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

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I'm depending on the Google translator. and the translation reads

  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

Therefore, I want to ask

  1. whether the translation of the above theorem is correct.
  2. whether theorems similar to the one above are still significant in some branch of group theory. If so, I hope to know some recent paper or book.
  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?

I'm afraid that the question might be peripheral, but I hope to read the paper of Frobenius steadily even though it is old and in German.

Thank you very much !

  • $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Mar 9 '21 at 16:36
  • $\begingroup$ @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. $\endgroup$
    – gualterio
    Mar 9 '21 at 16:56
  • $\begingroup$ 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 $1$. $\endgroup$
    – Derek Holt
    Mar 9 '21 at 17:21
  • $\begingroup$ @DerekHolt Now I can understand the statements of the theorem! Thank you very much. $\endgroup$
    – gualterio
    Mar 9 '21 at 22:51

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