(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Let [tex]\rho[/tex] [tex]\in[/tex]Sym(n), p be prime, r be the remainder when n is divided by p (so 0[tex]\leq[/tex]r<p and n=qp+r for some integer q).

1. Show that [tex]\rho[/tex]^^{p}= [tex]\iota[/tex] iff the cycles of [tex]\rho[/tex] all have lengths 1 or p.

2. Show that if [tex]\rho[/tex]^^{p}= [tex]\iota[/tex] then |Supp([tex]\rho[/tex])| is a multiple of p and |Fix([tex]\rho[/tex])|[tex]\equiv[/tex] r(mod p).

2. Relevant equations

Fix([tex]\rho[/tex]) := {x|x[tex]\rho[/tex] = x}

Supp([tex]\rho[/tex]) := {x|x[tex]\rho[/tex] [tex]\neq[/tex] x}

3. The attempt at a solution

I really don't have many ideas on these at all.

1) If all cycles have length 1 then it is clear that [tex]\rho[/tex]^{p}is the identity.

I don't know what I can deduce from all cycles having length p. The other way around, I can see if we have the identity that all cycles could be length 1, but I don't know how to go about getting length p.

2) I have no idea how to start this.

Thanks.

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# Homework Help: Permutations and cycles - linear algebra

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