My knowledge on group theory is limited only up to the Lagrange Theorem, and it seems that the proofs I found here use more advanced theorems. What are the facts that I should know to prove this one?
I am trying to come up with an element $c \in G$ such that $\langle c\rangle = G$. I'm thinking of letting $c=ab$. Do you think this will lead to the desired result?
On
I'll try to sketch a simple proof. Let $c=ab$ and consider the product $cc...c=abab...ab$. Since $G$ is abelian, $ab=ba$ and so $abab...ab=a^nb^m$.
Let $|ab|=r$. Then $(ab)^r=a^r b^r=1$, hence $a^r=b^{-r}$. So $a^r$ is an inverse for $b^r$ and hence $a^r∈〈b〉$. But by the Lagrange’s Theorem $〈a〉 ⋂ 〈b〉=1$, what forces $a^r=1$ and hence $|a|=p \ |\ r$. Analogously from $a^r b^r=1$ follows $a^{-r}=b^r$, so $b^r$ is an inverse for $a^r$, hence $b^r∈〈a〉$ and since $〈a〉 ⋂ 〈b〉=1$ it must be $b^r=1$ and hence $|b|=q \ | \ r$. Thus $r$ is divisible by both $p,q$ and since we’re looking for minimal such $r$ (as an order of $ab$), it follows that $r=lcm(p,q)$. Since $gcd(p,q)=1$, we have $lcm(p,q)=pq$.
Thus $|ab|=pq$ and since $|G|=pq$, it follows that $ab$ is a generator of $G$. So, $G=〈ab〉=〈c〉$ is cyclic.
First, we show $\langle a\rangle\cap \langle b\rangle=\{ e\}$. Assume this is not true, so exsits $x\neq e, x\in \langle a\rangle\cap \langle b\rangle$. Let $d=|\langle x\rangle|\Rightarrow d\ge 2$. Since $\langle x\rangle\le \langle a\rangle$ and $\langle x\rangle\le \langle b\rangle$, then we have $d|p$ and $d|q$, hence $1=\gcd(p, q)\ge d\ge2$, contradiction.
Next, it is easy to check $c=ab, c^{pq}=e$, so we need to show $|\langle c\rangle|=pq$. Assume, this is not true, namely, exsits $k<pq$, such that $c^k=e$. Then we get
$$a^k=b^{-k}=y\Rightarrow y\in \langle a\rangle\cap \langle b\rangle\Rightarrow y=e $$
$$\Rightarrow a^k=b^k=e\Rightarrow p|k, q|k$$
Since $\gcd(p, q)=1$, we get $$pq|k\Rightarrow k\ge pq$$
Contradiction.