For p,q primes such that q|p−1, show there is only one non-abelian group of order pq, up to isomorphism

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( 4 months ago )

This question has been answered a lot of times on this site, but I'm looking for an approach that does not use Sylow theory, since this is not covered in my syllabus. All answers I read this far used material that I did not yet learn. My level this far is up to automorphisms, group actions, and the isomorphism theorems.

My syllabus uses the following construction of a non-abelian group of order pqpq where q|p1q|p−1. Let N=CpN=Cp such that Aut(N)Aut⁡(N) has order p1p−1. From Cauchy's theorem we deduce that there exists a subgroup HAut(N)H⊂Aut(N) of order qq. Let τ:HAut(N)τ:H→Aut⁡(N) be the identity map. Then NτHN⋊τH has order pqpq and is non-abelian.

This far I can follow, but now I have to show that this group is the only non-abelian group of order pqpq. A hint for this exercise is to use that (Z/pZ)(Z/pZ)∗ is cyclic if pp is prime.

My attempt (it is not really an attempt, I just looked what I could deduce, but it led me nowhere): Let

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