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Direct products Finitely generated abelian groups

Section 11 – Direct products and finitely generated abelian groups

Instructor: Yifan Yang Fall 2006

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Outline

1

Direct products

2

Finitely generated abelian groups

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Cartesian product

Definition The Cartesian product of sets S1, S2, . . . , Sn is the set of

• rdered n-tuples (a1, a2, . . . , an), where ai ∈ Si for

i = 1, 2, . . . , n. It is denoted by either S1 × S2 × . . . × Sn

• r

n

• i=1

Si.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Direct products

Theorem Let G1, G2, . . . , Gn be groups. For (a1, a2, . . . , an) and (b1, b2, . . . , bn) in n

i=1 Gi define (a1, a2, . . . , an)(b1, b2, . . . , bn)

to be (a1b1, a2b2, . . . , anbn). Then n

i=1 Gi is a group under this

binary operation. Proof.

• Straightforward. See the textbook.

Definition The group n

i=1 Gi defined above is called the direct product of

the groups Gi.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Direct products

Theorem Let G1, G2, . . . , Gn be groups. For (a1, a2, . . . , an) and (b1, b2, . . . , bn) in n

i=1 Gi define (a1, a2, . . . , an)(b1, b2, . . . , bn)

to be (a1b1, a2b2, . . . , anbn). Then n

i=1 Gi is a group under this

binary operation. Proof.

• Straightforward. See the textbook.

Definition The group n

i=1 Gi defined above is called the direct product of

the groups Gi.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Direct products

Theorem Let G1, G2, . . . , Gn be groups. For (a1, a2, . . . , an) and (b1, b2, . . . , bn) in n

i=1 Gi define (a1, a2, . . . , an)(b1, b2, . . . , bn)

to be (a1b1, a2b2, . . . , anbn). Then n

i=1 Gi is a group under this

binary operation. Proof.

• Straightforward. See the textbook.

Definition The group n

i=1 Gi defined above is called the direct product of

the groups Gi.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Direct sums

When the groups Gi are all abelian with operation +, we sometimes use the notation G1 ⊕ G2 · · · ⊕ Gn = ⊕n

i=1Gi

instead of G1 × G2 × · · · Gn = n

i=1 Gi. Also, we call the group

⊕n

i=1Gi the direct sum of the groups Gi.

Remark The changing of the order of the factors in a direct product yields a group isomorphic to the original one. For example, define φ : G1 × G2 → G2 × G1 by φ((g1, g2)) = (g2, g1). It is easy to verify that φ is an isomorphism.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Direct sums

When the groups Gi are all abelian with operation +, we sometimes use the notation G1 ⊕ G2 · · · ⊕ Gn = ⊕n

i=1Gi

instead of G1 × G2 × · · · Gn = n

i=1 Gi. Also, we call the group

⊕n

i=1Gi the direct sum of the groups Gi.

Remark The changing of the order of the factors in a direct product yields a group isomorphic to the original one. For example, define φ : G1 × G2 → G2 × G1 by φ((g1, g2)) = (g2, g1). It is easy to verify that φ is an isomorphism.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Direct sums

When the groups Gi are all abelian with operation +, we sometimes use the notation G1 ⊕ G2 · · · ⊕ Gn = ⊕n

i=1Gi

instead of G1 × G2 × · · · Gn = n

i=1 Gi. Also, we call the group

⊕n

i=1Gi the direct sum of the groups Gi.

Remark The changing of the order of the factors in a direct product yields a group isomorphic to the original one. For example, define φ : G1 × G2 → G2 × G1 by φ((g1, g2)) = (g2, g1). It is easy to verify that φ is an isomorphism.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Example

Consider Z2 × Z3. It has 6 elements, namely (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), and (1, 2). Since it is abelian, it must be isomorphic to Z6. (A group of order 6 is isomorphic to either Z6

• r S3.) In fact, if we set g = (1, 1), then we have

2g = (1, 1) + (1, 1) = (0, 2) 3g = 2g + (1, 1) = (0, 2) + (1, 1) = (1, 0) 4g = 3g + (1, 1) = (1, 0) + (1, 1) = (0, 1) 5g = 4g + (1, 1) = (0, 1) + (1, 1) = (1, 2) 6g = 5g + (1, 1) = (1, 2) + (1, 1) = (0, 0). In other words, Z2 × Z3 is cyclic of order 6, and hence isomorphic to Z6.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Example

Consider Z2 × Z3. It has 6 elements, namely (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), and (1, 2). Since it is abelian, it must be isomorphic to Z6. (A group of order 6 is isomorphic to either Z6

• r S3.) In fact, if we set g = (1, 1), then we have

2g = (1, 1) + (1, 1) = (0, 2) 3g = 2g + (1, 1) = (0, 2) + (1, 1) = (1, 0) 4g = 3g + (1, 1) = (1, 0) + (1, 1) = (0, 1) 5g = 4g + (1, 1) = (0, 1) + (1, 1) = (1, 2) 6g = 5g + (1, 1) = (1, 2) + (1, 1) = (0, 0). In other words, Z2 × Z3 is cyclic of order 6, and hence isomorphic to Z6.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Example

Consider Z2 × Z3. It has 6 elements, namely (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), and (1, 2). Since it is abelian, it must be isomorphic to Z6. (A group of order 6 is isomorphic to either Z6

