Proof
(related to Theorem: Classification of Finite Groups with the Order of a Prime Number)
 By hypothesis, $(G,\ast)$ is a finite group with a group order $G=p,$ where $p$ is a prime number.
 Since finite order of an element equals group order, $\langle a\rangle$ is a subgroup of $G$ with a finite order $\langle a\rangle$ for any $a\in G.$
 According to the Lagrange's theorem, the $\langle a\rangle$ is a divisor of $p.$
 By definition of prime numbers, either $\langle a\rangle=1$ or $\langle a\rangle=p.$
 Because $\langle a\rangle$ contains at least two elements, the neutral element $e$ and $a,$ we have $\langle a\rangle\neq 1.$
 Therefore, $\langle a\rangle=p=G.$
 Since all subgroups of cyclic groups are cyclic, $\langle a\rangle$ is a cyclic group.
 Therefore, $G$ is a cyclic group.
 By the classification of cyclic groups, $(G,\ast)$ is isomorphic to the additive subgroup of integers $(\mathbb Z_p, + ).$
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References
Bibliography
 Modler, Florian; Kreh, Martin: "Tutorium Algebra", Springer Spektrum, 2013