I was reading the following proof for that question (Joanpemos' answer)-
How to prove a group of order $144$ is not simple using Normalizers of Sylow intersections.
And I understood it well up to the part where he says that the intersection of the sylow subgroups $P,Q$ is of size $3$.
Why is that?
Also, why the fact that these groups are abelian, means that the normalizer has at least $6+9$ elements?
Here's a completely elementary solution. By Sylow there are 1, 4, or 16 3-Sylow subgroups. It can't be 1 (then the 3-Sylow would be normal), and it can't be 4 (then there would be a non-trivial homomorphism to $S_4$ given by the conjugation action on the 4 3-Sylows, and the kernel would be a non-trivial normal subgroup). So there are 16 3-Sylows.
If any two of them intersect in a subgroup $H$ of size 3, then the centralizer $C$ of $H$ contains both of those 3-Sylows, and thus, by Sylow, either 4 or 16 3-Sylows. If $C$ contains 16 3-Sylows, it's the whole group, and $H$, being central is a non-trivial normal subgroup. Thus, $C$ contains 4 3-Sylows. But then $36\le |C|<144$, so $|C|$ is 72 or 36, and there is a non-trivial homomorphism from $G$ to $S_2$ or $S_4$, given by the action of $G$ on cosets of $C$. The kernel is a non-trivial normal subgroup.
Thus, all the 3-Sylows have trivial pairwise intersections. That means there are $8\cdot16=128$ elements of order 3 or 9. The remaining 16 elements thus must be the unique 2-Sylow, which is therefore normal. Contradiction.