• r S3.) In fact, if we set g = (1, 1), then we have

2g = (1, 1) + (1, 1) = (0, 2) 3g = 2g + (1, 1) = (0, 2) + (1, 1) = (1, 0) 4g = 3g + (1, 1) = (1, 0) + (1, 1) = (0, 1) 5g = 4g + (1, 1) = (0, 1) + (1, 1) = (1, 2) 6g = 5g + (1, 1) = (1, 2) + (1, 1) = (0, 0). In other words, Z2 × Z3 is cyclic of order 6, and hence isomorphic to Z6.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Example

Consider Z2 × Z3. It has 6 elements, namely (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), and (1, 2). Since it is abelian, it must be isomorphic to Z6. (A group of order 6 is isomorphic to either Z6

• r S3.) In fact, if we set g = (1, 1), then we have

2g = (1, 1) + (1, 1) = (0, 2) 3g = 2g + (1, 1) = (0, 2) + (1, 1) = (1, 0) 4g = 3g + (1, 1) = (1, 0) + (1, 1) = (0, 1) 5g = 4g + (1, 1) = (0, 1) + (1, 1) = (1, 2) 6g = 5g + (1, 1) = (1, 2) + (1, 1) = (0, 0). In other words, Z2 × Z3 is cyclic of order 6, and hence isomorphic to Z6.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Example

Consider Z2 × Z3. It has 6 elements, namely (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), and (1, 2). Since it is abelian, it must be isomorphic to Z6. (A group of order 6 is isomorphic to either Z6

• r S3.) In fact, if we set g = (1, 1), then we have

2g = (1, 1) + (1, 1) = (0, 2) 3g = 2g + (1, 1) = (0, 2) + (1, 1) = (1, 0) 4g = 3g + (1, 1) = (1, 0) + (1, 1) = (0, 1) 5g = 4g + (1, 1) = (0, 1) + (1, 1) = (1, 2) 6g = 5g + (1, 1) = (1, 2) + (1, 1) = (0, 0). In other words, Z2 × Z3 is cyclic of order 6, and hence isomorphic to Z6.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Example

Consider Z2 × Z2. It has 4 elements. Thus, it is isomorphic to either Z4 or Z×

8 = {1, 3, 5, 7}. Now for all (a, b) ∈ Z2 × Z2, we

have 2(a, b) = (0, 0). Thus, there is no element having order 4. The group must be isomorphic to Z×

8 .

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Example

Consider Z2 × Z2. It has 4 elements. Thus, it is isomorphic to either Z4 or Z×

8 = {1, 3, 5, 7}. Now for all (a, b) ∈ Z2 × Z2, we

have 2(a, b) = (0, 0). Thus, there is no element having order 4. The group must be isomorphic to Z×

8 .

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Example

Consider Z2 × Z2. It has 4 elements. Thus, it is isomorphic to either Z4 or Z×

8 = {1, 3, 5, 7}. Now for all (a, b) ∈ Z2 × Z2, we

have 2(a, b) = (0, 0). Thus, there is no element having order 4. The group must be isomorphic to Z×

8 .

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Example

Consider Z2 × Z2. It has 4 elements. Thus, it is isomorphic to either Z4 or Z×

8 = {1, 3, 5, 7}. Now for all (a, b) ∈ Z2 × Z2, we

have 2(a, b) = (0, 0). Thus, there is no element having order 4. The group must be isomorphic to Z×

8 .

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Example

Consider Z2 × Z2. It has 4 elements. Thus, it is isomorphic to either Z4 or Z×

8 = {1, 3, 5, 7}. Now for all (a, b) ∈ Z2 × Z2, we

have 2(a, b) = (0, 0). Thus, there is no element having order 4. The group must be isomorphic to Z×

8 .

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

When is a direct sum of abelian groups cyclic?

Theorem (11.5) The group Zm × Zn is cyclic and isomorphic to Zmn if and only if m and n are relatively prime. Proof. Let (a, b) be an element in Zm × Zn. The order of a in Zm divides m, and the order of b in Zn divides n. Thus, if k is a multiple of lcm(m, n), then we have k(a, b) = (0, 0). If lcm(m, n) = mn, i.e., if gcd(m, n) = 1, then Zm × Zn cannot be cyclic. Conversely, assume that gcd(m, n) = 1. Consider the element (1, 1). If k(1, 1) = (0, 0), then k must be both a multiple of m and a multiple of n, i.e., lcm(m, n)|k. Since gcd(m, n) = 1, we have lcm(m, n) = mn. Thus |(1, 1)| = mn = |Zm × Zn|. Therefore Zm × Zn is cyclic and isomorphic to Zmn.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

When is a direct sum of abelian groups cyclic?

Theorem (11.5) The group Zm × Zn is cyclic and isomorphic to Zmn if and only if m and n are relatively prime. Proof. Let (a, b) be an element in Zm × Zn. The order of a in Zm divides m, and the order of b in Zn divides n. Thus, if k is a multiple of lcm(m, n), then we have k(a, b) = (0, 0). If lcm(m, n) = mn, i.e., if gcd(m, n) = 1, then Zm × Zn cannot be cyclic. Conversely, assume that gcd(m, n) = 1. Consider the element (1, 1). If k(1, 1) = (0, 0), then k must be both a multiple of m and a multiple of n, i.e., lcm(m, n)|k. Since gcd(m, n) = 1, we have lcm(m, n) = mn. Thus |(1, 1)| = mn = |Zm × Zn|. Therefore Zm × Zn is cyclic and isomorphic to Zmn.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

When is a direct sum of abelian groups cyclic?

