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 lets 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?

  • $\begingroup$ Slightly different context, but essentially a duplicate of math.stackexchange.com/questions/2536201/… $\endgroup$ – Eric Wofsey Dec 8 '18 at 23:00
  • 2
    $\begingroup$ Of course they are exhaustive: (b) is the negation of (a). If two different subgroups of order 9 do not intersect trivially, then they necessarily intersect in a 3-element subgroup, as a proper nontrivial subgroup in a 9-element group must have 3 elements by the Lagrange theorem. $\endgroup$ – A. Pongrácz Dec 8 '18 at 23:04

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