As an excercise I am investigating the symmetric group $S_n$ beginning from its conjugacy classes and then taking their union to form normal subgroups.
Since conjugacy classes $C_m$ contain elements of the same order $m$ we see that normal subgroups $N_m = \langle C_m \rangle$ of $S_n$ are formed by taking the union $$N_m = \bigcup_{i \in I_m \cup \{1\}} C_i \text{ with } I_m \text{ containing the even numbers up to and including some } m\leq n .$$ Namely, we can always multiply products of $C_m$ until we get some element of order $p \geq 2$ , where $p$ can assume all even orders up to $m$, so the subgroup $\langle C_m \rangle$ must include these elements. But then $N_m$ must also contain the whole conjugacy class $C_m$ since a normal subgroup is a union of conjugacy classes.
My question now is: is there a elegant/easy way to compute the index $[S_n,N_m] = \#(S_n/N_m)$?
This question comes down to determining the cardinality of the subset $A_m \subset S_m \setminus N_m$ given such that every pair $a_1,a_2 \in A_m$ satisfies $a_1a_2^{-1} \notin N_m$. This follows from the equivalence $$a_1N_m = a_1N_m \iff a_1a_2^{-1} \in N_m .$$ I find my method of determining $\# A_m =[S_n,N_m]$ cumbersome and was wondering if there is some simpler/more elegant method.
Suggestions are welcome.
Supplement: can we easily generalise this problem to an arbitrary group $G$ instead of $S_n$?
As far as I can see, for a general group, the only way to proceed to construct all normal subgroups is to choose a collection of conjugacy classes $C_{1},C_{2},\ldots C_{n},$ and if their union does not already form a subgroup, adjoin all conjugacy classes which meet the product $C_{i}C_{j}$ for $1 \leq i,j \leq n.$ Then continue until you reach a situation where no new classes appear in the products obtained. This soon becomes impractical without very detailed knowledge of the group.