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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?

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How do you get 334? How many objects do you need to be able to permute to get a 167-cycle? What is the minimal difference between an odd permutation and an even permutation? – Mark Bennet Apr 1 '12 at 14:14
Is the OP talking about permutation groups and their order. Got confused :S – funktor Apr 1 '12 at 14:38
For the second question, I think one can prove that an element $x\in S_6$ commutes with $(12)(34)$ iff $x\in\langle(12),(34),(56)\rangle =\langle(12)\rangle\oplus\langle(34)\rangle\oplus\langle(56)\rangle$ which has order $8$. – bgins Apr 1 '12 at 14:47
Well I figured out that I'm wrong with my 334. Simply because I forgot to consider that $lcm(167,3,4,2) = 2004$ and therefore 176 is enough. Because a permutation with a 167 cycle, 3 cycle, 4 cycle, 2 cycle where all the cycles are disjoint has an order 2004. Now my problem is to prove that it is the smallest number. – bemyguest Apr 1 '12 at 14:50
To prove it's the smallest, note that if $x\in A_n$ has order $2004$, then $x^12$ has order $167$ (a prime), so $\langle x^{12}\rangle$ is a cyclic (Sylow) subgroup, which must therefore be a cycle. Can you take it from there? Hint: try some other powers of $x$. – bgins Apr 1 '12 at 15:28
up vote 1 down vote accepted

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 $S_6$.

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