4
$\begingroup$

What is the smallest number n such that $A_{n}$ contains a permutation of order 2004?

I calculated it to 334, but the answer is 176 and I can't see why? I first noticed that $2004 = 2^{2}\cdot 3 \cdot 167$ then I looked at the elements in $A_{n}$ as they were written in a form of disjoint cycles. Then I realized that in order for an element to have order 2004 the element must have disjoint cycles in such a way that the least common multiple of the orders of the disjoint cycles is 2004. I then tried to figure out how small I can pick a number n so $x,y .. \leq n$ and $lcm(x,y..) = 2004$ and still have a permutation with disjoint cycles of orders $x,y ..$ that is even.

How many permutations in $S_{6}$ commute with $(1 2) (3 4)$?

Is there any other way than using bruteforce?

  • 0
    Because I need a permutation in $A_{n}$. If I have a permutation with a 167- cycle, 3-cycle and only a 4-cycle it would be an odd permutation. Therefore I add an extra 2-cycle in order to get it even.2012-04-02

1 Answers 1

2

For the second question, are you familiar with the result that says that (the number of elements of a group $G$ commuting with an element $g$) times (the number of elements conjugate to $g$) equals the order of $G$? and the result that says that in $S_n$ elements are conjugate if and only if they have the same cycle structure? and can you work out how many elements of $S_6$ have the same cycle structure as $(12)(34)$? Oh, I suppose you also have to know how many elements there are in $S_6$.