Permutations and cycles - linear algebra

[Kate2010]

Homework Statement

Let \rho \inSym(n), p be prime, r be the remainder when n is divided by p (so 0\leqr<p and n=qp+r for some integer q).

1. Show that \rho^^p = \iota iff the cycles of \rho all have lengths 1 or p.

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

Homework Equations

Fix(\rho) := {x|x\rho = x}
Supp(\rho) := {x|x\rho \neq x}

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 \rho^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.

 

[JSuarez]

When you write a permutation \rho, of length n, in terms of cycles, you have:

\rho = \left(a_1...a_{k_1}\right)\left(a_{k_1}...a_{k_2}\right)...\left(a_{k_i}...a_{k_j}\right)

The cycles' lengths must sum to n and, if \rho^d = I, then all cycles must have undergone a full rotation after d iterations.

But this implies that d must divide your prime p. Now, what are the possible divisors of p?

As for (2), it's just a translation of the division n = qp + r in terms of the cycles: the cycles oh length p correspond to a factor in qp, and r = 1+1+...+1 (r times) are the cycles of length one.

 

[Kate2010]

Thank you so much! I've just started with permutations and as of yet I haven't really grasped it, but I do understand that now.

Another question:

If t1 and t2 are transpositions in Sym(n), then show that t1t2 = I, (t1t2)² = I or (t1t2)³ = I (I is the identity permutation).

What condition on t1 and t2 determines which of these occurs?

What is the cycle structure of t1t2 in each case?

 

[JSuarez]

A transposition is a permutation that exchanges just two elements; in terms of cycles, a transposition has a cycle of length 2, and all the others of length 1.

It's immediate that any transposition t verify t^2=I; it's less immediate that they generate the symmetric group, but they do: any permutation is the product of transpositions.

For any two transpositions t1 and t2, your three cases correspond to:

(1) t1 = t2; this gives t1t2=t2t1=I.

(2) t1 and t2 are disjoint (their 2-cycles don't affect each other); in this case t1t2 is a permutation with exactly two 2-cycles (the others are 1-cycles).

(3) The 2-cycles of t1 and t2 share one common element, so t1t2 has one 3-cycle (and the rest are 1-cycles).

 

[Kate2010]

Thank you again!

One last question:

Let \rho be in Sym(n). For which values of n is it true that if \rho is even then \rho^m = I for some odd integer m?

Is this true for n=km where k is a natural number?

:)