I'm trying to understand why a group $G$ of order $pq$ is solvable, for $p,q \in \mathbb{N}. p\neq q$ prime.
Proof: Let $G$ act on its subsets by conjugation: $$\cdot: G \times \mathbb{P}(G)\longrightarrow \mathbb{P}(G)$$ $$(g,A) \longmapsto g \cdot A=\{gag^{-1}:g\in G \}$$ Then clearly $N_G(H)=Stab_G(H)$ for all subgroups $H<G$. Let $H_p$ be a $p$ Sylow-subgroup. If $N(H_p)=G$, then we are done. Else $N(H_p)=H_p$ has index $q$ in $G$. Now the part that confuses me. By the orbit stabilizer theorem: $G/Stab_G(H_p)\cong G \cdot H_p$. Since $Stab_G(H_p)=N_G(H_p)=H_p$ this means $|G \cdot H_p|=|G/H_p|=q$. Hence the amount of $p$ Sylow-subgroups is $|Syl_p(G)|=q $ Further are these subgroups conjugated to each other and have trivial intersection (since $H_p$ is cyclic).
On the other hand, there is the third Sylow-theorem. I guess its proof uses a similar argument so I wish to fully understand it. Applying this theorem to $G$ we get: $|Syl_p(G)| \equiv 1 \, mod \, p$ which to me looks like a contradiction.
I'd be really thankful if someone could enlighten me.