If a group 𝐺 has prime order 𝑝=|𝐺|,
then it must be cyclic, #m/thm/group
i.e. isomorphic to ℤ+𝑝.
Proof
Let 𝐺 be a group of prime order 𝑝.
By Lagrange's theorem,
𝐺 only has subgroups of order 1 and 𝑝.
Since 𝑝>1,
there exists 𝑎∈𝐺 such that 𝑎≠𝑒.
Then |𝑎|=𝑝
and therefore ⟨𝑎⟩=𝐺.