Let G be an abelian group of order $pq$ with $(p,q)=1$. If G contains elements $a$ and $b$ of order $p$ and $q$ respectively, show that G is cyclic.

Question

I have found a proof:

Let $x=ab$ with $|x|=n$. If we can show that $n=pq$ then we will have a generator for G. Since G is abelian, $x^{pq} = (ab)^{pq} = (a^p)^q (b^q)^p=e$. This only tells us that $n|pq$. Now, $a^n = a^n b^n b^{-n}=(ab)^n(b^{-1})^n=(b^{-1})^n$. We also know that $|b^{-1}| = |b|$.

Now, $|a^n| = \frac{p}{gcd(p,n)}$ (why is this always true? I understand that this is valid when $a$ is a generator of G, but $a$ is not a generator in this case). Also, $|b^{-n}|=|b^n|=\frac{q}{gcd(q,n)}=|a^n|$ and therefore $p\cdot gcd(q,n) = q \cdot gcd(p,n) $ and so $p | gcd(p,n)$ because $gcd(p,q)=1$. Thus $p|n$ (again, why is this true? We haven't assumed $p$ is prime, so if $p|gcd(p,n)=\alpha p + \beta n$ and therefore $p|\beta n$ then it is not necessary that $p|n$). Similarly, if $q|gcd(q,n)\Rightarrow q|n$ and because $gcd(p,q)=1$ then $pq|n$ and so $n=pq$. Thus $x$ is a generator of G.

Answer 1

First we prove that $\langle a\rangle\cap\langle b\rangle={e}.$ If $x=a^{m}=b^{n}$, then $\mathrm{ord}(x)|gcd(p,q)=1,$ thus $x=e.$

Then we show that $G$ is a cyclic group generated by $ab$. Let $\mathrm{ord}(ab)=t.$ We have $e=(ab)^{t}=a^{t}b^{t},$ hence $a^{t}=b^{-t}.$ However $\langle a\rangle\cap\langle b\rangle={e},$ $a^{t}=b^{-t}=e.$ $p|t,q|t\Rightarrow pq=t.$ Clearly,$(ab)^{pq}=e,$ so $G$ is cyclic whose generator is $ab$.

Answer 2

$|a^n| = \frac{p}{gcd(p,n)}$ is true by the proposition (Dummit and Foote page 57):

Proposition. Let $G$ be a group, let $x\in G$ and let $a\in \mathbb{Z}-\{0\}$. If $|x|=n<\infty$, then $|x^{a}|=\frac{n}{(n,a)}$.

$p|n$ is true because $p|(p,n)$ and $(p,n)|n$.