Show that a group of order $pq$ has subgroups of orders $p$ and $q$ without using Sylow’s and Cauchy’s theorem 
If $o(G)$ is $pq$, $p>q$ are primes, prove that $G$ has a subgroup of order $p$ and a subgroup of order $q$.

This question is from Herstein and it comes before Sylow’s and Cauchy’s theorem. So I’m expecting an answer without using any of these.
Here’s what I got so far:
If $G$ is cyclic, then we are done, otherwise we can assume that it is not cyclic, which means every non-identity element must be of order $p$ or $q$.
Case (1): if there exists $a \in G$ such that $o(a) = p$ and if there also exists an element of order $q$, then we are done. So we can assume that every non-identity element is of order $p$. Now pick $b \in G$ such that $b \notin \langle a \rangle$ then $o(b) = p$ and $\langle a \rangle \cap \langle b \rangle = \langle e \rangle$.
So we have $\langle a \rangle \langle b \rangle \subset G$ but $o(\langle a \rangle \langle b \rangle) = \dfrac {o(\langle a \rangle) o(\langle b \rangle)}{o(\langle a \rangle \cap \langle b \rangle)} = p^2$ but $p^2 > pq$ [since $p>q$] so we got a contradiction.
Give me a hint for the second case and correct me if my argument for the first case is wrong.
 A: Assume that every non-identity element generates a cyclic group of order $q$, the smaller of the primes.
Conjugacy is an equivalence relation on a group. So, we should be able to partition the group into its equivalence classes. The size of the equivalence class an element belongs to is the index of the centralizer of the element. Why? Fix $x\in G$. Make a homomorphism from $G \rightarrow G$ by sending $g \rightarrow xgx^{-1}$. The size of the equivalence class is the order of the image. What is the kernel of this map?
If the centralizer is of order $p$ or $pq$, we are done. Assume every centralizer is of order $q$, the index of the centralizer is $pq/q=p$. Every element would belong in a equivalence class of size $p$, except for the identity element.
A simple cardinality calculation shows that $pq= kp+1$, where represents the number of equivalence classes. However, this is absurd and therefore, not every subgroup is of order $q$.
A: Firstly, if all the non-trivial elements of $G$ have order $pq$, then, for any given $g\in G\setminus\{e\}$, we get $e=g^{pq}=(g^p)^q$ and hence $o(g^p)\mid q$; but $g^p\in G\setminus\{e\}$, so, by assumption, $o(g^p)=pq$ and we end up with $pq\mid q$: contradiction. Therefore, there must be elements of order $p$ and/or elements of order $q$. If $G$ is Abelian, then your argument works for the "second case" either, because then every subgroup is normal and hence $\langle a\rangle\langle b\rangle$ is a subgroup of $G$ such that $o(\langle a \rangle \langle b \rangle) = \dfrac {o(\langle a \rangle)o(\langle b \rangle)}{o(\langle a\rangle \cap \langle b\rangle)} = q^2\nmid pq$: contradiction (Lagrange). So, there must be elements of both orders, $p$ and $q$. If $G$ is nonabelian, then it has trivial center (see here) and the Class Equation reads:
$$pq=1+k_pp+k_qq$$
where $k_i$ are the number of conjugacy classes of size $i$. None of the $k_i$ can be zero, because neither $p\nmid 1$ nor $q\nmid 1$. Say $O_p$ a conjugacy class of size $p$ and $O_q$ a conjugacy class of size $q$, which as said both exist; for $g\in O_p$, we get $|C_G(g)|=q$ (Orbit-Stabilizer Theorem) and hence $\langle g\rangle=C_G(g)$ (because in general $\langle g\rangle\le C_G(g)$) and finally $o(g)=q$. Likewise, for $h\in O_q$, we get $|C_G(h)|=p$ and hence $\langle h\rangle=C_G(h)$ and $o(h)=p$.
A: Case $G$ abelian. If $G$ has all the nontrivial elements of one same order, say $p$, then there are $k=\frac{pq-1}{p-1}\ge2$ distinct subgroups of order $p$. Call $H$ and $K$ two of them. The set $HK$ is a subgroup of $G$ of order $p^2$: contradiction, because $p\nmid q$. Same conclusion if $G$ has all the nontrivial elements of order $q$, as $q\nmid p$. So, $G$ has elements of both orders $p$ and $q$.
Case $G$ nonabelian. Then $pq=1+kp+mq$ is the only possible class equation, for some positive integers $l$ and $m$, and hence $G$ has subgroups (centralizers, actually) of both orders $p$ and $q$.
