Math 403 Chapter 4: Cyclic Groups

1. Introduction: The simplest type of group (where the word “type” doesn’t have a clear
meaning just yet) is a cyclic group.
2. Definition: A group G is cyclic if there is some g ∈ G with G = hgi. Here g is a generator
of the group G. Recall that hgi means all “powers” of g which can mean addition if that’s the
operation.
(a) Example: Z6 is cyclic with generator 1. Are there other generators?
(b) Example: Zn is cyclic with generator 1.
(c) Example: Z is cyclic with generator 1.
(d) Example: R∗ is not cyclic.
(e) Example: U (10) is cylic with generator 3.

3. Important Note: Given any group G at all and any g ∈ G we know that hgi is a cyclic
subgroup of G and hence any statements about cyclic groups applies to any hgi.
4. Properties Related to Cyclic Groups Part 1:
(a) Intuition: If |g| = 10 then hgi = {1, g, g 2 , ..., g 9 } and the elements cycle back again. For
example we have g 2 = g 12 and in general g i = g j iff 10 | (i − j).
(b) Theorem: Let G be a group and g ∈ G.
(i) If |g| = ∞ then g i = g j iff i = j.
(ii) if |g| = n then hgi = {1, g, g 2 , , , , .g n−1 } and g i = g j iff n | (i − j).
(iii) In both cases we have |g| = |hgi|.
Proof:
For (i), if |g| = ∞ then by definition we never have g i = e unless i = 0. Thus g i = g j iff
g i−j = e iff i − j = 0.
For (ii), If |g| = n < ∞ first note that {1, g, g 2 , ..., g n−1 } ⊆ hgi by definition of the right
side. To show that hgi ⊆ {1, g, g 2 , ..., g n−1 }, suppose g k ∈ hgi. Write k = qn + r with
0 ≤ r < n and then g k = (g n )q g r = eq g r = g r so g k is one of those elements.
Now for the iff. If g i = g j then g i−j = e. Write i − j = qn + r with 0 ≤ r < n. Then
e = g qn g r = g r . Since n (the order) is the least positive but r < n we must have r = 0
and so n | (i − j).
If n | (i − j) then i − j = qn and then g i = g j g qn = g j .
For (iii), it follows immediately.
QED
i
(c) Corollary: For any g ∈ G with |g| = n, g = e iff n | i.
Proof: This is the theorem with j = 0. QED
i
Example: If |g| = 10 then if g = e then 10 | i, meaning we only get e when the powers
are multiples of 10.
5. Properties Related to Cyclic Groups Part 2:


(a) Intuition: If |g| = 30 then if we examine something like g 24 we find:

g 24 = g 24

2
g 24 = g 48 = g 18

3
g 24 = g 72 = g 12

4
g 24 = g 96 = g 6

5
g 24 = g 120 = g 0 = e



We then see that g 24 = {e, g 6 , g 12 , g 18 , g 24 } = g 6 . which is a bit nicer since the 6 is
easier to work with. Note that 6 = gcd (30, 24).
   
From this we also see g 24  = g gcd (30,24) .
Likewise we can easily compute the order of g 24 . We see it cycles every 5, just like g 6 ,
and observe that 5 = 30/gcd (30, 24).
(b) Theorem: Let g ∈ G with |g| = n and let k ∈ Z+ then



(i) g k = g gcd (n,k)
   
(ii) g k  = g gcd (n,k) 
(iii) |g k | = n/gcd (n, k)
Proof: For (i) since gcd (n, k) | k we know that α gcd (n, k) = k for some α ∈ Z and so
 α D E
g k = g gcd (n,k) ∈ g gcd (n,k)

and so:
k  D gcd (n,k) E
g ⊆ g

Then write gcd (n, k) = αn + βk and observe that

g gcd (n,k) = (g n )α + (g k )β = (g k )β ∈ g k



so that D E

g gcd (n,k) ⊆ g k

Thus the two are equal.

so that:
n
|g gcd (n,k) | ≤
gcd (n, k)
 On the other hand if we had |g gcd (n,k) | = b < n/gcd (n, k) then we have e = (g gcd (n,k) )b =
g b gcd (n,k) with b gcd (n, k) < n, contradicting |g| = n. Thus we have:
n
|g gcd (n,k) | =
gcd (n, k)

