Let n012and nZnkkZ Prove that nZ is a subgroup of Z Show tha

Let n=0,1,2,...and nZ={nk:kZ}. Prove that nZ is a subgroup of Z. Show that these subgroups are the only subgroups of Z.

Solution

Let n= 0, 1, 2,....

nZ is a subgroup of Z

By the subgroup test, it suffices to show that nZ contains 0 and is closed under addition and additive inverses.since 0 = n*0, nZ contains 0. Since nx + ny = n(x+y) and -(nx) = n(-x) for all x, y in Z , colsure under addition and additive inverses are satisfied.

2. the only subgroups of Z are of the form nZ, where n is a non negative integer.

By 1), the sets nZ(n = 0, 1, 2...) are subgroups of Z . conversely, suppose H is a subgroup of Z. without loss of generality, assue H is nonzero. Consider S = {|m| : m in H\\{0}}. Then S is a nonempty subset of the positive integers. Hence, by the well- ordering principles, S has a least element, n. Since n= +/- m for some nonzero m in H and H is a subgroup of Z, n is in H. so the cyclic subgroup nZ is contained in H. On theother hand , given x in H, the division algorithm yields a representation x= nq + r, where q and r are integers with 0 <= r<n. since x and nq are in H, r= x - nq is in H( since H is a subgroup of Z) . if r is nonzero, then r is an element of S of lesser value than n, a contradiction. Hence r= 0 and x =nq is an element of nZ. Since x was arbitary, H is contained in nZ. thus H= nZ.