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I am trying to find the smallest positive integer $n$ such that $S_n$ contains an element of order 60. I know that every permutation in $S_n$ can be expressed as the product of disjoint cycles, and I also know that the order of a permutation is the least common multiple of the lengths of the cycles in its disjoint cycle product.

So, suppose $\sigma \in S_n$ such that $|\sigma|=60$. Then $60=lcm \left\{|c_1|, \ldots , |c_m| \right\}$, where $c_i$ is one of the cycles in the disjoint cycle product of $\sigma$. The prime factorization of 60 is $2^2\cdot 3 \cdot 5$, but the least common multiple of 2, 3, and 5 is 30, so that doesn't help. However, the least common multiple of 3, 4, and 5 is 60. I just don't know where to go from here.

Also, what is meant by a "disjoint cycle pattern"? Thanks for your help!!

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You are definitely on the right track. As you know that $lcm(3,4,5)=60$, you know that the following permutation has order $60$:

(1 2 3)(4 5 6 7)(8 9 10 11 12)

So $S_{12}$ definitely has an element of order $60$.

However, $3$, $4$, $5$ isn't the only set of numbers with an lcm of $60$. For example, $5$, $12$ also works. But that would give an element of $S_{17}$, which isn't any better.

You just need to do a little bit of explaining of various cases to show that any other set of numbers with an LCM of $60$ can't have a sum of less than $12$.

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