Cyclic Groups Lecture Notes
Cyclic Groups
M. Velrajan
Consider the group G = {1, 𝜔, 𝜔2, … , 𝜔n-1} of all
nth roots of unity under multiplication,
where 𝜔 = e2𝝅i/n, 𝜔n = 1.
Here 𝜔n+1 = 𝜔, 𝜔n+2 = 𝜔2, and 𝜔n+k = 𝜔k for any k ≥ 1.
Also 𝜔-1 = 𝜔n-1, 𝜔-2 = 𝜔n-2, 𝜔-n = 𝜔n-n = 1, 𝜔-(n+1) = 𝜔n-1, ….
We note that all integer powers of 𝜔 are in G.
Such groups are called cyclic groups.
Before defining a cyclic group, we prove the following.
Theorem Let G be a group and a ∈ G.
Then H = {an / n ∈ Z} is a subgroup of G.
Proof Clearly H is a nonempty subset of G.
For x = am and y = an ∈ H,
xy-1 = am(an)-1 = am a-n = am-n ∈ H.
Hence H is a subgroup of G.
Definition Let G be a group and a ∈ G. Then the subgroup
{an / n ∈ Z} is called the cyclic subgroup of G generated by
a and it is denoted by (a) or <a>.
i.e. (a) = {an / n ∈ Z}.
A subgroup H of G is called a cyclic subgroup of G
if H = (a) for some a ∈ G.
Note For any a ∈ G, (a) = {an / n ∈ Z} is the smallest subgroup
of G containing a.
For, Clearly a ∈ (a). Let K be any subgroup of G containing a.
Then a, a-1 ∈ K, hence an ∈ K for all n ∈ Z and hence (a) ⊆ K.
Definition A group G is said to be cyclic if there exists an element
a ∈ G such that G = (a) = {an / n ∈ Z}.
And the element a is called a generator of G.
Note In a group G, for any a ∈ G, (a) = (a-1).
Hence if a is a generator of a cyclic group G then a-1 is also a generator of G.
For, (a-1) = {(a-1)n / n ∈ Z} = {a-n / n ∈ Z} = {an / n ∈ Z} = (a).
Theorem Let G be a group and a ∈ G. Then the order of the
cyclic group generated by a is the same as the order of a.
Proof Suppose the order of a ∈ G is n.
Then n is the smallest positive integer such that an = e.
Hence e = a0, a, a2, a3, . . . , an-1 are distinct
(for, if ar = as for 0 ≤ r < s < n then
as - r = asa-r = ara-r = a0= e and 0 < s - r < n, a contradiction).
And for any m ∈ Z, by division algorithm,
m = nq + r for some q, r ∈ Z and 0 ≤ r < n.
Hence am = (an)q ar = (e)qar = ear = ar.
Therefore, (a) = {am / m ∈ Z} = {e = a0, a, a2, a3, . . . , an-1}.
Hence the order of (a) is n.
[And in (a), aras = ar + s = ar ⊕ s
(since r + s = qn + (r ⊕n s) for some integer q,
ar + s = (an)qar ⊕ s = eqar ⊕ s = ear ⊕ s = ar ⊕ s)]
Suppose the order of a ∈ G is infinite.
Then there is no positive integer n such that an = e.
Suppose am = an and m ≥ n.
Then am - n = ama-n = ana-n = a0 = e and m - n ≥ 0.
Hence, m - n = 0. i.e m = n.
Therefore for any m, n ∈ Z, am = an if and only if m = n.
Hence the order of (a) = {am/m ∈ Z} is infinite.
Therefore, the order of the cyclic group generated by a is
the same as the order of a.
Note Let G be a group, a ∈ G and the order of a = n.
Then n | O(G).
(since the order of the cyclic subgroup (a) is n,
by the Lagrange's theorem, n | O(G))
Examples 1. In (Z6, ⊕6), (0) = {0},
(1) = {0, 1, 2, 3, 4, 5} = Z6,
(2) = {0, 2, 4}, (3) = {0, 3}, (4) = {0, 4, 2} = (2),
(5) = {0, 5, 4, 3, 2, 1} = Z6 = (1).
