Question #144684
Assume that (G,*) is a group, and that |G|=p, where p is prime. Prove that G is cyclic.
1
Expert's answer
2020-11-17T07:39:33-0500

Let that (G,)(G,*) be a group of orderG=p|G|=p , where pp is prime. Let aGa\in G and aea\ne e where ee is the identity element of (G,)(G,*). According to Lagrange's theorem, the order of cyclic subgroup a\langle a\rangle divides G=p|G|=p. Therefore, a{1,p}|\langle a\rangle|\in\{1,p\}. Taking into account that the identity ee is a unique element of order 1, we conclude that a>1|\langle a\rangle|>1, and therefore, a=p|\langle a\rangle|=p. Since aG\langle a\rangle\subseteq G and a=p=G|\langle a\rangle|=p=|G|, we conclude that a=G.\langle a\rangle=G. Consequently, G is a cyclic group generated by aa.


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