Let p be prime and Ha subgroup of Z(p∞).
(a) Every element of Z(p∞ ) has finite order pⁿ for some n ≥ 0.
(b) If at least one element of H has order p^k and no element of H has order greater than p^k, then His the cyclic subgroup generated bywhence H ≅ Z_pk•(c) If there is no upper bound on the orders of elements of H, then H = Z(p∞).
(d) The only proper subgroups of Z(p∞) are the finite cyclic groups(n = 1,2, ... ). Furthermore, (0) = Co< C1 < C2 < Ca<···.
(e) Let x₁,x₂, ... be elements of an abelian group G such that lx₁I = p, px₂ = x₁, px₃ = x₂, ... , px_n+1 = x_n, . . . . The subgroup generated by the X_i (i ≥ 1) is isomorphic toZ(p∞).

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1/pk,