Theorem (11.5) The group Zm × Zn is cyclic and isomorphic to Zmn if and only if m and n are relatively prime. Proof. Let (a, b) be an element in Zm × Zn. The order of a in Zm divides m, and the order of b in Zn divides n. Thus, if k is a multiple of lcm(m, n), then we have k(a, b) = (0, 0). If lcm(m, n) = mn, i.e., if gcd(m, n) = 1, then Zm × Zn cannot be cyclic. Conversely, assume that gcd(m, n) = 1. Consider the element (1, 1). If k(1, 1) = (0, 0), then k must be both a multiple of m and a multiple of n, i.e., lcm(m, n)|k. Since gcd(m, n) = 1, we have lcm(m, n) = mn. Thus |(1, 1)| = mn = |Zm × Zn|. Therefore Zm × Zn is cyclic and isomorphic to Zmn.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

When is a direct sum of abelian groups cyclic?

Theorem (11.5) The group Zm × Zn is cyclic and isomorphic to Zmn if and only if m and n are relatively prime. Proof. Let (a, b) be an element in Zm × Zn. The order of a in Zm divides m, and the order of b in Zn divides n. Thus, if k is a multiple of lcm(m, n), then we have k(a, b) = (0, 0). If lcm(m, n) = mn, i.e., if gcd(m, n) = 1, then Zm × Zn cannot be cyclic. Conversely, assume that gcd(m, n) = 1. Consider the element (1, 1). If k(1, 1) = (0, 0), then k must be both a multiple of m and a multiple of n, i.e., lcm(m, n)|k. Since gcd(m, n) = 1, we have lcm(m, n) = mn. Thus |(1, 1)| = mn = |Zm × Zn|. Therefore Zm × Zn is cyclic and isomorphic to Zmn.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

When is a direct sum of abelian groups cyclic?

Theorem (11.5) The group Zm × Zn is cyclic and isomorphic to Zmn if and only if m and n are relatively prime. Proof. Let (a, b) be an element in Zm × Zn. The order of a in Zm divides m, and the order of b in Zn divides n. Thus, if k is a multiple of lcm(m, n), then we have k(a, b) = (0, 0). If lcm(m, n) = mn, i.e., if gcd(m, n) = 1, then Zm × Zn cannot be cyclic. Conversely, assume that gcd(m, n) = 1. Consider the element (1, 1). If k(1, 1) = (0, 0), then k must be both a multiple of m and a multiple of n, i.e., lcm(m, n)|k. Since gcd(m, n) = 1, we have lcm(m, n) = mn. Thus |(1, 1)| = mn = |Zm × Zn|. Therefore Zm × Zn is cyclic and isomorphic to Zmn.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

When is a direct sum of abelian groups cyclic?

Theorem (11.5) The group Zm × Zn is cyclic and isomorphic to Zmn if and only if m and n are relatively prime. Proof. Let (a, b) be an element in Zm × Zn. The order of a in Zm divides m, and the order of b in Zn divides n. Thus, if k is a multiple of lcm(m, n), then we have k(a, b) = (0, 0). If lcm(m, n) = mn, i.e., if gcd(m, n) = 1, then Zm × Zn cannot be cyclic. Conversely, assume that gcd(m, n) = 1. Consider the element (1, 1). If k(1, 1) = (0, 0), then k must be both a multiple of m and a multiple of n, i.e., lcm(m, n)|k. Since gcd(m, n) = 1, we have lcm(m, n) = mn. Thus |(1, 1)| = mn = |Zm × Zn|. Therefore Zm × Zn is cyclic and isomorphic to Zmn.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

When is a direct sum of abelian groups cyclic?

Theorem (11.5) The group Zm × Zn is cyclic and isomorphic to Zmn if and only if m and n are relatively prime. Proof. Let (a, b) be an element in Zm × Zn. The order of a in Zm divides m, and the order of b in Zn divides n. Thus, if k is a multiple of lcm(m, n), then we have k(a, b) = (0, 0). If lcm(m, n) = mn, i.e., if gcd(m, n) = 1, then Zm × Zn cannot be cyclic. Conversely, assume that gcd(m, n) = 1. Consider the element (1, 1). If k(1, 1) = (0, 0), then k must be both a multiple of m and a multiple of n, i.e., lcm(m, n)|k. Since gcd(m, n) = 1, we have lcm(m, n) = mn. Thus |(1, 1)| = mn = |Zm × Zn|. Therefore Zm × Zn is cyclic and isomorphic to Zmn.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

When is a direct sum of abelian groups cyclic?

Theorem (11.5) The group Zm × Zn is cyclic and isomorphic to Zmn if and only if m and n are relatively prime. Proof. Let (a, b) be an element in Zm × Zn. The order of a in Zm divides m, and the order of b in Zn divides n. Thus, if k is a multiple of lcm(m, n), then we have k(a, b) = (0, 0). If lcm(m, n) = mn, i.e., if gcd(m, n) = 1, then Zm × Zn cannot be cyclic. Conversely, assume that gcd(m, n) = 1. Consider the element (1, 1). If k(1, 1) = (0, 0), then k must be both a multiple of m and a multiple of n, i.e., lcm(m, n)|k. Since gcd(m, n) = 1, we have lcm(m, n) = mn. Thus |(1, 1)| = mn = |Zm × Zn|. Therefore Zm × Zn is cyclic and isomorphic to Zmn.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

When is a direct sum of abelian groups cyclic?

