AATA Cyclic Subgroups

Section 4.1 Cyclic Subgroups

Often a subgroup will depend entirely on a single element of the group; that is, knowing that particular element will allow us to compute any other element in the subgroup.

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Example 4.1.

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Suppose that we consider

3 \in Z

and look at all multiples (both positive and negative) of

3 .

As a set, this is

3 Z = {\dots, - 3, 0, 3, 6, \dots} .

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It is easy to see that

3 Z

is a subgroup of the integers. This subgroup is completely determined by the element

3

since we can obtain all of the other elements of the group by taking multiples of

3 .

Every element in the subgroup is “generated” by

3 .

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Example 4.2.

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H = {2^{n} : n \in Z},

then

H

is a subgroup of the multiplicative group of nonzero rational numbers,

Q^{*} .

a = 2^{m}

and

b = 2^{n}

are in

H,

then

a b^{- 1} = 2^{m} 2^{- n} = 2^{m - n}

is also in

H .

By Proposition 3.31,

H

is a subgroup of

Q^{*}

determined by the element

2 .

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Theorem 4.3.

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Let

G

be a group and

a

be any element in

G .

Then the set

⟨ a ⟩ = {a^{k} : k \in Z}

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is a subgroup of

G .

Furthermore,

⟨ a ⟩

is the smallest subgroup of

G

that contains

a .

Proof.

The identity is in

⟨ a ⟩

since

a^{0} = e .

g

and

h

are any two elements in

⟨ a ⟩,

then by the definition of

⟨ a ⟩

we can write

g = a^{m}

and

h = a^{n}

for some integers

m

and

n .

g h = a^{m} a^{n} = a^{m + n}

is again in

⟨ a ⟩ .

Finally, if

g = a^{n}

⟨ a ⟩,

then the inverse

g^{- 1} = a^{- n}

is also in

⟨ a ⟩ .

Clearly, any subgroup

H

G

containing

a

must contain all the powers of

a

by closure; hence,

H

contains

⟨ a ⟩ .

Therefore,

⟨ a ⟩

is the smallest subgroup of

G

containing

a .

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Remark 4.4.

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If we are using the “+” notation, as in the case of the integers under addition, we write

⟨ a ⟩ = {n a : n \in Z} .

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For

a \in G,

we call

⟨ a ⟩

the cyclic subgroup generated by

a .

G

contains some element

a

such that

G = ⟨ a ⟩,

then

G

is a cyclic group. In this case

a

is a generator of

G .

a

is an element of a group

G,

we define the order of

a

to be the smallest positive integer

n

such that

a^{n} = e,

and we write

| a | = n .

If there is no such integer

n,

we say that the order of

a

is infinite and write

| a | = \infty

to denote the order of

a .

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Example 4.5.

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Notice that a cyclic group can have more than a single generator. Both

1

and

5

generate

Z_{6};

hence,

Z_{6}

is a cyclic group. Not every element in a cyclic group is necessarily a generator of the group. The order of

2 \in Z_{6}

3 .

The cyclic subgroup generated by

2

⟨ 2 ⟩ = {0, 2, 4} .

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The groups

Z

and

Z_{n}

are cyclic groups. The elements

1

and

- 1

are generators for

Z .

We can certainly generate

Z_{n}

with 1 although there may be other generators of

Z_{n},

as in the case of

Z_{6} .

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Example 4.6.

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The group of units,

U (9),

Z_{9}

is a cyclic group. As a set,

U (9)

{1, 2, 4, 5, 7, 8} .

The element 2 is a generator for

U (9)

since

\begin{aligned} 2^{1} & = 2 2^{2} = 4 \\ 2^{3} & = 8 2^{4} = 7 \\ 2^{5} & = 5 2^{6} = 1 . \end{aligned}

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Example 4.7.

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Not every group is a cyclic group. Consider the symmetry group of an equilateral triangle

S_{3} .

The multiplication table for this group is Figure 3.7. The subgroups of

S_{3}

are shown in Figure 4.8. Notice that every subgroup is cyclic; however, no single element generates the entire group.

The lattice of subgroups for S-3: the top is S-3, the second line is the identity, rho-1, rho-2; the identity, mu-1; the identity, mu-2; the identity, mu-3, and the bottom is the identity subgroup. — Figure 4.8. Subgroups of $S_{3}$

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Theorem 4.9.

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Every cyclic group is abelian.

Proof.

Let

G

be a cyclic group and

a \in G

be a generator for

G .

g

and

h

are in

G,

then they can be written as powers of

a,

say

g = a^{r}

and

h = a^{s} .

