by the structure theorem for finitely-generated abelian groups we may write:

, for primes p_{1},...,p_{k}.

since G is assumed non-cyclic, we cannot have gcd(p_{1},...,p_{r}) = 1 (by the chinese remainder theorem), so for some i ≠ j, p_{i}= p_{j}.

without loss of generality (we may re-arrange the direct sum as we like) take i = 1, j = 2.

now has a subgroup isomorphic to

(since each is cyclic and thus has a cyclic subgroup of order p_{i}).

but p_{1}= p_{2}, so setting p = p_{1}= p_{2}:

is isomorphic to a subgroup of G.