Thus we have:   n
|g k | = g gcd (n,k)  =
 
gcd (n, k)
QED
(c) Corollary: In a finite cyclic group the order of an element divides the order of a group.
Proof: Follows since every element looks like g k and we have |g k | gcd (n, k) = n. QED
Example: In a cyclic group of order 200 the order of every element must divide 200. In
such a group an element could not have order 17, for example.
(d) Corollary: Suppose g ∈ G and |g| = n < ∞. Then:

i
j 
a = a iff gcd (n, i) = gcd (n, j) iff |ai | = |aj |
Proof: Follows immediately. QED
Example: If |g| = 18 then the fact that gcd (18, 12) = 6 = gcd (18, 6) guarantees that
|g 12 | = |g 6 |.
(e) Corollary: Suppose g ∈ G and |g| = n < ∞. Then:


hai = aj iff gcd (n, j) = 1 iff |a| = |aj |
Proof: Follows immediately. QED


Example: If |g| = 32 then the fact that gcd (15, 32) = 1 guarantees that g 15 = hgi,
meaning g 15 is a generator of hgi.
(f) Corollary: Suppose g ∈ G and |g| = n < ∞. Then there are φ(n) generators of hgi.
Proof: The generators are g k with gcd (n, k) = 1. QED
(g) Corollary: An integer k ∈ Zn is a generator of Zn iff gcd (n, k) = 1.
Proof: Follows immediately. QED
Example: The generators of Z10 are 1, 3, 7, 9.
6. Classification of Subgroups of Cyclic Groups:
(a) Theorem (Fundamental Theorem of Cyclic Groups):
Suppose G = hgi is cyclic.
(i) Every subgroup of G is cyclic.
(ii) If |G| = n then the order of any subgroup of G divides n.


(iii) If |G| = n then for any k | n the subgroup g n/k is the unique subgroup of order k.
Proof:
(i) Let H ≤ G. If H = {e} then we’re done so assume H 6= {e}. Choose g m ∈ H with
minimal m ∈ Z+ by well-ordering. Clearly hg m i ⊆ H. If some g k ∈ H then put
k = qm + r with 0 ≤ r < m so r = k − qm and then g r = g k (g m )−q ∈ H and so r = 0
by minimality of m and so g k = (g m )q and hence g k ∈ hg m i.
(ii) Take a subgroup H ≤ G. We know H is cyclic by (i) with H = hg m i with minimal
m ∈ Z+ by well-ordering. Write n = qm + r with 0 ≤ r < m so r = n − qm and then
g r = g n (g m )−q ∈ H and so r = 0 by minimality of m and so n = qm and then
n n
|H| = |hg m i| = |g m | = =
gcd (n, m) m
and so m|H| = n and so |H| | n.
(iii) Observe first that for any k | n we have
D E 
 n/k   n/k 
 n n
 g  = g  = = =k
gcd (n, n/k) n/k


Thus certainly g n/k is a subgroup of order k. We must show that it is unique.
Let H ≤ G with |H| = k | n. Since H ≤ G by (i) and (ii) we have H = hg m i with
m | n. Then we have:
n n
k = |H| = |hg m i| = |g m | = =
gcd (n, m) m


Thus m = n/k and so H = hg m i = g n/k .
QED
Example: This categorizes cyclic groups completely. For example suppose a cyclic group
has order 20. Every subgroup is cyclic and there are unique subgroups of each order
1, 2, 4, 5, 10, 20. If G has generator g then generators of these subgroups can be chosen to
be g 20/1 = g 20 , g 20/2 = g 10 , g 20/4 = g 5 , g 20/5 = g 4 , g 20/10 = g 2 , g 20/20 = g respectively.
(b) Corollary: For each positive divisor k of n ∈ Z+ , the set hn/ki is the unique subgroup
of Zn of order k. Moreover these are the only subgroups of Zk .
Proof: Follows immediately. QED
Example: In Z10 = h1i the subgroup h1i is the unique subgroup of order 10/1 = 10, the
subgroup h2i is the unique subgroup of order 10/2 = 5, the subgroup h5i is the unique
subgroup of order 10/1 = 2, the subgroup h10i = h0i is the unique subgroup of order
10/10 = 1.
(c) Definition: Define φ(1) = 1 and for any n ∈ Z with n > 1 define φ(n) to be the number
of positive integers less than n and coprime to n.
Example: We have φ(20) = 8 since 1, 3, 7, 9, 11, 13, 17, 19 are coprime.
(d) Theorem: Suppose G is cyclic of order n. If d | n then there are φ(d) elements of order
d in G.
Proof: Every element of order d generates a cyclic subgroup of order d but there is only
one such cyclic subgroup, thus every element of order d is in that single cyclic subgroup
of order d. If that cyclic subgroup is hgi with |g| = d then note that the only elements of
order d in it are those g k with gcd (d, k) = 1 and there are φ(d) of those. QED
Example: In a cyclic group of order 100 noting that 20 | 100 we then know there are
φ(20) = 8 elements of order 20.
(e) Theorem: If G is a finite group then the number of elements of order d is a multiple of
φ(d).
Outline of Proof: Elements of order d can be collected φ(d) at a time into subgroups
of order d. QED
Example: If G is an arbitrary finite group then the number of elements of order 20 is a
multiple of 8. Keep in mind that this might be zero!