There is no simple group of order $144$ 
There is no simple group of order $144$

I have a question to the proof of the statement above (from the book J. Gallian, Contemporary abstract algebra), it is about the index theorem, so I give first the theorem

The Index Theorem:
If $G$ is a finite group and $H$ is a proper subgroup of $G$ such that $|G|$ does not divide $|G:H|!$, then $H$ contains a nontrivial normal subgroup of G. In particular, $G$ is not simple.

$\bullet 3$rd line: How does the index theorem eliminate the case $n_3=4$ ?

 A: If $n_3=4$, then a Sylow 3-subgroup $H$ would have a normalizer $N(H)$ with index 4 (the number of Sylow $p$-subgroups is equal to the index of a normalizer of a $p$-subgroup). But you cannot have $N(H)$ with index 4 since $|G|=144=2^4\cdot 3^2$ does not divide $[G:N(H)]!=4!=2^3\cdot 3$ (the index theorem would then imply that $G$ has a proper nontrivial normal subgroup contradicting the simplicity assumption).
Again if $[G:N(H \cap H')] \leq 4$, then $|G|$ does not divide $[G:N(H \cap H')]!$ ($=4!=24,3!=6,2!=2,$ or $1!=1$) so $G$ has a proper nontrivial normal subgroup (contradiction).
Addendum: Let $n_p$ be the number of Sylow $p$-subgroups. Let $H$ be a Sylow $p$-subgroup. I claim that $n_p=[G:N(H)]$ (the number of Sylow $p$-subgroups matches the index of the normalizer of a $p$-subgroup).
This follows immediately from the "orbit-stablizer" theorem. Notice that $G$ acts on the set of Sylow $p$-subgroups via conjugation: $g \cdot H = gHg^{-1}$ is a Sylow $p$-subgroup for any $g\in G$. Why? Because conjugation preserves orders so a conjugate subgroup of a Sylow $p$-subgroup is still a Sylow $p$-subgroup. 
Next, all Sylow $p$-subgroups lie in a single orbit. This is commonly referred to as the second  Sylow theorem (any two Sylow $p$-subgroups are conjugate). Thus the orbit of $H$ (under this action) is the set of all Sylow $p$-subgroups. So the size of the orbit of $H$ is $n_p$.
Notice that  $xHx^{-1}=H$ if and only if $x \in N(H)$. Thus the stabilizer of $H$ is exactly $N(H)$. 
Finally,  the orbit-stabilizer theorem says that the size of an orbit times the size of a stabilizer is equal to the size of the group. Thus $|G|=n_p \cdot |N(H)|$. But Lagrange's theorem says $|G|=[G:N(H)] \cdot |N(H)|$. Thus $n_p=[G:N(H)]$.
A: Let $P$ denote a 3-Sylow subgroup of $G.$ If $n_3 = 4,$ then $|G|$ does not divide 4!. So by Index theorem, $P$ contains a non-trivial normal subgroup of $G.$ This will force that $P$ is normal in $G,$ contradicting the fact that $[G : N_G(P)] = 4.$
