# Has this result about the number of permutations of a given cycle type (or centralizers) been proved?

I was playing with the cardinality of conjugacy classes of the symmetric groups, which we know is the number of permutations of a given cycle type and there is a natural one-to-one correspondence between partitions of $$n$$ and conjugacy classes of $$S_n$$. For example, take the partition $$\mu_6^{(3)}=(1,2,3)$$. Note that I am writting the $$6$$ and the $$3$$ to emphasize that it is a partition of $$6$$ and has $$3$$ cycles. Now, $$\mathcal{C}_{(1,2,3)}$$ is the conjugacy class of permutations of cycle type $$(1,2,3)$$, and $$|\mathcal{C}_{(1,2,3)}|$$ is the number of permutations of cycle type $$(1,2,3)$$. I can express this number as a sum over $$(1,2,3)$$ and the "smaller" partitions:

$$(1,1,3) \ , \ (1,2,2) \ , \ (1,1,2) \ , \ (1,1,1)$$

Note that all these partitions have the same cycle number $$c=3$$. So, I will verify that: $$\frac{|\mathcal{C}_{(1,2,3)}|}{6!}=\frac{|\mathcal{C}_{(1,1,1)}|}{3!}\times 6-\frac{|\mathcal{C}_{(1,1,2)}|}{4!}\times 6 +\frac{|\mathcal{C}_{(1,2,2)}|}{5!}\times 4+\frac{|\mathcal{C}_{(1,1,3)}|}{5!}\times 2-\frac{|\mathcal{C}_{(1,2,3)}|}{6!}$$

How I get the factors 6, 6, 4, 2 and 1? I will use the following notation and then I substarct 1 to each number and calculate the binomial coefficients, I permute the numbers from $$(1,2,3)$$ but the others are fixed. $$\binom{(1,2,3)}{(1,1,1)}=\binom{0}{0}\binom{1}{0}\binom{2}{0}+permutations of 0,1,2=\binom{0}{0}\binom{1}{0}\binom{2}{0}+\binom{1}{0}\binom{2}{0}\binom{0}{0}+\cdots=1+1+1+1+1+1=6$$

$$\binom{(1,2,3)}{(1,1,2)}=\binom{0}{0}\binom{1}{0}\binom{2}{1}+permutations of 0,1,2=\binom{0}{0}\binom{1}{0}\binom{2}{1}+\binom{0}{0}\binom{2}{0}\binom{1}{1}+\binom{1}{0}\binom{0}{0}\binom{2}{1}+\binom{2}{0}\binom{0}{0}\binom{1}{1}=2+1+2+1=6$$

$$\binom{(1,2,3)}{(1,2,2)}=\binom{0}{0}\binom{1}{1}\binom{2}{1}+permutations of 0,1,2=\binom{0}{0}\binom{1}{1}\binom{2}{1}+\binom{0}{0}\binom{2}{1}\binom{1}{1}=2+2=4$$

$$\binom{(1,2,3)}{(1,1,3)}=\binom{0}{0}\binom{1}{0}\binom{2}{2}+permutations of 0,1,2=\binom{0}{0}\binom{1}{0}\binom{2}{2}+\binom{1}{0}\binom{0}{0}\binom{2}{2}=1+1=2$$

$$\binom{(1,2,3)}{(1,2,3)}=\binom{0}{0}\binom{1}{1}\binom{2}{2}+permutations of 0,1,2=\binom{0}{0}\binom{1}{1}\binom{2}{2}=1$$

Now, using the formula for the number of permutations $$|\mathcal{C}_{(1,1,1)}|=\frac{3!}{1^3(3!)}=\frac{3!}{6}$$ $$|\mathcal{C}_{(1,1,2)}|=\frac{4!}{1^2(2!)2(1!)}=\frac{4!}{4}$$ $$|\mathcal{C}_{(1,2,2)}|=\frac{5!}{1(1!)2^2(2!)}=\frac{5!}{8}$$ $$|\mathcal{C}_{(1,1,3)}|=\frac{5!}{1^2(2!)3(1!)}=\frac{5!}{6}$$ $$|\mathcal{C}_{(1,2,3)}|=\frac{6!}{1(1!)2(1!)3(1!)}=\frac{6!}{6}$$ Therefore $$\frac{|\mathcal{C}_{(1,1,1)}|}{3!}\times 6-\frac{|\mathcal{C}_{(1,1,2)}|}{4!}\times 6 +\frac{|\mathcal{C}_{(1,2,2)}|}{5!}\times 4+\frac{|\mathcal{C}_{(1,1,3)}|}{5!}\times 2-\frac{|\mathcal{C}_{(1,2,3)}|}{6!}=$$ $$1-\frac{3}{2}+\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{1}{6}=\frac{|\mathcal{C}_{(1,2,3)}|}{6!}$$ I have verified this result with others conjugacy classes and I infer that the general expression is

$$\sum^\infty_{n=c}\sum_{\mu^{(c)}_n}\frac{(-1)^{n+c}|\mathcal{C}_{\mu^{(c)}_{n}}|}{n!}\binom{\mu^{(c)}_{m}}{\mu^{(c)}_{n}}= \frac{|\mathcal{C}_{\mu^{(c)}_{m}}|}{m!}$$ Then, I learned that $$\frac{m!}{|\mathcal{C}_{\mu^{(c)}_{m}}|}$$ is the formula for the size of the centralizers, so the result can be written as: $$\sum^\infty_{n=c}\sum_{\mu^{(c)}_n}\frac{(-1)^{n+c}}{z_{\mu^{(c)}_{n}}}\binom{\mu^{(c)}_{m}}{\mu^{(c)}_{n}}=\frac{1}{z_{\mu^{(c)}_{m}}}$$

I am sure that this result should be in some book but I cannot find it. I have checked the Bruce E. Sagan book about the symmetric group but I couldn't find it. So I need a reference on this specific result. I am not mathematician, so excuse me for wrong notation.