AATA Finite Abelian Groups

Section 13.1 Finite Abelian Groups

In our investigation of cyclic groups we found that every group of prime order was isomorphic to

Z_{p},

where

p

was a prime number. We also determined that

Z_{m n} ≅ Z_{m} \times Z_{n}

when

gcd (m, n) = 1 .

In fact, much more is true. Every finite abelian group is isomorphic to a direct product of cyclic groups of prime power order; that is, every finite abelian group is isomorphic to a group of the type

Z_{p_{1}^{α_{1}}} \times \dots \times Z_{p_{n}^{α_{n}}},

🔗

where each

p_{k}

is prime (not necessarily distinct).

🔗

First, let us examine a slight generalization of finite abelian groups. Suppose that

G

is a group and let

{g_{i}}

be a set of elements in

G,

where

i

is in some index set

I

(not necessarily finite). The smallest subgroup of

G

containing all of the

g_{i}

’s is the subgroup of

G

generated by the

g_{i}

’s. If this subgroup of

G

is in fact all of

G,

then

G

is generated by the set

{g_{i} : i \in I} .

In this case the

g_{i}

’s are said to be the generators of

G .

If there is a finite set

{g_{i} : i \in I}

that generates

G,

then

G

is finitely generated.

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Example 13.1.

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Obviously, all finite groups are finitely generated. For example, the group

S_{3}

is generated by the permutations

(1 2)

and

(1 2 3) .

The group

Z \times Z_{n}

is an infinite group but is finitely generated by

{(1, 0), (0, 1)} .

🔗

Example 13.2.

🔗

Not all groups are finitely generated. Consider the rational numbers

Q

under the operation of addition. Suppose that

Q

is finitely generated with generators

p_{1} / q_{1}, \dots, p_{n} / q_{n},

where each

p_{i} / q_{i}

is a fraction expressed in its lowest terms. Let

p

be some prime that does not divide any of the denominators

q_{1}, \dots, q_{n} .

We claim that

1 / p

cannot be in the subgroup of

Q

that is generated by

p_{1} / q_{1}, \dots, p_{n} / q_{n},

since

p

does not divide the denominator of any element in this subgroup. This fact is easy to see since the sum of any two generators is

p_{i} / q_{i} + p_{j} / q_{j} = (p_{i} q_{j} + p_{j} q_{i}) / (q_{i} q_{j}) .

🔗

Proposition 13.3.

🔗

Let

H

be the subgroup of a group

G

that is generated by

{g_{i} \in G : i \in I} .

Then

h \in H

exactly when it is a product of the form

h = g_{i_{1}}^{α_{1}} \dots g_{i_{n}}^{α_{n}},

🔗

where the

g_{i_{k}}

s are not necessarily distinct.

Proof.

Let

K

be the set of all products of the form

g_{i_{1}}^{α_{1}} \dots g_{i_{n}}^{α_{n}},

where the

g_{i_{k}}

s are not necessarily distinct. Certainly

K

is a subset of

H .

We need only show that

K

is a subgroup of

G .

If this is the case, then

K = H,

since

H

is the smallest subgroup containing all the

g_{i}

Clearly, the set

K

is closed under the group operation. Since

g_{i}^{0} = 1,

the identity is in

K .

It remains to show that the inverse of an element

g = g_{i_{1}}^{k_{1}} \dots g_{i_{n}}^{k_{n}}

K

must also be in

K .

However,

g^{- 1} = (g_{i_{1}}^{k_{1}} \dots g_{i_{n}}^{k_{n}})^{- 1} = (g_{i_{n}}^{- k_{n}} \dots g_{i_{1}}^{- k_{1}}) .

🔗

The reason that powers of a fixed

g_{i}

may occur several times in the product is that we may have a nonabelian group. However, if the group is abelian, then the

g_{i}

s need occur only once. For example, a product such as

a^{- 3} b^{5} a^{7}

in an abelian group could always be simplified (in this case, to

a^{4} b^{5}

🔗

Now let us restrict our attention to finite abelian groups. We can express any finite abelian group as a finite direct product of cyclic groups. More specifically, letting

p

be prime, we define a group

G

to be a

p

-group if every element in

G

has as its order a power of

p .

