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I'm trying to understand the calculation for the cycle index of a Symmetric Group $S_n$.

This is discussed and presented in mathematical format on the wikipedia here:

https://en.wikipedia.org/wiki/Cycle_index#The_symmetric_group_Sn

My first question is merely a sanity-check, to make sure I'm understanding what I'm reading: Wikipedia says:

"There are three steps: first partition the set of $n$ labels into subsets, where there are $j_k$ subsets of size $k$."

If I'm on the right track, this $j_k$ is a Stirling number of the first kind, $j_k = Stirling1(n,k)$. Is this correct?

My second question regards the non-recursive presentation for $Z(S_n)$, wikipedia presents:

$\frac{n!}{\prod_{k=1}^n k^{j_k}j_k!}$

If I'm right (above) that $j_k$ is the Stirling number of the first kind for $n,k$ then I can compute this fraction for $n$ and I will get a single number. But I expect that a cycle index $S_n$ is not a single number but a sum of dummy variables. I've obviously gone off track somewhere. Any advice is much appreciated!

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    $j_k$ are *not* Stirling numbers of the first kind they are the number of cycles length $k$. An example: consider the cycle index $z_1^1z_2^2z_3^1$; the permutations with 1 cycle length 1, 2 cycles length 2 and 1 cycle length 3, clearly $n=8$. Each permutation with this cycle index is represented as (●)(●●)(●●)(●●●) E.g. (1)(23)(45)(678) is the permutation "13254786". In this example $j_1=1$, $j_2=2$, $j_3=1$, the total no. of perms with this cycle index is found by replacing the ●s in $n!$ ways, dividing by cycle lengths $1^12^23^1$ and permutations of equal length $j_1!j_2!j_3!=1!2!1!$ – 2017-02-28
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    This definitely addresses my sanity check (failed!) :) I see now that Stirling numbers of the first kind are not the same thing. The Stirling numbers of the first kind give me the number of permutations with of n with exactly k partitions. This is not the same thing as the number of partitions of length k across all permutations. I think that I need a simpler example than n=8, however, because although my goal is to compute the cycle index for $S_n$ generally, but 8 is too large to work out on a paper. – 2017-02-28
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    Try all permutations of [4] so that the possible cycle indices are $z_1^4$ (●)(●)(●)(●) $\frac{4!}{1^44!}=1$, $z_1^2z_2^1$ (●)(●)(●●) $\frac{4!}{1^22^12!1!}=6$, $z_1^1z_3^1$ (●)(●●●) $\frac{4!}{1^13^11!1!}=8$, $z_2^2$ (●●)(●●) $\frac{4!}{2^22!}=3$, $z_4^1$ (●●●●) $\frac{4!}{4^11!}=6$. $1+6+8+3+6=24=4!$ takes care of all the permutations. You might want to list them, begin (1)(2)(3)(4)="1234"; next (1)(2)(34)="1243" (1)(3)(24)="1432", (1)(4)(23)="1324", (2)(3)(14)="4231", (2)(4)(13)="3214", (3)(4)(12)="2134" etc. Notice for e.g. permutations (1)(234)=(1)(342)=(1)(423)and (12)(34)=(34)(12). – 2017-02-28
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    A little more info: A permutation (1)(234) says "starting with 1234 replace 1 with 1, 2 with 3, 3 with 4, 4 with 2" so (1)(234) is the permutation "1342". Stirling numbers of the first kind are related here because, for example $s(n,k)=s(4,2)$ is the number of permutations with exactly 2 $z$s, i.e. The ones $z_1^1z_3^1$ (●)(●●●) and $z_2^2$ (●●)(●●) then $s(4,2)=8+3=11$ as you may confirm. Sometimes Stirling #s of the 1st kind are described as the ways to seat $n$ people at $k$ circular tables, this should now link up with permutations of $[n]$ into $k$ cycles. – 2017-02-28
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    Also $s(4,1)=6$ are those cycle type (I've been incorrectly referring to cycle types as "indices") $z_4^1$ (●●●●), $s(4,3)=6$ are those $z_1^2z_2^1$ (●)(●)(●●) and $s(4,4)=1$ is $z_1^4$ (●)(●)(●)(●). Of course we must then have $\sum_{k=1}^{n}s(n,k)=n!$ since every permutation has been accounted for. – 2017-02-28
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    @N.Shales thanks so much for the further clarifications! After your example with the factorials and exponents in the fractions, I was able to see how to compute the cycle index for a $S_n$ based on integer partitions of $n$. – 2017-02-28
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/54521/discussion-between-jawguychooser-and-n-shales). – 2017-02-28
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    Please consider typing your comments up as an answer, @N.Shales. It closes the question and makes the answer easier to find in search engines. – 2018-12-22

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