We are thinking about the symmetric group of degree $12$ here and I would like to find the exact number of elements whose order is $5$. I am more interested in understanding how to think about problems similar to this one, than in actually knowing the answer to this specific question.
Regardless of that, I think this problem will serve as a good representative of this class of problems, so I decided to solve that one.
To start it off, I'm pretty sure that all the permutations of order $5$ will be of one of these two forms.
- It can either be a single cycle of length $5$, so something along the lines of $(a \ b \ c \ d \ e)$
- Or it could be of the form $(a \ b \ c \ d \ e)(f \ g \ h \ i \ j)$ so essentially two connected cycles, both of length $5$.
We will first try to handle the first option. Since we have $12$ elements, we can use this formula to get all the possible cycles $$ \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{5} $$ Notice that we're dividing by 5 because we are double counting some cycles. For example, the cycle $ (a \ b \ c \ d \ e) $ is equivalent to the cycle $ (b \ c \ d \ e \ a) $. In other words, each of these $5$ elements can come first in the cycle, so we have to divide everything by $5$.
Assuming that this is correct for the first form, we should have $19,008$ elements of this form.
Now to try and tackle the second form of these elements. We should be able to keep using the same expression for finding the first cycle of the two, and then we can multiply that by the following $$ \frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3}{5} $$ We are doing this because we used up $5$ elements already on "building" the first cycle (so we start this one with $7$) and we again have to divide everything by 5 because each of these elements can come first in the cycle.
If my logic is correct so far, this should give us $$ 19,008 \cdot 504 = 9,580,032 $$
Now this does seem like a lot, considering that we should also add to this the number of elements of the first form, so we are left with $$ 9,580,032 + 19,008 = 9,589,040 $$ but considering that this group has $$ 12! = 479,001,600 $$ it probably isn't as bas as it looks.
Now obviously my question is whether this is correct. And if it is, is this the best way of approaching problems like this or could I have done something more efficient?
All help and suggestions are very much appreciated!
It's worth trying to work out the answer in general: how many permutations in $S_n$ have $c_i$ cycles of length $i$? Try showing as an exercise that the answer is
$$\frac{n!}{\prod i^{c_i} c_i!}$$
and deduce as a corollary the identity
$$\sum_{\sum i c_i = n} \frac{1}{i^{c_i} c_i!} = 1$$
where the terms on the LHS can be interpreted as probabilities that a random permutation has some cycle type.