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Cycle prime decomposition is closed under cartesian product (#624)
In this pull request, I wanted to prove that the cartesian product of the cycles prime decomposition of `n` and `m` is equal to the cycle prime decomposition of `mul-\bN n m`. Initially, I thought it would be easy by using the fundamental theorem of arithmetic but I encountered some unexpected problems that I had to solve ^^. The main idea was to 1. Show that if we take the sorted concatenation of the lists corresponding to the prime decomposition of `n` and `m` we obtain a prime decomposition of `mul-\bN n m`. This was harder than I thought. I had to show that `mul-list-\bN` is invariant by permuting the list. I made the general case, and to do it I had to show that every transposition (on Fin n) is a composition of adjacent transpositions. 2. Then I used the fundamental theorem of arithmetic to demonstrate that the sorted concatenation of the lists corresponding to the prime decomposition of `n` and `m` is equal to the prime decomposition of `mul-\bN n m`. 3. Finally, I expected to only need to permute the cycles in the cartesian product of the cycles prime decomposition associated to `n` and `m` with the permutation given by the sort by insertion. I thought that applying step 2. would be sufficient to conclude. But I had to show this property : `map-list f (permute-list p t)` is equal to `permute-list (map-list p f) t` and which was quite hard to prove. --------- Co-authored-by: Fredrik Bakke <fredrbak@gmail.com>
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