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 )

  • $\begingroup$ There certainly aren't precisely 5 3-Sylow subgroups! $\endgroup$ – C Monsour Nov 9 '18 at 19:32
  • $\begingroup$ I only chose a random example off the top of my head , I don't know enough about groups to have known that my choice was impossible. what would be a better example and i'll edit it ? $\endgroup$ – excalibirr Nov 9 '18 at 19:57
  • $\begingroup$ Just about any number that leaves a remainder of 1 when divided by 3. For example, 7 would work. $\endgroup$ – C Monsour Nov 9 '18 at 20:10
  • $\begingroup$ I don't see any reason why all pairs must intersect nontrivially if one pair does. This seems to be what you are claiming. $\endgroup$ – Matt Samuel Nov 9 '18 at 21:37
  • $\begingroup$ @MattSamuel No I have the case where one or more pairs may intersect non-trivially and the second case where all of them intersect non-trivially . But I meant these as being didtinct scenarios not as one implying the other . $\endgroup$ – excalibirr Nov 9 '18 at 21:45

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