In general, the group (Zn, ⊕n) is cyclic and
Zn = (1) = (n-1).
2. In (Z, +), (1) = {n1 / n ∈ Z} = Z,
(-1) = {n(-1) / n ∈ Z} = {-n / n ∈ Z} = Z,
(2) = {n2 / n ∈ Z} = 2Z.
In general (n) = {nm / m ∈ Z} = nZ = (-n).
The group (Z, +) is cyclic and Z = (1) = (-1).
Since (n) = nZ, for all n,
1, -1 are the only generators of (Z, +).
3. Any group of prime order is cyclic.
For, Let G be any group of order p, a prime.
Let a ≠ e ∈ G.
Consider the cyclic subgroup <a> generated by a.
By Lagrange's theorem O(<a>) | O(G) = p.
Since p is prime, O(<a>) = 1 or p.
Since a ≠ e and a, e ∈ <a>, O(<a>) ≠ 1, hence O(<a>) = p.
Since O(G) = p, and <a> is a subgroup of G, G = <a>.
Hence G is cyclic.
We note that in the group G of prime order,
all non-identity elements are generators G.
4. In a group G, {e} = (e) is a cyclic subgroup.
5. The group G = {1, 𝜔, 𝜔2, . . . , 𝜔n-1} of all nth roots
of unity, under multiplication, where 𝜔 = e2𝝅i/n, 𝜔n = 1,
is a cyclic group generated by 𝜔.
If n = 1, G = {1}.
If n = 2, G = {1, -1}= (-1).
If n = 3, G = {1, 𝜔, 𝜔2}, where 𝜔 = e2𝝅i/3.
And G = (𝜔) = (𝜔2).
If n = 4, G = {1, i, -1, -i} = (i) = (-i).
6. Klein’s 4 group V4 is not cyclic.
But all proper subgroups of V4 are cyclic.
For, The Klein 4 group V4 = {e, a, b, c},
where a2 = b2 = c2 = e and ab = c.
Hence (e) = {e}, (a) = {e, a}, (b) = {e, b}, (c) = {e, c}.
And hence no element generates V4 i.e. V4 is not cyclic.
Since O(V4) = 4, by Lagrange's theorem,
the order of any subgroup of V4 is 1 or 2 or 4.
Hence (e)= {e}, (a) = {e, a}, (b) = {e, b}, (c) = {e, c}
and V4 are the only subgroups of V4.
We note that all the proper subgroups of V4 are cyclic,
but V4 is not cyclic.
Since 2, 3 are prime, any group of order 2 or 3 is cyclic.
And V4 is a group of order 4.
Hence V4 is the smallest group which is not cyclic
but all the proper subgroups are cyclic.
7. In S3 = {e, (1 2), (1 3), (2 3), (1 2 3), (1 3 2) },
(1 2)2 = e = (1 3)2 = (2 3 )2 ,
(1 2 3)2 = (1 3 2), (1 2 3)3 = e, (1 3 2)2 = (1 2 3), (1 3 2)3 = e.
Hence <e> = {e}, <(1 2)> = {e, (1 2)}, <(1 3)> = {e, (1 3)},
<(2 3)> = {e, (2 3)},
<(1 2 3)> = {e, (1 2 3), (1 3 2)} = <(1 3 2)>.
Hence no element generates S3 i.e. S3 is not cyclic.
Since O(S3) = 6, by the Lagrange's theorem,
the order of any subgroup of S3 is 1 or 2 or 3 or 6.
Hence <e> = {e}, <(1 2)> = {e, (1 2)}, <(1 3)> = {e, (1 3)},
<(2 3)> = {e, (2 3)}, <(1 2 3)> = {e, (1 2 3),
(1 3 2)} = <(1 3 2)> and S3 are the only subgroups of S3.
We note that all the proper subgroups of S3 are cyclic,
but S3 is not cyclic.
