Problem session solutions
1. Let G be a cyclic group, with identity element e. Then G = hai for some a ∈ G. Let H be a
subgroup of G.
Prove:
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
The set S = {k ∈ N : ak ∈ H} is either empty, or contains a least element.
If S = ∅ then H = hei.
Now suppose that S 6= ∅, and let m be the least element of S. Then ham i ⊆ H.
If an ∈ H and n = mq + r where r, q ∈ Z, then ar ∈ H.
If an ∈ H and n = mq + r where r, q ∈ Z and 0 ≤ r < m, then r = 0.
H = ham i.
Every subgroup of a cyclic group is a cyclic subgroup.
The subgroups of (Z, +) are precisely the sets kZ = {kn : n ∈ Z} for k ∈ Z.
Solution (a) Either S = ∅, or S 6= ∅. If S 6= ∅ then S is a non-empty subset of N0 , so
contains a least element, by the least integer principle.
(b) Let x ∈ H. Then x ∈ G = hai, so x = an for some n ∈ Z. If n > 0 then n ∈ S, but S = ∅
so this is impossible. If n < 0 then x−1 = (an )−1 = a−n ∈ H by (SG2), so −n ∈ S, but S = ∅
so this is impossible. So n = 0 is the only possibility, so x = a0 = e. This shows that H ⊆ {e},
so H = ∅ or H = {e}. Since H 6= ∅ by (SG0), we have H = {e}. Since ek = e for all k ∈ Z, we
have hei = {e} = H.
(c) m is the least element of S, so m ∈ S, so am ∈ H. So (am )k ∈ H for all k ≥ 1 (you can
prove this by induction, if you like: it holds for k = 1, and if it holds for k then (am )k ∈ H and
am ∈ H =⇒ (am )k am = (am )k+1 ∈ H by (SG1), so it also holds for k + 1). Hence by (SG2),
(am )−k = ((am )k )−1 ∈ H for all k > 1, and since H is a subgroup of G we have e = (am )0 ∈ H.
So (am )k ∈ H for all k ∈ Z, so ham i = {(am )k : k ∈ Z} ⊆ H.
(d) We have a−mq = (am )−q ∈ ham i ⊆ H, by (c), so a−mq ∈ H. And an ∈ H by hypothesis.
Hence by (SG1), ar = an−mq = an a−mq ∈ H.
(e) We have ar ∈ H and 0 ≤ r < m. If r > 0 then r ∈ S and r < m, contradicting the fact
that m is the least element of S. So we are forced to conclude that r = 0.
(f) We know from (c) that ham i ⊆ H. So it suffices to show that H ⊆ ham i.
Let x ∈ H. Then x ∈ G = hai, so x = an for some n ∈ Z. By the Division Algorithm, we can
write n = mq + r for some q ∈ Z and an integer r with 0 ≤ r < m. By (e), we have r = 0, so
x = amq = (am )q ∈ ham i. Hence H ⊆ ham i, as required.
So H = ham i.
(g) We’ve shown that every subgroup H of a cyclic group G is either hei or is ham i for some
a ∈ G and some integer m ∈ N. Both of these are cyclic subgroups. So every subgroup of G is
cyclic.
(h) We have hki = kZ, so these sets are precisely the cyclic subgroups of Z. Since Z = h1i is
cyclic, there are no other subgroups by (g).