# (0, 0) : order 1; (0, 1) : order 4; (0, 2) : order 2; (0, 3) : order 4; (1, 0) : order 2; (1, 1) : order 4; (1, 2) : order 2; (1, 3) : order 4.

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11.01 List the elements of Z2×Z4. Find the order of each of the elements

is this group cyclic?

Solution: The elements ofZ2×Z4 are:

(0,0) : order 1; (0,1) : order 4; (0,2) : order 2; (0,3) : order 4; (1,0) : order 2; (1,1) : order 4; (1,2) : order 2; (1,3) : order 4.

This group is not cyclic since no element can generate the whole group. 11.02 List the elements of Z3×Z4. Find the order of each of the elements

is this group cyclic?

Solution: The elements ofZ3×Z4 are:

(0,0) : order 1; (0,1) : order 4; (0,2) : order 2; (0,3) : order 4; (1,0) : order 3; (1,1) : order 12; (1,2) : order 6; (1,3) : order 12; (2,0) : order 3; (2,1) : order 12; (2,2) : order 6; (2,3) : order 12.

This group is cyclic since it can be generated by either of the elements (1,1), (1,3), (2,1), and (2,3).

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11.13 Disregarding the order of the factors, write direct products of two or more groups of the formZnso that the resulting product is isomorphic

toZ60 in as many ways as possible.

Solution: There are 4 different ways:

Z60 = Z22 ×Z3×Z5 =Z4×Z3×Z5,

Z60 = Z223×Z5 =Z12×Z5,

Z60 = Z225×Z3 =Z20×Z3,

Z60 = Z22 ×Z3·5=Z4×Z15.

11.14 a. The cyclic subgroup ofZ24 generated by 18 has order 4.

b. Z3×Z4 is of order 12.

c. The element (4,2) ofZ12×Z8 has order 12.

d. The Klein 4-group is isomorphic toZ2×Z2.

e. Z2×Z×Z4 has 8 elements of finite order.

11.15 Find the maximum possible order for some element of Z4×Z6.

Solution: (1,1) inZ4×Z6 has the maximum order lcm(4,6) = 12.

11.18 Are the groups Z8×Z10×Z24 and Z4×Z12×Z40 isomorphic? Why

or why not?

Solution: We decompose both groups into indecomposible ones:

Z8×Z10×Z24 ' Z8×(Z2×Z5)×(Z8×Z3) =Z2×(Z8)2×Z3×Z5,

Z4×Z12×Z40 ' Z4×(Z4×Z3)×(Z8×Z5) = (Z4)2×Z8×Z3×Z5.

So they are not isomorphic.

11.26 How many abelian groups (up to isomorphism) are there of order 24? of order 25? of order (24)(25)?

Solution: 24 = 23·3 = 2·22·3 = 2·2·2·3. So there are 3 abelian groups of order 24:

Z23×Z3, Z2×Z22×Z3, Z2×Z2×Z2×Z3.

25 = 52 = 5·5. So there are 2 abelian groups of order 25:

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Because gcd(24,25) = 1, there are 3×2 = 6 abelian groups of order (24)(25):

Z23 ×Z3 × Z52,

Z23 ×Z3 × Z5×Z5,

Z2×Z22 ×Z3 × Z52,

Z2×Z22 ×Z3 × Z5×Z5,

Z2×Z2×Z2×Z3 × Z52,

Z2×Z2×Z2×Z3 × Z5×Z5.

11.32 Mark each of the following true or false:

a. (T) IfG1 andG2 are any groups, thenG1×G2 is always

isomor-phic toG2×G1.

b. (T) Computation in an external direct product of groups is easy if you know how to compute in each component group.

c. (F) Groups of finite order must be used to form an external direct product.

d. (T) A group of prime order could not be the internal direct prod-uct of two proper nontrivial subgroups.

e. (F)Z2×Z4 is isomorphic toZ8.

f. (F)Z2×Z4 is isomorphic to§8.

g. (F)Z3×Z8 is isomorphic to§4.

h. (F) Every element inZ4×Z8 has order 8.

i. (F) The order of Z12×Z15 is 60.

j. (T) Zm ×Zn has mn elements whether m and n are relatively

prime or not.

11.34 a. How many subgroups of Z5×Z6 are isomorphic toZ5×Z6?

Solution: No subgroup of Z5×Z6 is isomorphic to Z5×Z6.

b. How many subgroups of Z×Zare isomorphic toZ×Z?

Solution: There are infinite many subgroups ofZ×Zthat are isomorphic toZ×Z. They are of the formmZ×nZfor positive integersm and nwithm6= 1 orn6= 1.

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a. (T) Every abelian group of prime order is cyclic. b. (F) Every abeliang roup of prime power order is cyclic.

c. (F)Z8 is generated by{4,6}.

d. (T) Z8 is generated by{4,5,6}.

e. (T) All finite abelian groups are classified up to isomorphism by Theorem 11.12.

f. (F) Any two finitely generated abelian gruops witht he same Betti number are isomorphic.

g. (T) Every abelian group of order divisible by 5 contains a cyclic subgroup of order 5.

h. (F) Every abelian group of order divisible by 4 contains a cyclic subgroup of order 4.

i. (T) Every abelian group of order divisible by 6 contains a cyclic subgroup of order 6.

j. (T) Every finite abelian group has a Betti number of 0. 11.46 Prove that a direct product of abelian groups is abelian.

Solution: Suppose Gi are abelian groups. We prove that Qni=1Gi

is an abelian group. Let (a1,· · ·, an) and (b1,· · · , bn) be elements of Qn

i=1Gi. Then

(a1,· · ·, an)(b1,· · ·, bn) = (a1b1,· · ·, anbn)

= (b1a1,· · ·, bnan)

= (b1,· · ·, bn)(a1,· · · , an).

This shows that the binary operation on Qni=1Gi is commutative. So Qn

i=1Gi is an abelian group.

11.47 Let G be an abelian group. Let H be the subset of G consisiting of the identity e together with all elements of G of order 2. Show that H is a subgroup of G.

Solution: We show thatH meets the criteria of a subgroup ofG: 1. (Closed) Ifa, b∈H, thena2 =b2 =e. So (ab)2=a2b2=e. This

implies thatab∈H.

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3. (Inverse) Ifa∈H, thena2=eand so a−1 =a∈H. Therefore,H is a subgroup ofG.

11.53 Prove that if a finite abelian group has order a power of a prime p, then the order of every element in the group is a power ofp. Can the hypothesis of commutativity be dropped? Why, or why not?

Solution: Suppose a groupGhas the orderpk for some primepand some positive integer k. Then the order m of every element a of G divides the orderpk ofG. Som=prfor some integer 0≤r≤k. That is, the order ofais a power of p.

The hypothesis of commutativity can be dropped, since we do not use the commutativity in the above argument.

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