Also all the proper subgroups of S3 are abelian,
but S3 is non-abelian.
And S3 is the smallest non-abelian group
(all the groups of order < 6 are abelian, but S3 is the only
non-abelian group of order 6).
Note Any cyclic group G is abelian.
But the converse need not be true.
For, Let G be a cyclic group and
G = (a) = {an / n ∈ Z}. Then for any m,n ∈ Z,
aman = am + n = an + m = an am . Hence G is abelian.
The Klein 4 group V4 = {e, a, b, c}, a2 = b2 = c2 = e and ab = c,
is abelian, but V4 is not cyclic.
Theorem Let G be a group, a ∈ G and the order of a = n.
Then for any s ≥ 1, the order of as = n/d ,
where d = (n, s), g.c.d. of n and s.
Proof Let d = (n, s).
Then (as)n/d = as(n/d) = an(s/d) = (an)s/d = es/d = e
(since s(n/d) = n(s/d) and an = e).
Suppose (as)m =asm = e.
Then, since the order of a = n, n | sm.
i.e. sm = nx for some integer x.
Hence (s/d)m = (n/d)x and hence (n/d) | (s/d)m.
Since d = (n, s), (n/d, s/d) = 1. Hence (n/d) | m.
Therefore, the order of as = n/d .
Corollary Let G be a group and a ∈ G generate a
subgroup of order n. Then for any s ≥ 1, as generates
a group of order n/d , where d = (n, s)
(It follows from the theorem, since the order of the
cyclic group generated by a is the same as the order of a).
Theorem Let G be a cyclic group of order n and
G = (a) = {e = a0, a, a2, . . . , an-1}.
Then for any s ≥ 1, as is a generator of G if and only if (n,s) = 1
i.e. n and s are relatively prime.
Proof Since G is a cyclic group of order n and
G = (a), order of a = n.
For any s ≥ 1,
as is a generator of G if and only if G = (as)
if and only if O(as) = O(G) = n
if and only if n/d = n, where d = (n,s) (since O(as)= n/d )
if and only if d = 1
if and only if (n,s) = 1.
Corollary Let G be a cyclic group of order n.
Then the number of generators of G = 𝜑(n).
Proof Suppose G = (a) = {e = a0, a, a2, a3, . . . , an-1}.
Then for any s ≥ 1,
as is a generator of G if and only if n and s are relatively prime.
Hence, the number of generators of G
= the number of positive integers less than n
and relatively prime to n = 𝜑(n).
Theorem Any subgroup of a cyclic group is cyclic.
Proof Let G be a cyclic group and G = (a) = {an / n ∈ Z}.
Let H be a subgroup of G. If H = {e} then H=(e) is cyclic.
Suppose H ≠ {e}. Let x ≠ e ∈ H.
Then x ∈ G and x = an for some n ≠ 0 ∈ Z.
Since H is a subgroup, x-1 = (an)-1 = a-n ∈ H.
Hence an, a-n ∈ H and n ≠ 0.
Therefore ak ∈ H for some positive integer k.
Let m be the smallest positive integer such that am ∈ H.
Let x ∈ H. Then x ∈ G and x = an for some n ∈ Z.
By division algorithm, n = mq + r for some q, r ∈ Z
and 0 ≤ r < m.
And ar = an - mq = an(am)-q ∈ H
(since an, am ∈ H and H is a subgroup).
Since m is the smallest positive integer such that
am ∈ H, r = 0, otherwise, we get a contradiction.
Hence n = mq and x = amq = (am)q ∈ (am).
Therefore H ⊆ (am).
Since am ∈ H, (am) ⊆ H. Hence H = (am),
i.e. H is cyclic.
Remark nZ, n ∈ Z are the only subgroups of (Z, +).
For, We know that for any n ∈ Z, nZ is a subgroup of Z.
Since (Z, +) is cyclic, any subgroup H of Z is cyclic.
Hence H = (n) = nZ, for some n ∈ Z.
