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.
The answer to the new question. The $7$-Sylow subgroup $S_7$ is cyclic and normal. Let $P$ be a 3-Sylow subgroup of order $27$. Then $S_7P$ has 189 elements, so $S_7P=G$, the whole group. So $G$ is a semidirect product of $S_7$ and $P$ corresponding to some homorphism $\phi: P\to Aut(S_7)$. The target group of this homomorphism is of order $6$, so $|\phi(P)|=3$ (it must be a power of 3 and if it is 1, $G$ is the direct product $S_7\times P$). So $K=Ker(\phi)$ is of order 9. That subgroup $K$ centralizes $S_7$ and is normal in $P$. Therefore for every $g\in G$, $g^{-1}Kg=K$. By the Sylow theorem then every 3-Sylov subgroup of $G$ must contain $K$. So the order of intersection of any two 3-Sylow subgroups is at least 9.