Finding the structure of a group without using sylows theorem. 
If$ |G|=pq $ and $p $ doesnt divide $(q-1)$ and $p <q$ then $G$ is cyclic.

My proof stands as this .I have used the fact that
$(i)$ If $|G|=pq$ then I showed that there will only be one element of order $p$ and one element of order $q$.
My approach in proving this part has been to show that if there are two elements of order $p$ say $x_1$ and $x_2$ , then say $H_1$ is a group of order $p$ generated by $x_1$ and $H_2$ is a group of order $p$ generated by $x_2$ .We assume that the intersection is {e} if not then we can get $H_1$=$H_2$(by property of subgroup).
Proving that the elements are distinct .
We assume that the elements $(x_1)^{i}.(x_2)^j$ are not distinct then $(x_1)^{i}.(x_2)^j =(x_1)^{i'}(x_2)^{j'}$.From here we can arrive at a contradiction as $H_1 \cap H_2 =e$.So if there are $p^2$ elements then we can arrive at a contradiction as $p$ and $q$ are both primes.
Similar results will hold in the case of $q$.
$(ii)$ Now  there is only one subgroup of order $p$ and one subgroup of order $q$ so they are both normal
$(iii)$ let $H$ and $K$ be two subgroups of order $p$ and order $q$.Then we know that $H \cap K={e}$.$H$ and $K$ are both normal .Then I showed that $x^{-1}y^{-1}xy \in H \cap K$ and $xy=yx$. So the order of the element $xy$ is $pq$.Where am I going wrong in my proof and since I have not used the fact that $p $ doesnot $q-1$.

An argument close to the OP's idea could proceed as follows. Fill in the details.

*

*If $x$ and $y$ are elements of order $q$, then among the products $x^iy^j$, $0\le i,j<q$, there must be repetitions. This is because $q^2>|G|$. Show that this implies that the subgroup $H$ of order $q$ is unique. Let's fix a generator $x$ of $H$.

*If $z\notin H$ has order $pq$ then $G$ is cyclic. Therefore the remaining possibility is that all such elements $z$ have order $p$.

*Because $H$ is a unique subgroup of its order, $H\unlhd G$. Why does it follow that $zxz^{-1}=x^i$ for some $i, 1\le i<q$?

*Why do we have $z^pxz^{-p}=x$?

*On the other hand we also have $z^pxz^{-p}=x^{i^p}$, why? Why does this imply the congruence $$i^p\equiv1\pmod q?$$

*It follows that the coset of $i$ in the multiplicative group $\Bbb{Z}_q^*$ has either order $1$ or order $p$. Why?

*If the order of the coset of $i$ is equal to one, then $zx$ has order $pq$. Why?

*If the order of the coset of $i$ is equal to $p$, why does it follow that $p\mid q-1$?

