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How many subgroups of order 5 does $S_6$ have?

There must be some theorem that I can use to find out the answer to this question, but I am coming up blank. I don't just want to know the answer, I'd like to know how you can figure it out. Thank you!

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  • $\begingroup$ Do you know Sylow's theorems? $\endgroup$ – user217285 Mar 26 '17 at 16:24
  • $\begingroup$ No we learn that at the end of the course :) $\endgroup$ – jgcello Mar 27 '17 at 6:33
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First, note that every prime-ordered (sub)group is cyclic, so this is almost equivalent to finding the total number of order-$5$ elements in $S_6$. Each subgroup of order $5$ will have $4$ elements of order $5$ (per Lagrange's theorem) and will be disjoint sans the identity. How can we count the total number of elements of order $5$?

Every element in $S_n$ can be decomposed into a product of disjoint cycles. Suppose some $\sigma \in S_n$ decomposes into disjoint cycles $\pi_1 \circ \pi_2 \circ \cdots \circ \pi_m$ of lengths $k_1, k_2, ..., k_m$. Then one can prove that the order of $\sigma$ is equal to the least common multiple of the $k_i$'s.

Now let's think about the problem at hand. Suppose a $\sigma \in S_6$ has order $5$. This means, when decomposed into disjoint cycles, the least common multiple of the cycle lengths is $5$. Meditate on when this is possible, and you'll be well on your way to an answer.

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  • $\begingroup$ Nice answer, @Kaj Hansen. $\endgroup$ – Dean Young Mar 26 '17 at 16:31
  • $\begingroup$ I appreciate your answer, thank you! $\endgroup$ – jgcello Mar 27 '17 at 6:35
  • $\begingroup$ Just to kind of wrap this up: I determined that $S_6$ has 144 elements of order 5. My intuition believes that the answer is then 144 subgroups, but after reading whats you wrote I cannot definitively say that this is true or not. $\endgroup$ – jgcello Mar 27 '17 at 6:42
  • $\begingroup$ Careful @jgcello: some of those are going to belong to the same subgroup. If you look at $\mathbb{Z}_5$ for example, every nonzero element of that group will generate it. So if you want the total number of subgroups, finding the total number of order-$5$ elements is over-counting. To correct for that, you'll need to divide your answer by $4$. $\endgroup$ – Kaj Hansen Mar 27 '17 at 14:56

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