My only thought is, since it is the number of ways to permute items. Hence, since it is the number of ways to permute items (if you have items, but stabilize 1, then you are only permuting items).
Hi! It's me again with another counting problem.
I am trying to find the order of the following set (where is fixed): , the stabilizer of 1 in .
Now, I have calculated 4 of these:
n = 2: so
n = 3: so
n = 4: so
I also did n = 5 which gives:
The pattern seems to be that , but the examples don't help me to understand the counting in general, which was the intent of doing the examples in the first place.
Any thoughts?
-Dan
Yes, it is as you say. (sighs) I was hoping to get back before someone answered. If I had calculated this with the nth position (the stabilizer of the highest place) I would have seen it right away. I was thinking about it way too hard and I got lost in it.
Thanks!
-Dan
Sometimes this is called the "fixed group" (or isotropy group) of 1, that is: the elements of S_{n} that fix 1. It should be obvious this is, in fact a subgroup of S_{n}, because it is closed under composition (the multiplication of S_{n}).
There are a couple of way to approach this:
the first is to show that the mapping:
given by:
, for k = 1,2,...,n-1 is an isomorphism.
The second is to use the orbit-stabilizer theorem:
Now because we have the n-1 transpositions (1 k), along with the identity map, it is clear that |S_{n}(1)| = n (the orbit of 1 has size n, because S_{n} acts TRANSITIVELY on the n letters (the set of cardinality n...in this case {1,2,...,n}) it permutes). Thus:
|Stab(1)| = n!/n = (n-1)!.