For example, both

Z_{2} \times Z_{2}

and

Z_{4}

are

2

-groups, whereas

Z_{27}

is a

3

-group. We shall prove the Fundamental Theorem of Finite Abelian Groups which tells us that every finite abelian group is isomorphic to a direct product of cyclic

p

-groups.

🔗

Theorem 13.4. Fundamental Theorem of Finite Abelian Groups.

🔗

Every finite abelian group

G

is isomorphic to a direct product of cyclic groups of the form

Z_{p_{1}^{α_{1}}} \times Z_{p_{2}^{α_{2}}} \times \dots \times Z_{p_{n}^{α_{n}}}

🔗

here the

p_{i}

’s are primes (not necessarily distinct).

🔗

Example 13.5.

🔗

Suppose that we wish to classify all abelian groups of order

540 = 2^{2} \cdot 3^{3} \cdot 5 .

The Fundamental Theorem of Finite Abelian Groups tells us that we have the following six possibilities.

$Z_{2} \times Z_{2} \times Z_{3} \times Z_{3} \times Z_{3} \times Z_{5};$
$Z_{2} \times Z_{2} \times Z_{3} \times Z_{9} \times Z_{5};$
$Z_{2} \times Z_{2} \times Z_{27} \times Z_{5};$
$Z_{4} \times Z_{3} \times Z_{3} \times Z_{3} \times Z_{5};$
$Z_{4} \times Z_{3} \times Z_{9} \times Z_{5};$
$Z_{4} \times Z_{27} \times Z_{5} .$

🔗

The proof of the Fundamental Theorem of Finite Abelian Groups depends on several lemmas.

🔗

Lemma 13.6.

🔗

Let

G

be a finite abelian group of order

n .

p

is a prime that divides

n,

then

G

contains an element of order

p .

Proof.

We will prove this lemma by induction. If

n = 1,

then there is nothing to show. Now suppose that the lemma is true for all groups of order

k,

where

k < n .

Furthermore, let

p

be a prime that divides

n .

G

has no proper nontrivial subgroups, then

G = ⟨ a ⟩,

where

a

is any element other than the identity. By Exercise 4.5.39, the order of

G

must be prime. Since

p

divides

n,

we know that

p = n,

and

G

contains

p - 1

elements of order

p .

Now suppose that

G

contains a nontrivial proper subgroup

H .

Then

1 < | H | < n .

p ∣ | H |,

then

H

contains an element of order

p

by induction and the lemma is true. Suppose that

p

does not divide the order of

H .

Since

G

is abelian, it must be the case that

H

is a normal subgroup of

G,

and

| G | = | H | \cdot | G / H | .

Consequently,

p

must divide

| G / H | .

Since

| G / H | < | G | = n,

we know that

G / H

contains an element

a H

of order

p

by the induction hypothesis. Thus,

H = (a H)^{p} = a^{p} H,

and

a^{p} \in H

but

a \notin H .

| H | = r,

then

p

and

r

are relatively prime, and there exist integers

s

and

t

such that

s p + t r = 1 .

Furthermore, the order of

a^{p}

must divide

r,

and

(a^{p})^{r} = (a^{r})^{p} = 1 .

We claim that

a^{r}

has order

p .

We must show that

a^{r} \neq 1 .

Suppose

a^{r} = 1 .

Then

\begin{aligned} a & = a^{s p + t r} \\ = a^{s p} a^{t r} \\ = (a^{p})^{s} (a^{r})^{t} \\ = (a^{p})^{s} 1 \\ = (a^{p})^{s} . \end{aligned}

Since

a^{p} \in H,

it must be the case that

a = (a^{p})^{s} \in H,

which is a contradiction. Therefore,

a^{r} \neq 1

is an element of order

p

G .

🔗

Lemma 13.6 is a special case of Cauchy’s Theorem (Theorem 15.1), which states that if

G

is a finite group and

p

a prime such that

p

divides the order of

G,

then

G

contains a subgroup of order

p .

We will prove Cauchy’s Theorem in Chapter 15.

