# prove that for $G$ a multiplicative nonabelian group of order $pq$, where $p$ and $q$ are prime numbers, any proper subgroup of $G$ is abelian

I need to prove that for $$G$$ a multiplicative nonabelian group of order $$pq$$, where $$p$$ and $$q$$ are prime numbers, any proper subgroup of $$G$$ is abelian. I use that, from Lagrange's theorem, the order of the proper subgroups must divide the order of the group and so the subgroups have the orders $$p,q$$. If $$(p,q)=1$$, then there exist $$k,l\in\mathbb{Z}$$ such that $$pk+gl=1$$ and so, for an $$x$$ of a subgroup of order $$p$$ we obtain $$x=x^{lq}$$ and similarly, for an $$y$$ of order $$q$$ $$y=y^{pk}$$. Also, $$p|lq-1$$ and $$q|pk-1$$. This are the only valuable information I obtained. How can I prove this?

• You need just the first line. – the_fox Mar 14 '19 at 8:24
• Try to avoid unnecessary assumptions. You don't need to assume that the group $G$ is multiplicative, and you don't need to assume that it is nonabelian. – Derek Holt Mar 14 '19 at 8:38

By Lagrange's theorem we can say that order of any proper subgroup of $$G$$ is either of order $$p$$ or of order $$q.$$ Since $$p$$ and $$q$$ are primes and we know that any group of prime order is cyclic so any proper subgroup of the group $$G$$ is cyclic and hence abelian.