Theorem (11.5) The group Zm × Zn is cyclic and isomorphic to Zmn if and only if m and n are relatively prime. Proof. Let (a, b) be an element in Zm × Zn. The order of a in Zm divides m, and the order of b in Zn divides n. Thus, if k is a multiple of lcm(m, n), then we have k(a, b) = (0, 0). If lcm(m, n) = mn, i.e., if gcd(m, n) = 1, then Zm × Zn cannot be cyclic. Conversely, assume that gcd(m, n) = 1. Consider the element (1, 1). If k(1, 1) = (0, 0), then k must be both a multiple of m and a multiple of n, i.e., lcm(m, n)|k. Since gcd(m, n) = 1, we have lcm(m, n) = mn. Thus |(1, 1)| = mn = |Zm × Zn|. Therefore Zm × Zn is cyclic and isomorphic to Zmn.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

When is a direct sum of abelian groups cyclic?

Corollary (11.6) The group n

i=1 Zmi is cyclic and isomorphic to Zm1m2...mn if and

• nly if the numbers mi are pairwise coprime.

Example

1

The group Z30 is isomorphic to Z2 × Z3 × Z5, also to Z6 × Z5, Z2 × Z15, and so on.

2

The group Z12 × Z21 is isomorphic to Z84 × Z3 because Z12 × Z21 ≃ Z12 × Z7 × Z3 ≃ Z84 × Z3.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

When is a direct sum of abelian groups cyclic?

Corollary (11.6) The group n

i=1 Zmi is cyclic and isomorphic to Zm1m2...mn if and

• nly if the numbers mi are pairwise coprime.

Example

1

The group Z30 is isomorphic to Z2 × Z3 × Z5, also to Z6 × Z5, Z2 × Z15, and so on.

2

The group Z12 × Z21 is isomorphic to Z84 × Z3 because Z12 × Z21 ≃ Z12 × Z7 × Z3 ≃ Z84 × Z3.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

When is a direct sum of abelian groups cyclic?

Corollary (11.6) The group n

i=1 Zmi is cyclic and isomorphic to Zm1m2...mn if and

• nly if the numbers mi are pairwise coprime.

Example

1

The group Z30 is isomorphic to Z2 × Z3 × Z5, also to Z6 × Z5, Z2 × Z15, and so on.

2

The group Z12 × Z21 is isomorphic to Z84 × Z3 because Z12 × Z21 ≃ Z12 × Z7 × Z3 ≃ Z84 × Z3.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

When is a direct sum of abelian groups cyclic?

Corollary (11.6) The group n

i=1 Zmi is cyclic and isomorphic to Zm1m2...mn if and

• nly if the numbers mi are pairwise coprime.

Example

1

The group Z30 is isomorphic to Z2 × Z3 × Z5, also to Z6 × Z5, Z2 × Z15, and so on.

2

The group Z12 × Z21 is isomorphic to Z84 × Z3 because Z12 × Z21 ≃ Z12 × Z7 × Z3 ≃ Z84 × Z3.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Order of an element in a direct product

Theorem (11.9) Let (a1, a2, . . . , an) ∈ n

i=1 Gi. If ai is of finite order ri in Gi, then

the order of (a1, a2, . . . , an) in n

i=1 Gi is equal to the least

common multiple of all ri. Proof. Since (a1, a2, . . . , an)k = (ak

1, ak 2, . . . , ak n), if

(a1, a2, . . . , an)k = (e1, e2, . . . , en), then ri|k for all i. That is, k must be a multiple of lcm(r1, r2, . . . , rn). This is a necessary and sufficient condition. Therefore, the order of (a1, a2, . . . , an) is lcm(r1, r2, . . . , rn).

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Order of an element in a direct product

Theorem (11.9) Let (a1, a2, . . . , an) ∈ n

i=1 Gi. If ai is of finite order ri in Gi, then

the order of (a1, a2, . . . , an) in n

i=1 Gi is equal to the least

common multiple of all ri. Proof. Since (a1, a2, . . . , an)k = (ak

1, ak 2, . . . , ak n), if

(a1, a2, . . . , an)k = (e1, e2, . . . , en), then ri|k for all i. That is, k must be a multiple of lcm(r1, r2, . . . , rn). This is a necessary and sufficient condition. Therefore, the order of (a1, a2, . . . , an) is lcm(r1, r2, . . . , rn).

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Examples

1

Consider (9, 5, 7) in the group Z12 × Z15 × Z20. The element 9 in Z12 is of order 12/ gcd(9, 12) = 4, the element 5 in Z15 has order 15/ gcd(5, 15) = 3, and the element 7 in Z20 has order 20/ gcd(7, 20) = 20. The least common multiple of 4, 3, and 20 is 60. Thus, the order of (9, 5, 7) is 60.

2

Consider (8, 4, 10) in the group Z12 × Z60 × Z24. The element 8 in Z12 has order 3 since gcd(8, 12) = 4, the element 4 in Z60 has order 15, and element 10 in Z24 has

• rder 12. The least common multiple of 3, 15, and 12 is 60.

Thus, the order of (8, 4, 10) is 60.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Examples

1

Consider (9, 5, 7) in the group Z12 × Z15 × Z20. The element 9 in Z12 is of order 12/ gcd(9, 12) = 4, the element 5 in Z15 has order 15/ gcd(5, 15) = 3, and the element 7 in Z20 has order 20/ gcd(7, 20) = 20. The least common multiple of 4, 3, and 20 is 60. Thus, the order of (9, 5, 7) is 60.

2

Consider (8, 4, 10) in the group Z12 × Z60 × Z24. The element 8 in Z12 has order 3 since gcd(8, 12) = 4, the element 4 in Z60 has order 15, and element 10 in Z24 has

• rder 12. The least common multiple of 3, 15, and 12 is 60.

