If $G$ an abelian group with order $pq$, $(p,q) = 1$ and $a,b\in G$ with orders $p$ and $q$, is $\langle a\rangle\cap\langle b\rangle = \{e\}$?

Question

I need help with the following question.

The exercise says:

Let $G$ an abelian group with order $pq$, $(p,q) = 1$. Let $a,b \in G$ with orders $p$ and $q$ respectively. Prove that $G$ is cyclic.

(This exercise was asked in the forum, but my doubt is oriented elsewhere)

If I take $a$ and $b$ and I consider the cyclic group $\langle ab\rangle$, it is clear that $(ab)^{pq} = e$ Now, not necessarily $pq$ is the order of $ab$, then suppose that $\alpha$ is the order of $ab$. Then $(ab)^{\alpha} = e $ $\rightarrow$ $a^{\alpha}b^{\alpha} = e$ $\rightarrow$ $a^{\alpha} = b^{-\alpha}$. I would like to prove that isn't possible.

So, with the given hypothesis, hoe do I prove that $ \langle a\rangle \cap \langle b\rangle= \langle e\rangle$?

For example: Suppose that exist $g \in \langle a\rangle \cap \langle b\rangle$, $g \neq e$. Then, $g = a^i$ and $g = b^j$ with and $i < p$ or $i = p\lambda + r$ and $j<q$ or $j = p\lambda' + r'$ Doing this I don't see a clear answer.

I was able to solve the exercise in another way, but I am interested to see how to demonstrate this. I would like to receive ideas on how I could do it.

Answer 1

Note that $\langle a\rangle\cap\langle b\rangle$ is a subgroup of $\langle a\rangle$, so by Lagrange's Theorem $|\langle a\rangle\cap\langle b\rangle|\mid p=|\langle a\rangle|$. Also, $\langle a\rangle\cap\langle b\rangle$ is a subgroup of $\langle b\rangle$, so by Lagrange's Theorem we know that $|\langle a\rangle\cap\langle b\rangle|\;\mid \;q=|\langle b\rangle|$. Because $(p,q)=1$ we will conclude that $|\langle a\rangle\cap\langle b\rangle|=1$, which implies that $\langle a\rangle\cap\langle b\rangle=\{e\}$.

Answer 2

I'll try to answer using the example you construct:

You have $g=a^i=b^j$ where $i<p$ and $j<q$. Since $g\neq e$, it follows that $i$ and $j$ are nonzero. First, we have $b^{jp}=(b^j)^p=(a^i)^p=a^{ip}=e$ since $ip$ is a multiple of $p$. Since the order of $b$ is $q$, we must have $q$ divides $jp$. From $(p,q)=1$, it follows that $q$ does not divide $p$ and hence divides $j$ by Euclid's Lemma and therefore $j\geq q$. This contradicts the assumption that $j<q$.