TITLE: Distinct Sylow $p$-subgroups intersect only at the identity, which somehow follows from Lagrange's Theorem. Why?
QUESTION [26 upvotes]: It seems that often in using counting arguments to show that a group of a given order cannot be simple, it is shown that the group must have at least $n_p(p^n-1)$ elements, where $n_p$ is the number of Sylow $p$-subgroups.

It is explained that the reason this is the case is because distinct Sylow $p$-subgroups intersect only at the identity, which somehow follows from Lagrange's Theorem.

I cannot see why this is true. 
Can anyone quicker than I tell me why? I know it's probably very obvious.  
Note: This isn't a homework question, so if the answer is obvious I'd really just appreciate knowing why. 
Thanks!

REPLY [4 votes]: In some situations, to prove that groups of order $n$ cannot be simple, you can use the counting argument if all Sylow subgroups have trivial intersection, and a different argument otherwise.
For example let $G$ be a simple group of order $n=144 = 16 \times 9$. The number $n_3$ of Sylow 3-subgroups is 1, 4 or 16. If $n_3 = 1$ then there is normal Sylow subgroup and if $n_3= 4$ then $G$ maps nontrivially to $S_4$, so we must have $n_3 = 16$.
If all pairs of Sylow 3-subgroups have trivial intersection, then they contain in total $16 \times 8$ non-identity elements, so the remaining 16 elements must form a unique and hence normal Sylow 2-subgroup of $G$.
Otherwise two Sylow 3-subgroups intersect in a subgroup $T$ of order 3. Then the normalizer $N_G(T)$ of $T$ in $G$ contains both of these Sylow 3-subgroups, so by Sylow's theorem it has at least 4 Sylow 3-subgroups, and hence has order at least 36, so $|G:N_G(T)| \le 4$ and $G$ cannot be simple.