🔗

Lemma 13.7.

🔗

A finite abelian group is a

p

-group if and only if its order is a power of

p .

Proof.

| G | = p^{n}

then by Lagrange’s theorem, then the order of any

g \in G

must divide

p^{n},

and therefore must be a power of

p .

Conversely, if

| G |

is not a power of

p,

then it has some other prime divisor

q,

so by Lemma 13.6,

G

has an element of order

q

and thus is not a

p

-group.

🔗

Lemma 13.8.

🔗

Let

G

be a finite abelian group of order

n = p_{1}^{α_{1}} \dots p_{k}^{α_{k}},

where where

p_{1}, \dots, p_{k}

are distinct primes and

α_{1}, α_{2}, \dots, α_{k}

are positive integers. Then

G

is the internal direct product of subgroups

G_{1}, G_{2}, \dots, G_{k},

where

G_{i}

is the subgroup of

G

consisting of all elements of order

p_{i}^{r}

for some integer

r .

Proof.

Since

G

is an abelian group, we are guaranteed that

G_{i}

is a subgroup of

G

for

i = 1, \dots, k .

Since the identity has order

p_{i}^{0} = 1,

we know that

1 \in G_{i} .

g \in G_{i}

has order

p_{i}^{r},

then

g^{- 1}

must also have order

p_{i}^{r} .

Finally, if

h \in G_{i}

has order

p_{i}^{s},

then

(g h)^{p_{i}^{t}} = g^{p_{i}^{t}} h^{p_{i}^{t}} = 1 \cdot 1 = 1,

where

t

is the maximum of

r

and

s .

We must show that

G = G_{1} G_{2} \dots G_{k}

and

G_{i} \cap G_{j} = {1}

for

i \neq j .

Suppose that

g_{1} \in G_{1}

is in the subgroup generated by

G_{2}, G_{3}, \dots, G_{k} .

Then

g_{1} = g_{2} g_{3} \dots g_{k}

for

g_{i} \in G_{i} .

Since

g_{i}

has order

p_{i}^{α_{i}},

we know that

g_{i}^{p^{α_{i}}} = 1

for

i = 2, 3, \dots, k,

and

g_{1}^{p_{2}^{α_{2}} \dots p_{k}^{α_{k}}} = 1 .

Since the order of

g_{1}

is a power of

p_{1}

and

gcd (p_{1}, p_{2}^{α_{2}} \dots p_{k}^{α_{k}}) = 1,

it must be the case that

g_{1} = 1

and the intersection of

G_{1}

with any of the subgroups

G_{2}, G_{3}, \dots, G_{k}

is the identity. A similar argument shows that

G_{i} \cap G_{j} = {1}

for

i \neq j .

Next, we must show that it possible to write every

g \in G

as a product

g_{1} \dots g_{k},

where

g_{i} \in G_{i} .

Since the order of

g

divides the order of

G,

we know that

| g | = p_{1}^{β_{1}} p_{2}^{β_{2}} \dots p_{k}^{β_{k}}

for some integers

β_{1}, \dots, β_{k} .

Letting

a_{i} = | g | / p_{i}^{β_{i}},

the

a_{i}

’s are relatively prime; hence, there exist integers

b_{1}, \dots, b_{k}

such that

a_{1} b_{1} + \dots + a_{k} b_{k} = 1 .

Consequently,

g = g^{a_{1} b_{1} + \dots + a_{k} b_{k}} = g^{a_{1} b_{1}} \dots g^{a_{k} b_{k}} .

Since

g^{(a_{i} b_{i}) p_{i}^{β_{i}}} = g^{b_{i} | g |} = e,

it follows that

g^{a_{i} b_{i}}

must be in

G_{i} .

Let

g_{i} = g^{a_{i} b_{i}} .

Then

g = g_{1} \dots g_{k} \in G_{1} G_{2} \dots G_{k} .

Therefore,

G = G_{1} G_{2} \dots G_{k}

is an internal direct product of subgroups.

🔗

If remains for us to determine the possible structure of each

p_{i}

-group

G_{i}

in Lemma 13.8.

🔗

Lemma 13.9.

