Do we not then have access to every permutation? – JMoravitz May 04 '23 at 12:23
  • That would stand to reason. Does it follow that the trace of the product of n symmetric matrices is independent of the order of the product? Wikipedia claims otherwise https://en.wikipedia.org/wiki/Trace_(linear_algebra) – Arthur B. May 04 '23 at 20:26
  • I would say no. Note that the property was stated about the product of exactly three symmetric matrices, that if $A,B,C$ are symmetric then $\text{tr}(ABC)$ is equal to the trace of any other permutation of the matrices. When considering more than three factors this need not be true. Perhaps you were under the impression that you could "collapse" a product of several symmetric matrices into a product of just three matrices. While you could, those three resulting matrices might not be symmetric and so the trace need not be the same after permuting them. – JMoravitz May 04 '23 at 22:18
  • Perhaps if you add more conditions and could guarantee that all products of all matrices involved were also symmetric, then you could... as an example if all of the matrices were diagonal... but that sounds like a really strict condition that makes things boil down to the commutativity property of normal multiplication of numbers. – JMoravitz May 04 '23 at 22:20
  • Oh right, the product of two symmetric matrices is not necessarily symmetric, that tripped me – Arthur B. May 05 '23 at 17:43

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