Therefore, nZ, n ∈ Z are the only subgroups of (Z, +).
Remark 1. If G1 and G2 are two groups then G1 × G2
is a group under the operation (a1, a2)(b1, b2) = (a1b1, a2b2).
The group G1 × G2 is called the product of G1 and G2.
The identity element of G1 × G2 is (e1, e2) and
(a1, a2)-1 = (a1-1, a2-1). (verify)
2. Let G1 and G2 be two groups. Then
G1 × G2 is an abelian group if and only if G1 and G2 are abelian.
(verify)
3. The product of two cyclic groups need not be cyclic.
For example, (Z2, ⊕2) is cyclic.
But (Z2, ⊕2) × (Z2, ⊕2) is not cyclic.
Since in (Z2, ⊕2) × (Z2, ⊕2),
(0, 1)2 = (0 ⊕2 0, 1 ⊕2 1) = (0, 0), (1, 0)2 = (0, 0) and
(1, 1)2 = (0, 0), the order of all non identity elements are 2
and hence they do not generate (Z2, ⊕2) × (Z2, ⊕2).
Therefore, (Z2, ⊕2) × (Z2, ⊕2) is not cyclic.
4. Let G1 and G2 be two groups and the orders of
a1 ∈ G1 and a2 ∈ G2 are finite. Then the order of (a1, a2)
is the least common multiple of O(a1) and O(a2).
Proof For any positive integer n, (a1, a2)n = (a1n, a2n)
(verify by induction on n).
Hence
(a1, a2)n = (e1, e2) if and only if (a1n, a2n) = (e1, e2)
if and only if a1n = e1 and a2n = e2
if and only if O(a1) | n and O(a2) | n
if and only if n is a common multiple of O(a1) and O(a2).
Since in G1 × G2, the order of (a1, a2) is the
smallest positive integer n such that
(a1, a2)n = (e1, e2), the order of (a1, a2)
is the smallest common multiple of O(a1) and O(a2)
i.e. the least common multiple of O(a1) and O(a2).
5. Let G1 and G2 be two groups of order m and n.
Then G1 × G2 is a cyclic group if and only if (m, n) =1
i.e. m and n are relatively prime and G1 and G2 are cyclic.
Proof Since G1 and G2 are two groups of order m and n,
G1 × G2 is a group of order mn.
Suppose G1 × G2 is a cyclic group and
(a1, a2) is a generator of G1 × G2.
Then the least common multiple of O(a1) and O(a2)
= the order of (a1, a2) = mn…..(1).
Claim : G1 and G2 are cyclic groups generated by a1 and a2.
For, suppose not. Then either G1 ≠ (a1) or G2 ≠ (a2).
Since O(ai) ≤ O(Gi), i = 1, 2,
either O(a1) < m or O(a2) < n.
Hence O(a1)O(a2) < mn and hence
the lcm of O(a1) and O(a2) ≤ O(a1)O(a2) < mn,
which is a contradiction to (1).
Hence G1 = (a1) and G2 = (a2) and O(a1) = O(G1) = m
and O(a2) = O(G2) = n.
Therefore, by(1), lcm of m, n = mn.
Since lcm{m, n} × gcd(m, n) = mn,
we get gcd(m, n) = 1
i.e. m and n are relatively prime.
Conversely, suppose m and n are relatively prime
and G1 and G2 are cyclic.
Suppose G1= (a1) and G2 = (a2).
Then O(a1) = m and O(a2) = n and
O(a1) and O(a2) are relatively prime.
Hence, the order of (a1, a2)
= the least common multiple of O(a1) and O(a2)
= mn.
Hence the order of the cyclic subgroup generated
by (a1, a2) = mn = O(G1 × G2).
Hence G1 × G2 is a cyclic group generated by (a1, a2).
Note G1 × G2 is a cyclic group generated by (a1, a2)
if and only if G1 and G2 are cyclic groups generated
by a1 and a2 and O(G1) and O(G2) are relatively prime.
Comments
Post a Comment