Timeline for Does the 2-shift map have a root automorphism?
Current License: CC BY-SA 3.0
8 events
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| Apr 4, 2017 at 18:09 | history | edited | YCor | CC BY-SA 3.0 |
Added answer to this root question in general, due to Boyle
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| Jun 17, 2015 at 4:13 | vote | accept | EvaristoCarriego | ||
| Jun 17, 2015 at 4:12 | comment | added | EvaristoCarriego | yes, right, in the context of permutations that are not necessarily homeomorphisms you can make a $p$-root fix any $p'$-cycle (for $p'\neq p$) but I was almost sure that if $U$ is an homeomorphism (so that, by Hedlund-Curtis-Lyndon theorem it is induced by a block coding with a fixed block size) then this combinatorial constraint should prevent $U$ from fixing any $p'$-cycle in $T$ for infinitely many $p'$ (whence $p$ must divide the number of $p'$-cycles in $T$). Yet it seems that I can't prove it. | |
| Jun 5, 2015 at 17:43 | history | edited | YCor | CC BY-SA 3.0 |
Added that there is indeed such a 1093-root in the group of permutations.
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| Jun 5, 2015 at 17:13 | comment | added | YCor | @EvaristoCarriego: no. A necessary and sufficient condition for a permutation to be a $p$-power ($p$ prime) is that for every multiple $n$ of $p$, the number of $n$-cycles is divisible by $p$. For instance, a single $p'$-cycle is a $p$-power when $p'\neq p$ is prime. And actually if $p$ is a Wieferich prime, then it is not hard to check that this is fulfilled, so that the 2-sided shift on 2 letters is a $p$-power in the full group of permutations. | |
| Jun 5, 2015 at 15:22 | comment | added | EvaristoCarriego | Yves, your argument holds for general cycles (and not just for $p$-cycles), right? I mean, given a $p$-root candidate $U$ (in fact, given any shift-commuting $U$) it should act on $J_{p'}$, the union of $p'$-cycles for any prime $p'$ (and not just $p$). And then, since $U$ is a $p$-root (which means any orbit of the action on $J_{p'}$ has cardinal $p$), one has it that $p$ must divide the cardinal of this union, which is $\frac{2^{p'}-2}{p'}$. Then $p$ must satisfy $2^{p'-1}-2\equiv 0\quad (pp')$ for any $p'$, which implies that the 2-shift has no roots at all. | |
| Jun 4, 2015 at 6:15 | vote | accept | EvaristoCarriego | ||
| Jun 17, 2015 at 4:13 | |||||
| May 12, 2015 at 22:52 | history | answered | YCor | CC BY-SA 3.0 |