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Let $\Sigma$ be the set of all permutations in $S_n$. What is the minimum number of states in a DFA accepting the language of all words over $\Sigma$ which multiply to the identity permutation?

For example, if $n=2$, then $\Sigma$ consists of two mappings, the identity mapping $\iota$ and the transposition $\tau$. The language in this case consists of all words containing an even number of $\tau$'s.

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  • $\begingroup$ I explained more. Now is it clear? $\endgroup$ – Srestha Feb 14 '19 at 9:11
  • $\begingroup$ This function contains only n! bijective function. right? Now if we need to put it in a DFA, then is minimised DFA also contain n! states?? $\endgroup$ – Srestha Feb 14 '19 at 10:20
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You haven't stated what happens to the empty word – I'm assuming it's accepted.

It is easy to check that the equivalence classes of the Myhill–Nerode relation correspond to all words multiplying to a certain permutation. Therefore the minimal DFA contains $n!$ states.

If the empty word is not allowed, there are $n!+1$ equivalence classes, one of them consisting only of $\epsilon$.

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  • $\begingroup$ what about in case of injective mapping? $\endgroup$ – Srestha Feb 14 '19 at 13:33
  • $\begingroup$ The answer might depend on which direction we're composing the functions. I'm sure you can work it out on your own. Try the case $n=2$ first. $\endgroup$ – Yuval Filmus Feb 14 '19 at 13:34
  • $\begingroup$ thank you sir.. $\endgroup$ – Srestha Feb 14 '19 at 13:44
  • $\begingroup$ Can u give some good link for it, I have not seen such concept before $\endgroup$ – Srestha Feb 14 '19 at 13:47
  • $\begingroup$ Neither have I. The problems which are more interesting to solve are those that we haven't seen before. $\endgroup$ – Yuval Filmus Feb 14 '19 at 13:47

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