If $P \cap Q = \{1\}$ and $P, Q \triangleleft G$, then $G \cong P \times Q$

Question

I'm trying to prove that given $G$ a group of order $pq$, with $p,q$ primes such that $q \nmid p - 1$ and $p > q$, then $G$ is cyclic. For that, Cauchy's theorem guarantees us that there are $P, Q \leq G$ with $P = \langle a \rangle$ and $Q = \langle b \rangle$ where $o(a) = p$ and $o(b) = q$. To finish the proof, $\textrm{Sylow-III}$ says that $P, Q \triangleleft G$ and $P \cap Q = \{1\}$ so I just have to prove that $G \cong P \times Q$. Any ideas on how to solve this? My first thoughts were to find a morfism between $G$ and $P \times Q$ which had $\ker = P \cap Q$, but this led me to nothing.

Answer 1

Your subgroups $P,Q$ satisfy the conditions $P,Q\trianglelefteq G$, $P\cap Q=\{e\}$ and $G=PQ$. (the last condition holds because $P,Q\leq PQ$, so by Lagrange's theorem $|PQ|$ is divisible by both $p$ and $q$, hence by $pq$ as well. So we conclude that $PQ$ must be equal to the whole group $G$)

Ok, so now define an isomorphism $\varphi: P\times Q\to G$ by $\varphi(p,q)=pq$. It is clearly surjective, because $G=PQ$. Now we will show that it is also injective. Suppose $\varphi(p_1,q_1)=\varphi(p_2,q_2)$. Then $p_1q_1=p_2q_2$ and hence $p_2^{-1}p_1=q_2q_1^{-1}$. Note that the right hand side belongs to $Q$ and the left hand side belongs to $P$, so we conclude that the element $p_2^{-1}p_1=q_2q_1^{-1}$ belongs to $P\cap Q=\{e\}$, which implies $p_1=p_2$ and $q_1=q_2$. This proves $\varphi$ is injective.

Now to finish the proof we just have to show $\varphi$ is a homomorphism of groups. Note that for $p_1,p_2\in P$ and $q_1,q_2\in Q$ we have:

$\varphi((p_1,q_1)(p_2,q_2))=\varphi(p_1p_2,q_1q_2)=p_1p_2q_1q_2$

$\varphi(p_1,q_1)\varphi(p_2,q_2)=p_1q_1p_2q_2$

So if we will show that every element of $P$ commutes with every element of $Q$ then it will give us the equality $\varphi((p_1,q_1)(p_2,q_2))=\varphi(p_1,q_1)\varphi(p_2,q_2)$ which is exactly what we want. So now we are going to show that elements of $P$ commute with elements of $Q$. Let $p\in P$ and $q\in Q$. We want to show that $pq=qp$, or equivalently that $pqp^{-1}q^{-1}=e$. Now, since $Q$ is normal we have:

$pqp^{-1}q^{-1}=(pqp^{-1})q^{-1}\in Q$

And similarly, since $P$ is normal:

$pqp^{-1}q^{-1}=p(qp^{-1}q^{-1})\in P$

So indeed $pqp^{-1}q^{-1}\in P\cap Q=\{e\}$, which implies $pq=qp$.