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Without using Cauchy's or Sylow theorems, can we prove that every group of order $65$ is cyclic? Please help, thanks in advance (any technique of group homomorphisms and normal subgroups can be used).

pjs36
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1 Answers1

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Well known theorem that

groups of order $pq$ (primes, $p<q$) are cyclic except when $p | (q-1)$.

I think this is from looking at the conjugation action of a cyclic $p$ subgroup on the whole group. Or follow the chain of duplicates...

Proving $|G|=pq$ and $p>q$ , $q$ does not divide $p-1$ $\implies$ $G$ is cyclic , without using Cauchy's and Sylow's theorems