Thus, the order of (8, 4, 10) is 60.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Examples

1

Consider (9, 5, 7) in the group Z12 × Z15 × Z20. The element 9 in Z12 is of order 12/ gcd(9, 12) = 4, the element 5 in Z15 has order 15/ gcd(5, 15) = 3, and the element 7 in Z20 has order 20/ gcd(7, 20) = 20. The least common multiple of 4, 3, and 20 is 60. Thus, the order of (9, 5, 7) is 60.

2

Consider (8, 4, 10) in the group Z12 × Z60 × Z24. The element 8 in Z12 has order 3 since gcd(8, 12) = 4, the element 4 in Z60 has order 15, and element 10 in Z24 has

• rder 12. The least common multiple of 3, 15, and 12 is 60.

Thus, the order of (8, 4, 10) is 60.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Examples

1

Consider (9, 5, 7) in the group Z12 × Z15 × Z20. The element 9 in Z12 is of order 12/ gcd(9, 12) = 4, the element 5 in Z15 has order 15/ gcd(5, 15) = 3, and the element 7 in Z20 has order 20/ gcd(7, 20) = 20. The least common multiple of 4, 3, and 20 is 60. Thus, the order of (9, 5, 7) is 60.

2

Consider (8, 4, 10) in the group Z12 × Z60 × Z24. The element 8 in Z12 has order 3 since gcd(8, 12) = 4, the element 4 in Z60 has order 15, and element 10 in Z24 has

• rder 12. The least common multiple of 3, 15, and 12 is 60.

Thus, the order of (8, 4, 10) is 60.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Examples

1

Consider (9, 5, 7) in the group Z12 × Z15 × Z20. The element 9 in Z12 is of order 12/ gcd(9, 12) = 4, the element 5 in Z15 has order 15/ gcd(5, 15) = 3, and the element 7 in Z20 has order 20/ gcd(7, 20) = 20. The least common multiple of 4, 3, and 20 is 60. Thus, the order of (9, 5, 7) is 60.

2

Consider (8, 4, 10) in the group Z12 × Z60 × Z24. The element 8 in Z12 has order 3 since gcd(8, 12) = 4, the element 4 in Z60 has order 15, and element 10 in Z24 has

• rder 12. The least common multiple of 3, 15, and 12 is 60.

Thus, the order of (8, 4, 10) is 60.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Examples

1

Consider (9, 5, 7) in the group Z12 × Z15 × Z20. The element 9 in Z12 is of order 12/ gcd(9, 12) = 4, the element 5 in Z15 has order 15/ gcd(5, 15) = 3, and the element 7 in Z20 has order 20/ gcd(7, 20) = 20. The least common multiple of 4, 3, and 20 is 60. Thus, the order of (9, 5, 7) is 60.

2

Consider (8, 4, 10) in the group Z12 × Z60 × Z24. The element 8 in Z12 has order 3 since gcd(8, 12) = 4, the element 4 in Z60 has order 15, and element 10 in Z24 has

• rder 12. The least common multiple of 3, 15, and 12 is 60.

Thus, the order of (8, 4, 10) is 60.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Examples

1

Consider (9, 5, 7) in the group Z12 × Z15 × Z20. The element 9 in Z12 is of order 12/ gcd(9, 12) = 4, the element 5 in Z15 has order 15/ gcd(5, 15) = 3, and the element 7 in Z20 has order 20/ gcd(7, 20) = 20. The least common multiple of 4, 3, and 20 is 60. Thus, the order of (9, 5, 7) is 60.

2

Consider (8, 4, 10) in the group Z12 × Z60 × Z24. The element 8 in Z12 has order 3 since gcd(8, 12) = 4, the element 4 in Z60 has order 15, and element 10 in Z24 has

• rder 12. The least common multiple of 3, 15, and 12 is 60.

Thus, the order of (8, 4, 10) is 60.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Examples

1

Consider (9, 5, 7) in the group Z12 × Z15 × Z20. The element 9 in Z12 is of order 12/ gcd(9, 12) = 4, the element 5 in Z15 has order 15/ gcd(5, 15) = 3, and the element 7 in Z20 has order 20/ gcd(7, 20) = 20. The least common multiple of 4, 3, and 20 is 60. Thus, the order of (9, 5, 7) is 60.

2

Consider (8, 4, 10) in the group Z12 × Z60 × Z24. The element 8 in Z12 has order 3 since gcd(8, 12) = 4, the element 4 in Z60 has order 15, and element 10 in Z24 has

• rder 12. The least common multiple of 3, 15, and 12 is 60.

Thus, the order of (8, 4, 10) is 60.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Examples

1

Consider (9, 5, 7) in the group Z12 × Z15 × Z20. The element 9 in Z12 is of order 12/ gcd(9, 12) = 4, the element 5 in Z15 has order 15/ gcd(5, 15) = 3, and the element 7 in Z20 has order 20/ gcd(7, 20) = 20. The least common multiple of 4, 3, and 20 is 60. Thus, the order of (9, 5, 7) is 60.

2

Consider (8, 4, 10) in the group Z12 × Z60 × Z24. The element 8 in Z12 has order 3 since gcd(8, 12) = 4, the element 4 in Z60 has order 15, and element 10 in Z24 has

• rder 12. The least common multiple of 3, 15, and 12 is 60.

Thus, the order of (8, 4, 10) is 60.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Examples

1

Consider (9, 5, 7) in the group Z12 × Z15 × Z20. The element 9 in Z12 is of order 12/ gcd(9, 12) = 4, the element 5 in Z15 has order 15/ gcd(5, 15) = 3, and the element 7 in Z20 has order 20/ gcd(7, 20) = 20. The least common multiple of 4, 3, and 20 is 60. Thus, the order of (9, 5, 7) is 60.

