I've been trying to figure this out:
Let $G$ be a group of order 189. Show that any two distinct Sylow 3-subgroups intersect in 27 elements.
I know that $189=3^3\times 7$, so there is at least one Sylow 3-subgroup of order 27. Furthermore, I know that there is either one or 7 Sylow 3-subgroups.
So if I'm interpreting the question correctly, I'm just supposed to prove that there is only one Sylow 3-subgroup of order 27? If so, how do I go about proving that?
EDIT: I finally found out that this question had a typo. It is supposed to say "…intersect in 9 elements." I haven't figured this out yet either, so if anyone has hints I would appreciate it.
Best Answer
That was my answer to the original question. The claim is not true. According to this Web site there are exactly 8 groups of order 189 where the Sylow 3-subgroup is not normal Edit: A concrete example; Take any group of order 27. It has a quotient of order 3. Hence it is acting on the additive group of integers mod 7 (the automorphisms of that group include multiplication by 2 mod 7 which has order 3). The corresponding semidirect product is an example. In it, the subgroup of order 7 is not in the center so the group is not a direct product of a group of order 7 a a group of order 27.