A group $G$ of order $pq$ (for $p < q$), $p$ and $q$ are prime numbers. Prove: $G$ cannot have two different subgroups of order $q$. 
A group $G$ of order $pq$ (for $p < q$), $p$ and $q$ are prime numbers. Prove: $G$ cannot have two different subgroups of order $q$.

I know that the proof is obvious using Sylow theorem. But I haven't learned it and this question is in the exercise before Sylow theorem. Can anyone give me a solution without using Sylow theorem?
 A: Hint: A subgroup of index equal to least prime divisor of the group order is normal.
A: Assume that $H$ and $K$ are two different subgroups of $G$ of order $q$. It follows from Lagrange's Theorem that $H \cap K=1$. Hence $$pq= |G| \geq |HK|=\frac{|H| \cdot |K|}{|H \cap K|}=q^2$$ and this yields $p \geq q$, a contradiction.
A: Suppose there are two different subgroups of order $q$. For $q$ is a prime number, so this two groups are cyclic groups. We can denote these two subgroups as $<a>$ and $<b>$ ($a$ $\neq$ $b$). It is obvious that $<a>\cap<b>$ is a normal subgroup of $<a>$ and $<b>$. For $<a>$ and $<b>$ are single groups, $<a>\cap<b>=\{e\}$.
Let's see the set $<a><b>$. Suppose $x$ and $y$ are two elements of $<a><b>$. $x$ and $y$ can be expressed as $a^n$$b^m$ and $a^s$$b^t$ (n, m, s, t are integers between $0$ and $q-1$). If x = y, then $a^{n-s}$ = $b^{m-t}$. For $<a>\cap<b>=\{e\}$, then $a^{n-s}$ = $b^{m-t}$ = $\{e\}$. It implies that $n = s$ and $m = t$. 
So there are $q^2$ elements in the $<a><b>$ $\subset$ $G$. But G has only $pq$ elements.
