5
Cyclic groups
The aim of this section is to classify all cyclic groups up to isomorphism. We shall find a
collection of ‘model’ cyclic groups, whose structure we understand very well, and such that
any other cyclic group will be isomorphic to one of our model groups. One of these model
groups will be Z under addition. Other groups, Zn , are introduced in the first subsection.
5.1
The groups Zn
Below we will often refer to the following well known statement (see MAA141 Real and
Complex Numbers) often called the division algorithm for integers.
Proposition 5.1. Let n be a positive integer and let m be any integer. Then there exist
unique integers q and r such that
m = qn + r,
0 ≤ r < n.
Here q is called as the quotient and r the remainder when m is divided by n.
Definition 5.2. Let n be a fixed positive integer and let k and l be any integers. The
remainder r when k + l is divided by n is the sum of k and l modulo n. We shall sometimes
denote the sum of k and l modulo n by k ⊕ l. In this notation, one has
k + l = qn + k ⊕ l,
0 ≤ k ⊕ l < n.
Example 5.3. The sum of 23 and 31 modulo 45 is 9, since 23 + 31 = 54 = 45 × 1 + 9 and
0 ≤ 9 < 45.
Theorem 5.4. The set {0, 1, 2, . . . , n − 1} under addition modulo n is a group (denoted by
Zn ). This group is cyclic.
Proof. Let ⊕ denote the addition modulo n. Clearly, ⊕ is a binary operation on the set
{0, 1, 2, . . . , n − 1}. We need to check the three axioms of the group.
(G1) Let p, q, r ∈ Zn . Let us prove that both (p⊕q)⊕r and p⊕(q⊕r) equal to the
remainder of p + q + r when divided by n. Indeed,
p + q = mn + (p⊕q),
m ∈ Z,
(p⊕q) + r = ln + ((p⊕q)⊕r),
l ∈ Z.
Thus,
p + q + r = (l + m)n + ((p⊕q)⊕r),
which proves the claim for (p⊕q)⊕r. In the same way,
q + r = in + (q⊕r),
i ∈ Z,
p + (q⊕r) = jn + (p⊕(q⊕r)),
j ∈ Z.
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Thus,
p + q + r = (i + j)n + (p⊕(q⊕r)),
which proves the claim for p⊕(q⊕r).
Thus,
p⊕(q⊕r) = (p⊕q)⊕r
and so ⊕ is associative.
(G2) Clearly, 0 is the identity in Zn .
(G3) The inverse of k ∈ Zn is n − k ∈ Zn :
k + (n − k) = qn + k ⊕ (n − k),
k + (n − k) = n + 0,
thus, k ⊕ (n − k) = 0.
It remains to prove that Zn is cyclic. Let us compute 1. One has 0 ∈ 1, 1 ∈ 1,
1⊕1 = 2 ∈ 1, 2⊕1 = 3 ∈ 1, . . . , n − 1 ∈ 1, so Zn = 1.
Theorem 5.5. For any positive integer n, the groups Zn and Un are isomorphic.
Proof. Denote α = exp(2πi/n); note that αn = 1. Recall that, as sets,
Zn = {0, 1, 2, . . . , n − 1},
Un = {α0 , α1 , α2 , . . . , αn−1 }.
Let us define the function
f : Zn → Un ,
f (k) = αk .
Clearly, f is a bijection. It remains to check that f preserves the group structure, i.e., that
for any k, l ∈ {0, 1, . . . , n − 1}, one has f (k⊕l) = f (k)f (l) (where ⊕ is addition modulo n).
But k + l = mn + (k⊕l) for some m ∈ Z, and thus
f (k)f (l) = αk αl = αk+l = αmn+(k⊕l) = (αn )m αk⊕l = αk⊕l = f (k⊕l).
5.2
Classification of cyclic groups
Theorem 5.6. Every infinite cyclic group is isomorphic to Z under addition.
Proof. Let G be an infinite cyclic group with a generator a. Let us define the mapping
f : Z → G,
f (n) = an .
We will prove that f is an isomorphism.
1. Clearly, f is onto, since any element g ∈ G has the form an for some n ∈ Z.
2. Let us prove that f is one-to-one. Suppose to the contrary that for some m, n ∈ Z,
n > m, one has f (n) = f (m). Denote l = n − m. Then one has
al = an−m = an (am )−1 = f (n)(f (m))−1 = e,
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where e is the identity. We claim that G has at most l elements, namely,
G = {e, a, a2 , . . . , al−1 },
contrary to the assumption that G is infinite. (Note that the l elements listed above may
or may not be distinct.) Any element in G has the form ak for some k ∈ Z. Using the
division algorithm, we get
k = pl + r,
p ∈ Z,
0 ≤ r < l.
Thus,
ak = apl ar = (al )p ar = ar ∈ {e, a, a2 , . . . , al−1 },
as asserted.
3. It remains to prove that f preserves the group structure, i.e., f (m + n) = f (m)f (n).
But this is evident:
f (m + n) = am+n = am an = f (m)f (n).
Example 5.7. As we saw above, 3Z is an infinite cyclic group. It might seem odd that 3Z
is isomorphic to Z although 3Z is a subgroup of Z. But renaming 1 by 3, 2 by 6, 3 by 9 etc,
we can convert Z into 3Z, keeping its algebraic properties. Thus, Z and 3Z are structurally
the same.