🔗

Let

G

be a finite abelian

p

-group and suppose that

g \in G

has maximal order. Then

G

is isomorphic to

⟨ g ⟩ \times H

for some subgroup

H

G .

Proof.

By Lemma 13.7, we may assume that the order of

G

p^{n} .

We shall induct on

n .

n = 1,

then

G

is cyclic of order

p

and must be generated by

g .

Suppose now that the statement of the lemma holds for all integers

k

with

1 \leq k < n

and let

g

be of maximal order in

G,

say

| g | = p^{m} .

Then

a^{p^{m}} = e

for all

a \in G .

Now choose

h

G

such that

h \notin ⟨ g ⟩,

where

h

has the smallest possible order. Certainly such an

h

exists; otherwise,

G = ⟨ g ⟩

and we are done. Let

H = ⟨ h ⟩ .

We claim that

⟨ g ⟩ \cap H = {e} .

It suffices to show that

| H | = p .

Since

| h^{p} | = | h | / p,

the order of

h^{p}

is smaller than the order of

h

and must be in

⟨ g ⟩

by the minimality of

h;

that is,

h^{p} = g^{r}

for some number

r .

Hence,

(g^{r})^{p^{m - 1}} = (h^{p})^{p^{m - 1}} = h^{p^{m}} = e,

and the order of

g^{r}

must be less than or equal to

p^{m - 1} .

Therefore,

g^{r}

cannot generate

⟨ g ⟩ .

Notice that

p

must occur as a factor of

r,

say

r = p s,

and

h^{p} = g^{r} = g^{p s} .

Define

a

to be

g^{- s} h .

Then

a

cannot be in

⟨ g ⟩;

otherwise,

h

would also have to be in

⟨ g ⟩ .

Also,

a^{p} = g^{- s p} h^{p} = g^{- r} h^{p} = h^{- p} h^{p} = e .

We have now formed an element

a

with order

p

such that

a \notin ⟨ g ⟩ .

Since

h

was chosen to have the smallest order of all of the elements that are not in

⟨ g ⟩,

| H | = p .

Now we will show that the order of

g H

in the factor group

G / H

must be the same as the order of

g

G .

| g H | < | g | = p^{m},

then

H = (g H)^{p^{m - 1}} = g^{p^{m - 1}} H;

hence,

g^{p^{m - 1}}

must be in

⟨ g ⟩ \cap H = {e},

which contradicts the fact that the order of

g

p^{m} .

Therefore,

g H

must have maximal order in

G / H .

By the Correspondence Theorem and our induction hypothesis,

G / H ≅ ⟨ g H ⟩ \times K / H

for some subgroup

K

G

containing

H .

We claim that

⟨ g ⟩ \cap K = {e} .

b \in ⟨ g ⟩ \cap K,

then

b H \in ⟨ g H ⟩ \cap K / H = {H}

and

b \in ⟨ g ⟩ \cap H = {e} .

It follows that

G = ⟨ g ⟩ K

implies that

G ≅ ⟨ g ⟩ \times K .

🔗

The proof of the Fundamental Theorem of Finite Abelian Groups follows very quickly from Lemma 13.8 and Lemma 13.9. By Lemma 13.8,

G

is a product of

p

-groups. Suppose

G

is a

p

-group and let

g

be an element of maximal order in

G .

⟨ g ⟩ = G,

then we are done; otherwise,

G ≅ Z_{| g |} \times H

for some subgroup

H

contained in

G

by the Lemma 13.9. Since

| H | < | G |,

we can apply mathematical induction.

🔗

We now state the more general theorem for all finitely generated abelian groups. The proof of this theorem can be found in any of the references at the end of this chapter.

🔗

Theorem 13.10. The Fundamental Theorem of Finitely Generated Abelian Groups.

🔗

Every finitely generated abelian group

G

is isomorphic to a direct product of cyclic groups of the form

Z_{p_{1}^{α_{1}}} \times Z_{p_{2}^{α_{2}}} \times \dots \times Z_{p_{n}^{α_{n}}} \times Z \times \dots \times Z,

🔗

where the

p_{i}

’s are primes (not necessarily distinct).

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