Since

g h = a^{r} a^{s} = a^{r + s} = a^{s + r} = a^{s} a^{r} = h g,

G

is abelian.

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Subsection Subgroups of Cyclic Groups

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We can ask some interesting questions about cyclic subgroups of a group and subgroups of a cyclic group. If

G

is a group, which subgroups of

G

are cyclic? If

G

is a cyclic group, what type of subgroups does

G

possess?

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Theorem 4.10.

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Every subgroup of a cyclic group is cyclic.

Proof.

The main tools used in this proof are the division algorithm and the Principle of Well-Ordering. Let

G

be a cyclic group generated by

a

and suppose that

H

is a subgroup of

G .

H = {e},

then trivially

H

is cyclic. Suppose that

H

contains some other element

g

distinct from the identity. Then

g

can be written as

a^{n}

for some integer

n .

Since

H

is a subgroup,

g^{- 1} = a^{- n}

must also be in

H .

Since either

n

- n

is positive, we can assume that

H

contains positive powers of

a

and

n > 0 .

Let

m

be the smallest natural number such that

a^{m} \in H .

Such an

m

exists by the Principle of Well-Ordering.

We claim that

h = a^{m}

is a generator for

H .

We must show that every

h^{'} \in H

can be written as a power of

h .

Since

h^{'} \in H

and

H

is a subgroup of

G,

h^{'} = a^{k}

for some integer

k .

Using the division algorithm, we can find numbers

q

and

r

such that

k = m q + r

where

0 \leq r < m;

hence,

a^{k} = a^{m q + r} = (a^{m})^{q} a^{r} = h^{q} a^{r} .

a^{r} = a^{k} h^{- q} .

Since

a^{k}

and

h^{- q}

are in

H,

a^{r}

must also be in

H .

However,

m

was the smallest positive number such that

a^{m}

was in

H;

consequently,

r = 0

and so

k = m q .

Therefore,

h^{'} = a^{k} = a^{m q} = h^{q}

and

H

is generated by

h .

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Corollary 4.11.

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The subgroups of

Z

are exactly

n Z

for

n = 0, 1, 2, \dots .

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Proposition 4.12.

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Let

G

be a cyclic group of order

n

and suppose that

a

is a generator for

G .

Then

a^{k} = e

if and only if

n

divides

k .

Proof.

First suppose that

a^{k} = e .

By the division algorithm,

k = n q + r

where

0 \leq r < n;

hence,

e = a^{k} = a^{n q + r} = a^{n q} a^{r} = e a^{r} = a^{r} .

Since the smallest positive integer

m

such that

a^{m} = e

n,

r = 0 .

Conversely, if

n

divides

k,

then

k = n s

for some integer

s .

Consequently,

a^{k} = a^{n s} = (a^{n})^{s} = e^{s} = e .

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Theorem 4.13.

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Let

G

be a cyclic group of order

n

and suppose that

a \in G

is a generator of the group. If

b = a^{k},

then the order of

b

n / d,

where

d = gcd (k, n) .

Proof.

We wish to find the smallest integer

m

such that

e = b^{m} = a^{k m} .

By Proposition 4.12, this is the smallest integer

m

such that

n

divides

k m

or, equivalently,

n / d

divides

m (k / d) .

Since

d

is the greatest common divisor of

n

and

k,

n / d

and

k / d

are relatively prime. Hence, for

n / d

to divide

m (k / d)

it must divide

m .

The smallest such

m

n / d .

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Corollary 4.14.

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The generators of

Z_{n}

are the integers

r

such that

1 \leq r < n

and

gcd (r, n) = 1 .

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Example 4.15.

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Let us examine the group

Z_{16} .

The numbers

1,

3,

5,

7,

9,

11,

13,

and

15

are the elements of

Z_{16}

that are relatively prime to

16 .

Each of these elements generates

Z_{16} .

For example,

\begin{aligned} 1 \cdot 9 & = 9 & 2 \cdot 9 & = 2 & 3 \cdot 9 & = 11 \\ 4 \cdot 9 & = 4 & 5 \cdot 9 & = 13 & 6 \cdot 9 & = 6 \\ 7 \cdot 9 & = 15 & 8 \cdot 9 & = 8 & 9 \cdot 9 & = 1 \\ 10 \cdot 9 & = 10 & 11 \cdot 9 & = 3 & 12 \cdot 9 & = 12 \\ 13 \cdot 9 & = 5 & 14 \cdot 9 & = 14 & 15 \cdot 9 & = 7 . \end{aligned}

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