# Computing the index $[G:H]$ with $H \triangleleft G$.

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 $S_n$?

• I do not understand your construction. It is certainly true that a normal subgroup is a unio of conjugacy classes, but it is usually difficult to decide which unions of conjugcy classes form subgroups. The symmetric group $S_{n}$ only has one proper non-identity normal subgroup for $n > 4.$ – Geoff Robinson Mar 29 '14 at 20:34
• @GeoffRobinson I extended it to the case of $n$ to make it more general, although I was working on $S_4$. In which case we have $N_3 = C_1 \cup C_2 \cup C_3$ and $N_4 = C_1 \cup C_2 \cup C_4$. Can you comment on the $n = 4$ case? – Mussé Redi Mar 29 '14 at 20:54
• And for $n >4,$ the only normal subgroup other than the identity and the whole group is that containing all conjugacy classes of even permutations- but I do not see how your description makes that clear. – Geoff Robinson Mar 29 '14 at 20:57
• Given the knowledge that $N$ is a normal subgroup of $G$. Can we use this to compute the index $[G,N]$? – Mussé Redi Mar 29 '14 at 21:19

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.
• Consider $N_3$, according to my notation. Is there a easy way to compute the index $[S_4,N_3]$? And for $[S_4,N_4]$? – Mussé Redi Mar 29 '14 at 21:06
By Lagrange's theorem we have $$[G:H] = \frac{\# G}{\#H},$$ with $H$ a subgroup of $G$.