I The cases in proving that group of order 90 is not simple

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I am trying to attempt this problem, but I am wondering why exactly these are the two cases the problem is split into. I can understand the first case, since that let's us count elements and get a contradiction, but why is the second case there? In other words, why do these two cases exhaust all possibilities?

EDIT: Actually, I think that I see it now. Since the intersection of subgroups is a subgroup, by Lagrange we must have that the negation of two distinct Sylow 3-subgroups intersecting trivially is intersecting with 3 elements.
 
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see pages 65-66 of these class notes:

http://alpha.math.uga.edu/%7Eroy/843-1.pdf
 
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