>>10320238I'm sorry, my bad i made a chaos with copying and pasting. Here i corrected the post:
Consider this:
I have a group G with cardinality 20 that is a subgroup of S6 and it has this structure:
- 1 normal subgroup of elements of order 5 (and cardinality 5)
- 5 cyclic subgroup of elements of order 2 and each subgroup has cardinality 4( thus the whole set has cardinality 15,they are disjointed, without the identity)
Now i want to study the number of orbits of the action of G on the set X={1,2,3,4,5,6}
and i will use this formula for that (
https://groupprops.subwiki.org/wiki/Orbit-counting_theorem, the first form).
MY PROBLEM IS NOW THAT:
if i count the elements that are fixed by the 5 order elements' subgroup i should get 4 (i'm not counting the identity), because they are cycle of 5 numbers and the number that is fixed by them is the one that doesn't appear in the cycle, thus:
4?1
Same reasoning applied to the 5 cyclic subgroups of order 2 i should get 4 fixed numbers for each of them, thus:
thus:
15?4
Now i count the elements fixed by the identity i get the whole set:
6
Summing all of them i get: 4+60+6=70
which it is not divisible by 20, which is an absurd (according to the formula above)
Where is my mistake?