Let $S$ and $T$ be distinct $3$-Sylow subgroups of the symmetric group $S_6$. Prove that $S$ and $T$ intersect trivially.
Here are my thoughts so far:
I figured the Sylow Theorems could give us some insight here. Let $G = S_6$. Then $G$ has order $6! = 720 = 2^4 \cdot 3^2 \cdot 5$. It follows that any $3$-Sylow subgroup of $G$ must have order $9$, and that, denoting $n_3$ by the number of $3$-Sylow subgroups of $G$, $n_3 | 80$ and $n_3 \equiv 1$ (mod $3$) $\Rightarrow$ $n_3 = 1, 4, 10, 16, 40$.
I'm not sure how to proceed from here. There's a long list of possibilities of possible orders for the $5$-Sylow and $2$-Sylow subgroups 0f $G$ -- so it doesn't seem like we can get away with a counting argument here, showing that if $3$-Sylow subgroups of $G$ intersect non-trivially, we end up with more than $720$ elements in our group, contradicting the order of $G$. How can I reach a contradiction?
I appreciate all the help. Thanks!