Group of order $24$ is not simple How can we show that a group of order $24$ is not simple? This is my attempt:
$|G| = 24 = 3\cdot 2^3$ so $n_G(2) \in \{1,3\}$ where $n_G(p)$ donotes the number of Sylow $p$-groups. If $n_G(2) = 1$ we are done. Otherwise consider the action (by conjugation) of $G$ on Syl$_2(G)$ the Sylow 2-groups. This action is transitive since all Sylow $2-$groups are conjugated. This also means that the normalisator of each Sylow $2$-group has order 8. Now define $$\phi:G \rightarrow Sym(Syl_2(G)): g \mapsto \theta_g$$ with
$$
\theta_g(P) := g(P) = gPg^{-1}
$$ This is a homomorphism since we have a well defined action here.  But for $P \in Syl_2(G)$ there are 8 elements which fix $P$ and these elements are contained in $\ker \phi \lhd G$ so $G$ contains a non-trivial normal subgroup and thus $G$ is not simple. 
Since $|im(\phi)| \leq 6$ we have $| \ker \phi | \geq 8$ so $G$ is not simple.
Is everything correct here? 

Let $|G|=2^n \cdot 3, \ \ n \geqslant 2$ and $H$ Sylow 2-subgroup of $G$. Then $|G:H|=3$, so $|G:\operatorname{Core}(H)| | 3!=6$. So, $\operatorname{Core}(H)$ is non-trivial group, because $4| |G|$ and $4 \nmid6.$ $\operatorname{Core}(H)$ is normal, and she is not whole group because $\operatorname{Core}(H) \le H \lneq G$. So, $G$ is not simple group.