2

Consider (8, 4, 10) in the group Z12 × Z60 × Z24. The element 8 in Z12 has order 3 since gcd(8, 12) = 4, the element 4 in Z60 has order 15, and element 10 in Z24 has

• rder 12. The least common multiple of 3, 15, and 12 is 60.

Thus, the order of (8, 4, 10) is 60.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Examples

1

Consider (9, 5, 7) in the group Z12 × Z15 × Z20. The element 9 in Z12 is of order 12/ gcd(9, 12) = 4, the element 5 in Z15 has order 15/ gcd(5, 15) = 3, and the element 7 in Z20 has order 20/ gcd(7, 20) = 20. The least common multiple of 4, 3, and 20 is 60. Thus, the order of (9, 5, 7) is 60.

2

Consider (8, 4, 10) in the group Z12 × Z60 × Z24. The element 8 in Z12 has order 3 since gcd(8, 12) = 4, the element 4 in Z60 has order 15, and element 10 in Z24 has

• rder 12. The least common multiple of 3, 15, and 12 is 60.

Thus, the order of (8, 4, 10) is 60.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Examples

1

Consider (9, 5, 7) in the group Z12 × Z15 × Z20. The element 9 in Z12 is of order 12/ gcd(9, 12) = 4, the element 5 in Z15 has order 15/ gcd(5, 15) = 3, and the element 7 in Z20 has order 20/ gcd(7, 20) = 20. The least common multiple of 4, 3, and 20 is 60. Thus, the order of (9, 5, 7) is 60.

2

Consider (8, 4, 10) in the group Z12 × Z60 × Z24. The element 8 in Z12 has order 3 since gcd(8, 12) = 4, the element 4 in Z60 has order 15, and element 10 in Z24 has

• rder 12. The least common multiple of 3, 15, and 12 is 60.

Thus, the order of (8, 4, 10) is 60.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

In-class exercises

1

Determine whether Z18 × Z15 × Z12 is isomorphic to Z9 × Z45 × Z8.

2

Determine whether Z18 × Z15 × Z12 is isomorphic to Z9 × Z10 × Z36.

3

Find the order of (2, 6, 3) in Z12 × Z9 × Z15.

4

Find the order of (3, 4, 8) in Z12 × Z18 × Z20.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Finitely generated abelian groups

Definition An abelian group is finitely generated if it can be generated by a finite number of elements. Theorem (Fundamental theorem of finitely generated abelian groups) Suppose that G is a finitely generated abelian group. Then G is isomorphic to a direct product of cyclic groups in the form Zp

e1 1 × Zp e2 2 × . . . × Zpen n × Z × Z × . . . × Z,

where pi are primes, not necessarily distinct. The direct product is unqiue except for possible rearrangement of the factors.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Finitely generated abelian groups

Definition An abelian group is finitely generated if it can be generated by a finite number of elements. Theorem (Fundamental theorem of finitely generated abelian groups) Suppose that G is a finitely generated abelian group. Then G is isomorphic to a direct product of cyclic groups in the form Zp

e1 1 × Zp e2 2 × . . . × Zpen n × Z × Z × . . . × Z,

where pi are primes, not necessarily distinct. The direct product is unqiue except for possible rearrangement of the factors.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Finitely generated abelian groups

Definition Let G be a finitely generated abelian group. The number of copies of Z in the above theorem is the Betti number of G. In some occasions, it is also called the rank of G.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Example

Example Using the above theorem, we find that the only abelian groups

• f order 360 = 23325, up to isomorphism, are

Z23 × Z32 × Z5 (≃ Z360), Z22 × Z2 × Z32 × Z5, Z2 × Z2 × Z2 × Z32 × Z5, Z23 × Z3 × Z3 × Z5, Z22 × Z2 × Z3 × Z3 × Z5, Z2 × Z2 × Z2 × Z3 × Z3 × Z5.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Finitely generated abelian groups

In general, if the prime factorization of a positive integer n is n = pe1

1 pe2 2 . . . pek k , where pi are distinct primes, then the number

• f all abelian groups of order n, up to isomorphism, is equal to

p(e1)p(e2) . . . p(en), where p(e) is the number of ways to write an integer e as a sum of positive integers. (For example, 4 = 4 = 3 + 1 = 2 + 2 = 2 + 1 + 1 = 1 + 1 + 1 + 1. Thus p(4) = 5.)

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Example

Consider n = 200 = 2352. We have 3 = 3 = 2 + 1 = 1 + 1 + 1, and 2 = 2 = 1 + 1. Thus, there are 3 × 2 non-isomorphic abelian groups of order 200. They are Z23 × Z52 (≃ Z200) Z23 × Z5 × Z5 Z22 × Z2 × Z52 Z22 × Z2 × Z5 × Z5 Z2 × Z2 × Z2 × Z52 Z2 × Z2 × Z2 × Z5 × Z5.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Example

Consider n = 200 = 2352. We have 3 = 3 = 2 + 1 = 1 + 1 + 1, and 2 = 2 = 1 + 1. Thus, there are 3 × 2 non-isomorphic abelian groups of order 200. They are Z23 × Z52 (≃ Z200) Z23 × Z5 × Z5 Z22 × Z2 × Z52 Z22 × Z2 × Z5 × Z5 Z2 × Z2 × Z2 × Z52 Z2 × Z2 × Z2 × Z5 × Z5.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Example

