Permutation group - Groups and Symmetries

Question

How many elements of order $k$ are in $S_n$ for $k<n$.

$S_n$ is the permutation group of order $n$.

Attempt: I thought it's just the number of ways you can pick $k$ objects out of $n$ objects so it's $n$ choose $k$. Is this right? I think the answer should not exceed the order of $S_n$ which is $n!$ so this makes sense to me. Please correct me if I'm wrong. Thanks!

Not exactly, since $(123)(45) $ has order $6 $ but only involves $5 $ elements. Another example is $(12)(34) $ which has order $2 $ but involves $4 $ elements. — Uncountable, Feb 21 '15 at 18:30
You've only counted permutations that consist of a single cycle. — paw88789, Feb 21 '15 at 18:46
This is actually a tricky question. It is discussed in Number of permutations of order k. Note that the restriction $k<n$ is nontrivial; for example, $S_5$ contains elements of order 6 and $S_8$ contains elements of orders 12 and 15, but not 13 or 14. — MY USER NAME IS A LIE, Feb 21 '15 at 20:51

pjs36 · Answer 1 · 2015-02-21T19:17:42.990

Your method gives you a lower bound on the number of elements of order $k$ in $S_n$, where $1 < k \leq n$. As has been pointed out, this counts elements that are a single cycle, in cycle notation.

So, in $S_6$, the permutation $\pi = (123)(456)$ is an element of order 3, but it won't have been counted by the above argument. Additionally, permutations $(1234)$ and $(1243)$ are distinct, but would presumably have lumped together, in your combinatorial argument.

Permutation group - Groups and Symmetries

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