Theorem 5.8. If G is a cyclic group of a finite order n, then G is isomorphic to Zn .
Proof. Let a be a generator of G and e be the identity.
1. First let us prove that G consists of the following distinct elements:
S := {e, a, a2 , . . . , an−1 }.
Suppose that some of these elements coincide: ak = al for k, l ∈ Z, k > l. Then ak−l = e
and then by the same argument as in Theorem 5.6, we see that G consists of at most k − l
distinct elements e, a, a2 , . . . , ak−l−1 . Since k − l < n, this contradicts to the assumption
|G| = n.
Since S ⊂ G and S contains n elements, we see that G = S.
2. Let us define the mapping
f : Zn → G,
f (k) = ak .
By the first part of the proof, f is bijective. As in the proof of Theorem 5.6, it is evident
that f (n + m) = f (n)f (m).
Corollary 5.9. (i) A cyclic group generated by an element g is finite if and only if g n = e
for some n = 0.
(ii) The order of a finite cyclic group generated by an element g coincides with the
minimal positive integer n such that g n = e.
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Proof. (i) If the group g is finite, then, as we saw in the proof of Theorem 5.8, one has
g n = e. If, on the other hand, g is infinite, then, as we saw in the proof of Theorem 5.6,
g n = e for any n = 0.
(ii) Follows from the first part of the proof of Theorem 5.8.
We have almost completed the programme formulated in the preamble to this section.
Each cyclic group G has been proven to be isomorphic either to Z, if G is infinite, or to Zn ,
if |G| = n is finite. It remains to study the structure of these ‘model’ groups Z and Zn .
5.3
Subgroups of cyclic groups
Theorem 5.10. A subgroup of a cyclic group is cyclic.
Proof. Let G be a cyclic group generated by a and e be the identity. Let H be a subgroup
of G. If H = {e}, then H = e is cyclic. If H = {e}, then an ∈ H for some n ∈ N. Let m
be the smallest positive integer such that am ∈ H.
We claim that c = am generates H; that is, H = c. The inclusion c ⊂ H holds
because c is the smallest subgroup containing c (i.e., it belongs to any other subgroup
containing c).
We must show that H ⊂ c. In other words, we have to prove that every b ∈ H is a
power of c. Since b ∈ H and H is a subgroup of G, we have b = an for some n. Find q and
r such that
n = mq + r,
0 ≤ r < m.
Then
an = amq+r = (am )q ar ,
so
ar = (am )−q an .
Now since an ∈ H and am ∈ H, it follows that ar ∈ H. Since m was the smallest positive
integer such that am ∈ H and 0 ≤ r < m, we must have r = 0. Thus n = qm and
b = an = (am )q = cq , so b is a power of c.
Corollary 5.11. The subgroups of Z under addition are precisely the groups nZ under
addition for n ∈ Z.
Proof. 1. It is clear that for any n ∈ Z, nZ is a cyclic subgroup of Z, generated by the
element n.
2. If H is a subgroup of Z, then, by Theorem 5.10, H must be cyclic. Let n ∈ Z be the
generator of H. Then H = nZ.
Theorem 5.12. Let G be a cyclic group with n elements and generated by a. Let b ∈ G and
let b = as , s = 0. Then the cyclic subgroup b of G generated by b contains n/d elements,
where d is the greatest common divisor of n and s.
Before proving this theorem, we recall a statement known from MAA141 Real and
Complex Numbers.
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Proposition 5.13. Let r, s, m be positive integers. If r and s are relatively prime and r
divides sm, then r divides m.
Proof of Theorem 5.12. By the same argument as in Theorem 5.6 and Theorem 5.8, we see
that b has as many distinct elements as the smallest positive power m of b that gives the
identity. Now b = as , and bm = e if and only if (as )m = e, or if and only if n divides ms.
What is the smallest positive integer m such that n divides ms? We wish to find the
smallest positive m such that
ms
m(s/d)
=
is an integer.
n
(n/d)
Since d is the g.c.d. of s and n, the numbers s/d and n/d are relatively prime (see MAA141
Real and Complex Numbers).
From Proposition 5.13, we conclude that n/d must divide m, so the smallest such m is
n/d. Thus the order of b is n/d.
Example 5.14. Consider Z12 , with the generator a = 1. The gcd of 3 and 12 is 3. Thus,
3 generates a subgroup of 12/3 = 4 elements, namely
3 = {0, 3, 6, 9}.
The gcd of 8 and 12 is 4. Thus, 8 generates a subgroup of 12/4 = 3 elements, namely
8 = {0, 4, 8}.
The gcd of 12 and 5 is 1. Thus, 5 generates a subgroup of 12/1 = 12 elements, namely the
whole group Z12 .
The last example illustrates the following corollary of Theorem 5.12.
Corollary 5.15. (i) In the group Zn , all the generators are those numbers r ∈ {1, . . . , n−1}
which are relatively prime to n.
(ii) If p is a prime number, then any non-zero element of the group Zp generates this
group.
(iii) If p is a prime number, then the group Zp has no subgroups.
(i) follows directly from Theorem 5.12 and (ii) follows from (i). To prove (iii), note that
all the subgroups of Zn are cyclic, and, by (ii), all cyclic subgroups of Zp coincide with Zp .
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