If P is an nxn permutation matrix, is there an upper bound on k such that $P^k = I$?

Asked Oct 04 '18 at 14:55

Active Oct 04 '18 at 14:55

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So if P is an nxn permutation matrix, then because there are only finitely many ways to permute finitely many elements, we know that the sequence

$P, P^2, P^3, ...$

eventually has to repeat, between say $P^i$ and $P^j$, which implies $P^{j-i} = I$. Since we know there are only n! different permutation matrices, by the pigeonhole principle, we know that we must have $P^k = I$ for k<=n!.

But is there a tighter bound?

Any thoughts appreciated.

Thanks.

asked Oct 04 '18 at 14:55

user49404

The order of elements in this group, which is isomorphic to $S_n$, is given by the lcm of disjoint cycles. This gives much better bounds, see this duplicate. – Dietrich Burde Oct 04 '18 at 14:57
@DietrichBurde ah, ok. I read about basic algebra a long time ago and had forgotten about this... thanks. – user49404 Oct 04 '18 at 15:41

If P is an nxn permutation matrix, is there an upper bound on k such that $P^k = I$?

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