Proof easy version of Burnside's Lemma

Question

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.

Answer 1

Actually in this easy case, no Sylow theorems are necessary. Say $p < q$, and let $H$ be the $q$-Sylow. Since $p$ is the smallest prime dividing the order of $G$, and $[G:H]=p$, it is now a standard argument that $H$ is normal.

To spell it out, $G$ acts on $G/H$ on the left, and so there is a homomorphism $\phi : G\to S_p$, where $S_p$ is the symmetric group, and $\ker \phi\le \text{Stab}(H) = H$. Then $\ker \phi$ is either trivial or equal to $H$. But $\phi$ is not injective since $pq$ does not divide $p!$, so $\ker \phi = H$. This means that $H$ acts trivially on $G/H$, so for all $g\in G$ we have $HgH = gH \implies Hg=gH$. This shows that $H$ is normal, so you're done.

As far as your Sylow theorems argument goes, it's basically correct as far as I can tell. One way to finish would be to WLOG assume $q < p$. Then $q\equiv 1\mod p$ is indeed a contradiction. I was a bit unclear what you were asking for here; if you want to understand the content/proof of the third Sylow theorem, any book on abstract algebra would have that information (e.g. Dummit and Foote).

Another way to finish would be to observe that if there are $q$ of the $p$-Sylow subgroups, these account for $pq-q+1$ elements of $G$, meaning there is a unique $q$-Sylow accounting for the other $q-1$ elements. This $q$-Sylow must be normal in $G$.

It is important to realize that there is only necessarily a normal subgroup of the larger prime order. For example in a dihedral group of order $2q$, a subgroup of order $2$ generated by a reflection is not normal.