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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?

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  • $\begingroup$ You need just the first line. $\endgroup$ – the_fox Mar 14 at 8:24
  • $\begingroup$ 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. $\endgroup$ – Derek Holt Mar 14 at 8:38
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Hint: use Lagrange's Theorem, the order of a subgroup divides the order of the whole group. And, a group of prime order must be cyclic (also follows from Lagrange).

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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.

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