Intersection of Sylow 3-subgroups in a group of order 189.

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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.

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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.

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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.