When is it valid to argue using my 'trivial intersection -> so many distinct elements'? How do I know if I can make counting statements for sylow theorems that rely on intersecting trivially? I.e it is a common argument that I use in which I rely on we have $x$ amount of sylow $p$-groups that are cyclic, if they intersect trivially they give $y$ number of non-trivial elements and this forces $n_q=1$ which gives us a normal subgroup, simple-non simple yada yada, or this normal subgroup is cyclic, abelian, solvable.
When is it valid to argue using my 'trivial intersection -> so many distinct elements'?

Example for clarity: $|G|=36=3\times12=3^2\times 2^2$, so in the case that $n_3=4$, we have $4$ sylow $3$-groups of order $9$, if they intersect trivially we have $(9-1)\times 4$ non-trivial elements i.e. $32$ non trivial elements, we need to obtain $3$ more non-trivial elements, this forces $n_2=1$, which gives us a normal subgroup(since $n_2=1$) and this is cyclic of order $2$, and cyclic means abelian which means solvable. 
 A: In the example you give, the Sylow subgroups might not intersect trivially, so you also have to consider the situation when they do not.
In that case, two Sylow $3$-subgroups $P_1$ and $P_2$ would intersect in a subgroup $Q$ of order $3$. We can analyse that case by considering the centralizer $C_G(Q)$ of $Q$ in $G$. Since it contains both $P_1$ and $P_2$, $C_G(Q)$ has more than one Sylow $3$-subgroup, so it must have at least $4$, but then we get $|C_G(Q)| \ge 36 = |G|$, so $C_G(Q) = G$, and hence $Q \lhd G$ and we can carry on the analysis by studying $G/C_G(Q)$.
A: If you have $x$ $p$-Sylows $S_1$,...,$S_x$ of order $p^k$ then using inclusion exclusion principle you get that :
$$|S_1\cup...\cup S_x|\geq p^kx-\frac{x(x-1)}{2}(p^k-p^{k-1}) $$
Because the number of elements in the union is greater than the sum of the number of elements in each set ($xp^k$) minus the maximum of number of elements in the intersection of two different $p$-Sylows ($p^k-p^{k-1}$) multiplied by the number of ordered pairs of different $p$-Sylows ($\frac{x(x-1)}{2}$). 
$$|S_1\cup...\cup S_x|\geq p^{k-1}x(p-\frac{(x-1)(p-1)}{2}) $$
This lower bound is effective if and only if $p>>x$. 
In general, if you have a better upper bound for the number of elements in the intersection of two different $p$-Sylows : $M$. You get :
$$|S_1\cup...\cup S_x|\geq p^kx-\frac{x(x-1)}{2}M $$
Remark that the only way to have, a priori, that the intersection of two different $p$-Sylows is trivial is that their cardinal is prime. 
