# Reference for the Netto's theorem on the permutation groups which was mentioned in the paper of Frobenius

I'm trying to read 'Uber die Charaktere der mehrfach transitiven Gruppen' written by Frobenius. There he mentioned some theorems of Netto.

1. If one multiplies the number of cycles of length $$s$$ that occur in all permutations of a group of order $$h$$ by the number $$s$$, one obtains a multiple of $$h$$, and if the group is $$s$$-fold transitive, the number $$h$$ itself.
2. If you multiply the number of combinations of $$x$$ cycles of length 1, $$y$$ cycles of length 2, $$z$$ cycles of length 3, etc., which occur in all permutations of a group of order $$h$$, by the number $$s=1^{x}x!2^{y}y!3^{z}z!\ldots$$, you get a multiple of $$h$$, and if the group is $$r=x+2y+3z+\cdots$$-fold transitive, the number $$h$$ itself.

But the theorem does not make sense to me. Because for example if we take the cyclic subgroup of order 3 of $$S_{3}$$. Then $$h=3$$. If we take $$s=2$$, then I don't know how I should understand the theorem.

The paper of Frobenius is available in the website. https://www.e-rara.ch/doi/10.3931/e-rara-18850

3. I especially want to understand the concept of $$\textbf{dimension}$$ of an irreducible representation $$\chi^{\lambda}$$ of the symmetric group $$S_{n}$$ associated to the partition $$\lambda=(\lambda_{1}, \lambda_{2}, \ldots )$$. The dimension is defined to be $$n - \lambda_{1}$$ and it is the smallest integer $$r$$ for which the character $$\chi^{\lambda}$$ appears as a constituent of the permutation character of $$S_{n}$$ on $$[n]^{r}$$, the $$r$$-fold cartesian product of the set of $$n$$-elements. Is there a reference for the proof of the equality of the two numbers?
• The translation looks roughly right to me. I believe I understand the first statement, at least in the case when the group is $s$-fold transitive. In your example of $S_3$ with $s=2$, the total number of cycles of length $2$ in all elements of the group is $3$, and if you multiply that by $2$, you get $6$, which is the order of the group. The same example works with $s=1$ or with $s=3$. With $s=3$ there are $2$-cycles of length $3$ and $2 \times 3 = 6$. I haven't attempted to understand part 2. Mar 9 '21 at 16:36
• @DerekHolt Thank you very much for the comment. In your demonstration for the case s=2, did you take the whole symmetric group as the group? By the subgroup of order 3, I have meant the subgroup of $S_{3}$ generated by $(1,2,3)$, in that case, in my opinion, each non-trivial elements have cycle type 3, thus the number of cycles of length 2 is 0. Here is where I got stuck. Mar 9 '21 at 16:56
• Oh sorry, yes I misread your example and took the whole group $S_3$. Yes, if you take its cyclic subgroup of order $3$, you get $0$ which, as the statement claims, is a multiple of the group order $3$. This group is not $2$-transitive, so it is not obliged to be equal to $3$. Your group is $1$-transitive, and the count does indeed come to $3$ in this case, because the identity element has $3$ cycles of length $1$. Mar 9 '21 at 17:21