I am trying to use the Sylow theorems to prove the following.
Let $G$ be a group of order $2pq$, where $p>q>2$ are primes. Show that $G$ is solvable.
Now, I realize that If I can show either $n_p=1$ or $n_q=1$, then I can find a subgroup of order $pq$ which is normal (index 2).
Now, I use Sylow to show that either $n_q = 1$ or $n_q=2p$.
So we assume $n_q=2p$. We also learn from the Sylow theorems that $n_q=2p\equiv 1\mod q$
Now, my idea is to somehow use this to show $n_p=1$, but I am not seeing how to do this.
I realize that this question has been posted before, namely here:Suppose $|G| = 2pq$. Does $G$ have a subgroup of order $pq$?
but I cannot understand this final step involving the modular arithmetic.
Thanks in advance!
Best Answer
We will prove more generarily the following result.
Lemma: Let $G$ be a group with $2n$ elements. If $n$ is odd, then $G$ has a subgroup of order $n$.
Proof: This is a standard result and proof, one of the simplest and cutest application of Cayley theorem.
Recall the proof of the Cayley theorem:
For each $a \in G$ define $\phi_a : G \to G$ via $\phi_a(x)=ax$.
The mapping $a \to \phi_a$ defines an isomorphism between $G$ and the subgorup $$H:= \{ \phi_a : a \in G \} \leq S_{2n}$$ of $S_{2n}$.
By standard group theory, $G$ has an element $b$ of order $2$ [this follows from Cauchy's theorem, but it has a very standard proof using only elementary means].
Now, look at $\phi_b$. This needs to be an element on $S_{2n}$ of order $2$, thus it is the product of pairwise disjoint two-cycles. Since $\phi_b: G \to G$ has no fixed points, $\phi_b$ must be the product of exactly $n$ two-cycles. Finally, as $n$ is odd, it follows that $\phi_b$ is the product of an odd number of two-cycles, and hence an odd permutation.
The claim follows now immediately:
Consider the sign $$\epsilon : H \to \{ \pm 1 \}$$ This is an onto (here is why we needed to argue that $H$ has an odd permutation) and hence by the first Isomorphism theorem, $\ker(\epsilon) = H \cap A_{2n}$ is a subgroup of $H$ of index 2.
Since $G$ is isomorphic to $H$ the claim follows.