Consider n = 200 = 2352. We have 3 = 3 = 2 + 1 = 1 + 1 + 1, and 2 = 2 = 1 + 1. Thus, there are 3 × 2 non-isomorphic abelian groups of order 200. They are Z23 × Z52 (≃ Z200) Z23 × Z5 × Z5 Z22 × Z2 × Z52 Z22 × Z2 × Z5 × Z5 Z2 × Z2 × Z2 × Z52 Z2 × Z2 × Z2 × Z5 × Z5.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Example

Consider n = 200 = 2352. We have 3 = 3 = 2 + 1 = 1 + 1 + 1, and 2 = 2 = 1 + 1. Thus, there are 3 × 2 non-isomorphic abelian groups of order 200. They are Z23 × Z52 (≃ Z200) Z23 × Z5 × Z5 Z22 × Z2 × Z52 Z22 × Z2 × Z5 × Z5 Z2 × Z2 × Z2 × Z52 Z2 × Z2 × Z2 × Z5 × Z5.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Applications

Theorem (11.16) If m divides the order of a finite abelian group G, then G has a subgroup of order m. Proof. By the fundamental theorem of finitely generated abelian groups, we have G ≃ Zp

e1 1 × Zp e2 2 × · · · × Zpen n . If m

• |G|, then

m = pf1

1 . . . pfn n for some non-negative integers fi. (There may be

more than one way to write m since pi are not necessarily distinct.). Then the subgroup pe1−f1

1

× pe2−f2

2

× · · · × pen−fn

n

• has order pf1

1 . . . pfn n = m.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Applications

Theorem (11.16) If m divides the order of a finite abelian group G, then G has a subgroup of order m. Proof. By the fundamental theorem of finitely generated abelian groups, we have G ≃ Zp

e1 1 × Zp e2 2 × · · · × Zpen n . If m

• |G|, then

m = pf1

1 . . . pfn n for some non-negative integers fi. (There may be

more than one way to write m since pi are not necessarily distinct.). Then the subgroup pe1−f1

1

× pe2−f2

2

× · · · × pen−fn

n

• has order pf1

1 . . . pfn n = m.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Remark

The statement in Theorem 11.16 in general is false if G is not

• abelian. For example, A4 has 12 elements, but it does not have

a subgroup of order 6. To see this, suppose that H < A4 has

• rder 6. Then it has an element τ of order 2 and an element σ
• f order 3 with τστ −1 = σi, where i = 1 or 2. Without loss of

generality, we may assume that σ = (1, 2, 3). Now there are 3 elements of order 2 in A4, namely, τ1 = (1, 2)(3, 4), τ2 = (1, 3)(2, 4), and τ3 = (1, 4)(2, 3). However, τ1στ −1

1

= (1, 2)(3, 4)(1, 2, 3)(3, 4)(1, 2) = (1, 4, 2) τ2στ −1

2

= (1, 3)(2, 4)(1, 2, 3)(2, 4)(1, 3) = (1, 3, 4) τ3στ −1

3

= (1, 4)(2, 3)(1, 2, 3)(2, 3)(1, 4) = (2, 4, 3). Thus, there is no subgroup of order 6 in A4.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Remark

The statement in Theorem 11.16 in general is false if G is not

• abelian. For example, A4 has 12 elements, but it does not have

a subgroup of order 6. To see this, suppose that H < A4 has

• rder 6. Then it has an element τ of order 2 and an element σ
• f order 3 with τστ −1 = σi, where i = 1 or 2. Without loss of

generality, we may assume that σ = (1, 2, 3). Now there are 3 elements of order 2 in A4, namely, τ1 = (1, 2)(3, 4), τ2 = (1, 3)(2, 4), and τ3 = (1, 4)(2, 3). However, τ1στ −1

1

= (1, 2)(3, 4)(1, 2, 3)(3, 4)(1, 2) = (1, 4, 2) τ2στ −1

2

= (1, 3)(2, 4)(1, 2, 3)(2, 4)(1, 3) = (1, 3, 4) τ3στ −1

3

= (1, 4)(2, 3)(1, 2, 3)(2, 3)(1, 4) = (2, 4, 3). Thus, there is no subgroup of order 6 in A4.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Remark

The statement in Theorem 11.16 in general is false if G is not

• abelian. For example, A4 has 12 elements, but it does not have

a subgroup of order 6. To see this, suppose that H < A4 has

• rder 6. Then it has an element τ of order 2 and an element σ
• f order 3 with τστ −1 = σi, where i = 1 or 2. Without loss of

generality, we may assume that σ = (1, 2, 3). Now there are 3 elements of order 2 in A4, namely, τ1 = (1, 2)(3, 4), τ2 = (1, 3)(2, 4), and τ3 = (1, 4)(2, 3). However, τ1στ −1

1

= (1, 2)(3, 4)(1, 2, 3)(3, 4)(1, 2) = (1, 4, 2) τ2στ −1

2

= (1, 3)(2, 4)(1, 2, 3)(2, 4)(1, 3) = (1, 3, 4) τ3στ −1

3

= (1, 4)(2, 3)(1, 2, 3)(2, 3)(1, 4) = (2, 4, 3). Thus, there is no subgroup of order 6 in A4.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Remark

The statement in Theorem 11.16 in general is false if G is not

• abelian. For example, A4 has 12 elements, but it does not have

a subgroup of order 6. To see this, suppose that H < A4 has

• rder 6. Then it has an element τ of order 2 and an element σ
• f order 3 with τστ −1 = σi, where i = 1 or 2. Without loss of

generality, we may assume that σ = (1, 2, 3). Now there are 3 elements of order 2 in A4, namely, τ1 = (1, 2)(3, 4), τ2 = (1, 3)(2, 4), and τ3 = (1, 4)(2, 3). However, τ1στ −1

