In my group theory class, when we are asked questions like
Determine the groups of order 30
or
Prove no group of order 90 is simple
We can do this, using a combination of Sylow's theorems and a bunch of little tricks. One of the tricks that comes up again and again involves the principle of inclusion/exclusion. I don't quite get it, though.
Given a group of order 30, we know there are either 1 or 6 Sylow 5-subgroups and 1 or 10 Sylow 3-subgroups. If we assume there to be both 6 5-subgroups and 10 3-subgroups simultaneously, we run into a problem. That's because the Sylow p-subgroups for a given p intersect on the identity, so we can easily say there are 6 * 4 = 24 elements of order 5 and 10 * 2 elements of order 3. That's too many elements.
The problem I have is when the Sylow p-subgroups are of prime powered order. For example, in the answer to the second question above we see the following argument:
if we pick two distinct Sylow 3-subgroups K1 , K2 , then |K1 ∩ K2 | = 1 or 3, so |K1 ∪ K2 | ≥ 9 + 9 − 3 = 15 and K1 ∪ K2 \ {1} consists of elements of orders 3 or 9. Thus G has at least 14 elements of order 3 or 9.
What is in the intersection of K1 and K2? Why are we we taking the union of just two elements? Why does their union consist of elements of order 3 and 9? If we took a case where the Sylow 3-subgroups had order 27 instead of 9, how would we proceed? Is there a situation where we need to take a union or 3 or more groups and use the more general inclusion/exclusion?
Thank you in advance.