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I'm trying to prove it.

So I assume there is more than one sylow-3 subgroup, each has order 9, we have 4 of them, now their intersection is either e or has order 3.

if it is e, then we have 32 elements in these subgroups besides e (33rd)

then assume we have more than one sylow-2 subgroup, we should have 3, but then their intersection has order 1, or 2.

So now it works for all cases except when the intersection of the sylow 3 subgroups is 3 and that of the sylow 2-subgrps is 1, then I get the magic number 36.

What should I be looking for in this proof. I was advised there is a way to prove it using this way without the appeal to homomorphisms to Sn.

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# A group of order 36 is simple

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