1

= (1, 2)(3, 4)(1, 2, 3)(3, 4)(1, 2) = (1, 4, 2) τ2στ −1

2

= (1, 3)(2, 4)(1, 2, 3)(2, 4)(1, 3) = (1, 3, 4) τ3στ −1

3

= (1, 4)(2, 3)(1, 2, 3)(2, 3)(1, 4) = (2, 4, 3). Thus, there is no subgroup of order 6 in A4.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Remark

The statement in Theorem 11.16 in general is false if G is not

• abelian. For example, A4 has 12 elements, but it does not have

a subgroup of order 6. To see this, suppose that H < A4 has

• rder 6. Then it has an element τ of order 2 and an element σ
• f order 3 with τστ −1 = σi, where i = 1 or 2. Without loss of

generality, we may assume that σ = (1, 2, 3). Now there are 3 elements of order 2 in A4, namely, τ1 = (1, 2)(3, 4), τ2 = (1, 3)(2, 4), and τ3 = (1, 4)(2, 3). However, τ1στ −1

1

= (1, 2)(3, 4)(1, 2, 3)(3, 4)(1, 2) = (1, 4, 2) τ2στ −1

2

= (1, 3)(2, 4)(1, 2, 3)(2, 4)(1, 3) = (1, 3, 4) τ3στ −1

3

= (1, 4)(2, 3)(1, 2, 3)(2, 3)(1, 4) = (2, 4, 3). Thus, there is no subgroup of order 6 in A4.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Remark

The statement in Theorem 11.16 in general is false if G is not

• abelian. For example, A4 has 12 elements, but it does not have

a subgroup of order 6. To see this, suppose that H < A4 has

• rder 6. Then it has an element τ of order 2 and an element σ
• f order 3 with τστ −1 = σi, where i = 1 or 2. Without loss of

generality, we may assume that σ = (1, 2, 3). Now there are 3 elements of order 2 in A4, namely, τ1 = (1, 2)(3, 4), τ2 = (1, 3)(2, 4), and τ3 = (1, 4)(2, 3). However, τ1στ −1

1

= (1, 2)(3, 4)(1, 2, 3)(3, 4)(1, 2) = (1, 4, 2) τ2στ −1

2

= (1, 3)(2, 4)(1, 2, 3)(2, 4)(1, 3) = (1, 3, 4) τ3στ −1

3

= (1, 4)(2, 3)(1, 2, 3)(2, 3)(1, 4) = (2, 4, 3). Thus, there is no subgroup of order 6 in A4.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Remark

The statement in Theorem 11.16 in general is false if G is not

• abelian. For example, A4 has 12 elements, but it does not have

a subgroup of order 6. To see this, suppose that H < A4 has

• rder 6. Then it has an element τ of order 2 and an element σ
• f order 3 with τστ −1 = σi, where i = 1 or 2. Without loss of

generality, we may assume that σ = (1, 2, 3). Now there are 3 elements of order 2 in A4, namely, τ1 = (1, 2)(3, 4), τ2 = (1, 3)(2, 4), and τ3 = (1, 4)(2, 3). However, τ1στ −1

1

= (1, 2)(3, 4)(1, 2, 3)(3, 4)(1, 2) = (1, 4, 2) τ2στ −1

2

= (1, 3)(2, 4)(1, 2, 3)(2, 4)(1, 3) = (1, 3, 4) τ3στ −1

3

= (1, 4)(2, 3)(1, 2, 3)(2, 3)(1, 4) = (2, 4, 3). Thus, there is no subgroup of order 6 in A4.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Remark

The statement in Theorem 11.16 in general is false if G is not

• abelian. For example, A4 has 12 elements, but it does not have

a subgroup of order 6. To see this, suppose that H < A4 has

• rder 6. Then it has an element τ of order 2 and an element σ
• f order 3 with τστ −1 = σi, where i = 1 or 2. Without loss of

generality, we may assume that σ = (1, 2, 3). Now there are 3 elements of order 2 in A4, namely, τ1 = (1, 2)(3, 4), τ2 = (1, 3)(2, 4), and τ3 = (1, 4)(2, 3). However, τ1στ −1

1

= (1, 2)(3, 4)(1, 2, 3)(3, 4)(1, 2) = (1, 4, 2) τ2στ −1

2

= (1, 3)(2, 4)(1, 2, 3)(2, 4)(1, 3) = (1, 3, 4) τ3στ −1

3

= (1, 4)(2, 3)(1, 2, 3)(2, 3)(1, 4) = (2, 4, 3). Thus, there is no subgroup of order 6 in A4.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Exercises

In-class exercises

1

Let p(e) be the number of ways to write a positive integer e as a sum of positive integers. Find p(8).

2

Find all abelian groups of order 144, up to isomorphism.

3

Find all abelian groups of order 216, up to isomorphism. Homework Problems 6, 8, 10, 12, 16, 18, 24, 26, 29, 38, 39, 47 of Section 11.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups

Direct products Finitely generated abelian groups

Exercises

In-class exercises

1

Let p(e) be the number of ways to write a positive integer e as a sum of positive integers. Find p(8).

2

Find all abelian groups of order 144, up to isomorphism.

3

Find all abelian groups of order 216, up to isomorphism. Homework Problems 6, 8, 10, 12, 16, 18, 24, 26, 29, 38, 39, 47 of Section 11.

Instructor: Yifan Yang Section 11 – Direct products and finitely generated abelian groups