Say you have some collection of Sylow $p$-subgroups. For example lets say their are 7 Sylow $3$-subgroups. and that each has order $9$ There are three cases( right ?)
i) any pair $P_1\neq P_2 $ intersect non-trivially
ii) every pair $P_1\neq P_2$ intersect non trivially
iii) every pair of $P_1 \neq P_2$ intersect trivially
My questions: (Note: trivial intersection then we know that there are 7(9-1)=56 elements of order nine)
a) If we say that any pair $P_1,P_2$ intersect non trivially does this mean that there could be , for instance , just two p-sylow subgroups which intersected non- trivially so they shared three elements one of which was shared by all the other sylow 3 subgroups. like wise we could then have three sylow 3 subgroups which shared 3 elements etc. etc. Is this correct (/ possible in a Sylow p-subgroup ) ?
b) In my notes (regarding showing whether a group was simple) my lecturer didn't consider case two which led me to believe that perhaps case i) and ii) as I have stated them might be in fact the same thing . Is there any merit to this thought ?
c)If all of the P's intersect non trivially then we know that the size of there subgroup is 3 but the number of total elements of order 3,9 is 7(9-3)=42 ( as opposed to the trivial case. see note ). But if cases i) and ii) are in fact distinct then this only works for ii). I don't want the actual equation just to know... is there a similarly simplistic formula as the two i mentioned here, for calculating the number of elements , that can be applied to case i) ( conceptual explanations would be appreciated aswell )