Let $r(n) = \left| \left\{ \sigma \in S_4 : \mbox{ord} ( \sigma) = n \right\} \right|$. Is it true that:

  1. $r(2)>r(4)$
  2. $r(4) > r(3)$
  3. $r(1)+r(3) = r(2)$
  4. $r(5) = r(6)$

I can write all elements of $S_4$ but I suppose that is better and faster way to solve it.

  • 1
    $\begingroup$ The order of $\sigma$ must divide $4!=24$, so $5$ is impossible. We have $r(1)=1,r(2)=9,r(3)=8,r(4)=6$. $\endgroup$ – Dietrich Burde Aug 9 '13 at 18:15
  • 3
    $\begingroup$ Ok but how did you compute it? $\endgroup$ – Thomas Aug 9 '13 at 18:17
  • $\begingroup$ math.stackexchange.com/a/311688/8581 $\endgroup$ – mrs Aug 9 '13 at 20:34
  • $\begingroup$ @user2213654: I think what you already did and what user gave you is enough. Unless, you are looking for a computer based approach. $\endgroup$ – mrs Aug 10 '13 at 0:39

There is a formula that calculates the number of ways how a given integer can be split into parts, to be found for example here: Groupprops: Conjugacy class size formula in symmetric group:

Suppose $n$ is a natural number and $\lambda$ is an unordered integer partition of $n$ such that $\lambda$ has $a_j$ parts of size $j$ for each $j$. In other words, there are $a_1$ $1$s, $a_2$ $2$s, $a_3$ $3$s, and so on. Let $c$ be the conjugacy class in the symmetric group of degree $n$ comprising the elements whose cycle type is $\lambda$, i.e., those elements whose cycle decomposition has $a_j$ cycles of length $j$ for each $j$. Then: $$ \! |c| = \frac{n!}{\prod_j (j)^{a_j}(a_j!)} $$

The ways how to split $4$ are $1+1+1+1=2+1+1=2+2=3+1=4$, where each corresponds to conjugacy class of $S_4$, e.g. $2+1+1$ is a 2-cycle and two 1-cycles, The order of elements of a conjugacy class can be calculated by the least common multiple of the addends, e.g. $\operatorname{lcm}(2,1,1)=2$.

Let's do the first of your examples: $r(2)>r(4)$

$2+1+1$ and $2+2$ have $\operatorname{lcm}(2,1,1)=\operatorname{lcm}(2,2)=2$ and $4$ has $\operatorname{lcm}(4)=4$. Therefore $$ \! \frac{4!}{[(2)^1(1!)][(1)^2(2!)]} + \! \frac{4!}{(2)^2(2!)} > \frac{4!}{(4)^1(1!)}\\ 6 + 3 > 6 $$


Think about the elements of $S_4$. These are: the identity (order 1), the transpositions (order 2), the 3-cycles (order 3), the 4-cycles (order 4) and the products of two disjoint transpositions (order 2). Now let's count how many there are of each type: 6 transpositions, 8 3-cycles, 6 4-cycles, and 3 products of two disjoint transpositions.


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