Abelian group of order pq, two different prime numbers

Question

$G$ is an abelian group of order $pq$, two different prime numbers. Need to prove that there is an element of order $p$ and of order $q$.

The proof that I found goes like this:

By Lagrange, order of an element in finite group divides the order of the group. So it can be, then it is id. So let take an element which is not the id. So it's order can be $p$,$q$ or $pq$. If it is $pq$, then it is easy to show an element from each order $p$ and $q$. Ok, so let's take and element of an order that isn't pq. Then it is $p$ or $q$. Let it be $q$. Now we will show that there is an element from order $p$.

Then let us assume that there isn't such element. So for each other element it's order is $q$ (because if it pq it's fine and id there is only one). $q$ is prime, then the cyclic sub-group that he creates is from order $q$, but the id is already counted, so each such element generates $q-1$ elements. $$$$ But that can't be because $q-1$ doesn't divides $q(p-1)$.

Can somebody explain to me if there is any reason that $q-1$ doesn't divide $p-1$ for two different prime numbers?

And secondly,

Is there any more algebraic way to prove this ?

Answer 1

As you saw in the comments, your conjecture is false. Here is a way you can prove your result, just off the top of my head, without even assuming $G$ is abelian. There probably exists a cleaner proof somewhere.

Say you have an element $a$ of order $q$, generating a subgroup $C_q\subset G$. Now by Lagrange's theorem $C_q$ is the unique subgroup of order $q$ in $G$, hence it is normal. To see this, note that for any $g\in G$, the conjugate subgroup $gC_qg^{-1}$ also has order $q$, hence $gC_q g^{-1}=C_q$ for all $g\in G$. It follows that the quotient group $G/C_q$ is cyclic of order $p$. Choose a generator $b$ of $G/C_p$ and consider once of its preimages $b'$ in $G$. Note that $b'$ is not in $C_q$. Then the image of $b'^p$ will be in $C_q$, hence $b'$ has order either $p$ or $pq$. If it is the latter, $b'^q$ has order $p$.

Answer 2

The only divisors of $pq$ are $1$, $p$, $q$, and $pq$. If $z$ is an element of order $pq$, then $z^p$ is of order $q$ and vice versa. So we may assume that you’ve taken an element $g$ of $G$ that’s of order $p$. It generates a subgroup of order only $p$, so there are other elements. If your next element is of order $pq$, you’re done, and if that next element is of order $q$, you’re also done. So the only remaining possibility is that this second element $h$ is of order $p$ as well, not in $\langle g\rangle$. But then the $p^2$ elements $g^ih^j$ for $0\le i,j<p$ would all be distinct, giving $p^2|pq$, a falsity. So this last possibility doesn’t occur.