MTH 211: HW#7 SOLUTIONS

Chapter 8

14. Suppose that G1∼= G2 and H1

∼= H2. Prove that G1 ⊕H1∼= G2 ⊕H2.

Proof. Let φ : G1 → G2 and ψ : H1 → H2 be isomorphisms. Then it is easy to check

that

(φ, ψ) : G1 ⊕H1 → G2 ⊕H2

is an isomorphism, where (φ, ψ)(g, h) = (φ(g), ψ(h)). �

20. The group S3 ⊕ Z2 is isomorphic to one of the following groups: Z12, Z6 ⊕ Z2, A4,

D6. Determine which one by elimination.

Proof. S3 ⊕ Z2 is non-abelian, since S3 is non-abelian. Therefore, S3 ⊕ Z2 is not

isomorphic to Z12 nor to Z6 ⊕ Z2. S3 ⊕ Z2 has elements of order 6 (for example,((123), 1

)has order 6), but A4 has only elements of order 1, 2 and 3. By elimination,

S3 ⊕ Z2∼= D6. �

22. Find a subgroup of Z4 ⊕ Z2 that is not of the form H ⊕K, where H is a subgroup

of Z4 and K is a subgroup of Z2.

Proof. The subgroups of Z4 are {0}, 〈2〉 and Z4. The only subgroups of Z2 are {0}

and itself. Therefore, the subgroups of the form H ⊕K, where H is a subgroup of

Z4 and K is a subgroup of Z2, are {(0, 0)}, {0} ⊕ Z2, 〈2〉 ⊕ {0}, 〈2〉 ⊕ Z2, Z4 ⊕ {0},

and Z4 ⊕ Z2. However, the cyclic subgroup 〈(1, 1)〉 = {(0, 0), (1, 1), (2, 0), (3, 1)} is a

subgroup of Z4 ⊕ Z2 that is not of one of these. �

42. Find an isomorphism from Z12 to Z4 ⊕ Z3.

Proof. A homomorphism from Z12 to Z4 ⊕ Z3 is determined by where 1 is sent. In

order to get an isomorphism it must be sent to an element of order 12. Define

2 MTH 211: HW#7 SOLUTIONS

φ : Z12 → Z4 ⊕ Z3 by φ(1) = (1, 1). (Therefore, φ(n) = (n mod 4, n mod 3).) Since

(1, 1) has order 12, we are done. �

46. Let G = {ax2 + bx + c | a, b, c ∈ Z3}. Add elements of G as you would polynomials

with integer coefficients, except use modulo 3 addition. Prove that G is isomorphic

to Z3 ⊕ Z3 ⊕ Z3. Generalize.

Proof. It is straightforward to check that the function φ : G→ Z3 ⊕Z3 ⊕Z3, defined

by φ(ax2+bx+c) = (a, b, c), is an isomorphism. More generally, Z3 can be replaced by

any group, where the addition of polynomials respects the operation of the coefficient

group. �