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 $q-1$.

If $|Z(G)|=p$, then the class equation yields: $pq=p+kq$, a contradiction because $q\nmid p$.
If $|Z(G)|=q$, then the class equation yields: $pq=q+lp$, again a contradiction because $p\nmid q$.
If $|Z(G)|=1$, then the class equation yields:
$$pq=1+mq+np \tag1$$
So, there are $np$ elements of order $q$, grouped in $n'$ subgroups of order $q$, and then:
$$np=n'(q-1) \tag2$$
Since by assumption $p\nmid q-1$, from $(2)$ follows $p\mid n'$, say $n'=n''p$, whence:
$$n=n''(q-1) \tag3$$
which plugged in $(1)$ gives: $$pq=1+mq+n''p(q-1)\tag4$$
But, from $(1)$:
$$1+mq=m'p \tag5$$
which plugged in $(4)$ yields:
$$q=m'+n''(q-1) \tag6$$ which implies $m'=n''=1$ and then, from $(5)$:
$$1+mq=p \tag7$$
a contradiction, because $q>p$.
Therefore $Z(G)=G$, namely $G$ is abelian, and hence (Cauchy) cyclic.
