Mathematics 351 – Abstract Algebra 1 Solutions for Problem Set 10 December 5, 2007

Section 7.7

4. First notice that G = ℤ_{4} × ℤ_{4} is abelian, so N and every other subgroup is normal. This implies that G/N has the structure of a group. We can determine this structure most directly by constructing the group table for the quotient group. First, the elements in N are N = {(0, 0), (3, 2), (2, 0), (1, 2)}. The distinct cosets are

(0, 0) + N, (0, 1) + N = {(0, 1), (3, 3), (2, 1), (1, 3),
(0, 2) + N = {(1, 0), (0, 2), (3, 0), (2, 2)}}, (0, 3) + N = {(1, 1), (0, 3), (3, 1), (2, 3)}.

Then the group table for the quotient group G/N is constructed using the coset addition operation (omitting +N’s for simplicity):

\begin{matrix}+ & (0, 0) & (0, 1) & (0, 2) & (0, 3) \\ (0, 0) & (0, 0) & (0, 1) & (0, 2) & (0, 3) \\ (0, 1) & (0, 1) & (0, 2) & (0, 3) & (0, 0) \\ (0, 2) & (0, 2) & (0, 3) & (0, 0) & (0, 1) \\ (0, 3) & (0, 3) & (0, 0) & (0, 1) & (0, 2) \end{matrix}

For example, (0, 2)+N +(0, 3)+N = (0, 1)+N from the above. By inspection, G/N is isomorphic to the cyclic group of order 4, ℤ_{4}, since the coset (0, 1) + N is a generator for the whole group.

5. We proceed as in 4 above. N = {(0, 0), (1, 1), (2, 0), (3, 1), (4, 0), (5, 1)} Hence (1, 1) is an element of order 6, and by Lagrange’s Theorem |G/N| = 2. There are two cosets of N in G, N = (0, 0) + N and (0, 1) + N = {(0, 1), (1, 0), (2, 1), (3, 0), (4, 1), (5, 0)}. The quotient group is isomorphic to ℤ_{2} (every group of order 2 is isomorphic to ℤ_{2} since 2 is prime).

6. This could also be done by constructing the group tables of U_{32}/N and U_{16} and comparing their structures. We will illustrate a different method in this solution. We have that U_{32} = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31} is an abelian group of order 16 under multiplication mod 32. The subgroup N = {1, 17} has 8 distinct multiplicative cosets, one for each k = 1, 3, 5, 7, 9, 11, 13, 15. This true because if k = 2ℓ + 1, then 17k = (16 + 1)(2ℓ + 1) = 32ℓ + 16 + (2ℓ + 1) ≡ 16 + k mod 32. For instance, this says N · 5 = {5, 21}, etc. We construct

φ : U_{32}/N → U_{16}
N · k ↦ [k]_{16}

This is well defined since if N · k = N · ℓ, then by the above k ≡ ℓ mod 16, so they define the same class in U_{16}. The mapping φ is one-to-one and onto by construction. It is a group homomorphism because

φ((N · k)(N · ℓ)) = φ(N · (kℓ)) = [kℓ]_{16} = [k]_{16}[ℓ]_{16} = φ(N · k)φ(N · ℓ).

Therefore U_{32}/N and U_{16} are isomorphic groups.

8. Let Nx ∈ G/N. Then by the definition of the coset product operation and the fact that x^{2} ∈ N, (Nx)^{2} = (Nx)(Nx) = Nx^{2} = N = Ne. Since Ne is the identity element in G/N. This shows that if Nx ≠ Ne, then Nx has order 2 in G/N.

9. N is automatically normal in G, so G/N is a group. Let Na and Nb be any two cosets. Then by the definition of the coset product and the fact that G is abelian,

(Na)(Nb) = N(ab) = N(ba) = (Nb)(Na).

This shows that G/N is an abelian group.

15. (a) Let p/q + ℤ ∈ ℚ/ℤ. The operation is addition, so

q · (p/q + ℤ) = (p/q + p/q + ⋯ + p/q) + ℤ = p + ℤ = ℤ = 0 + ℤ.

This shows that p/q + ℤ has finite order (equal to some divisor of the integer q).

(b) Let p/q be a rational number in lowest terms (that is, (p, q) = 1). Then when we form q(p/q + ℤ) = 0 + ℤ, no smaller multiple of p/q + ℤ equals 0 + ℤ since the smallest positive multiple of p that is also a multiple of q is pq. This shows that ℚ/ℤ contains elements of every order q ≥ 1.

16. The result in part (a) of Exercise 15 also shows that every element of the subgroup ℚ/ℤ ⊂ ℝ/ℤ has finite order in ℝ/ℤ. Now let r ∈ ℝ/ℤ and assume that there is some positive integer q such that 0 + ℤ = q(r + ℤ) = qr + ℤ. This is only true if qr = p ∈ ℤ. But then r = p/q ∈ ℚ. Hence the set of elements of finite order in ℝ/ℤ is contained in ℚ/ℤ. Since both containments hold, the two subgroups are equal.

23. (a) Let a, b ∈ G, then following the hint, for all g ∈ G,

g^{−1}aba^{−1}b^{−1}g = (g^{−1}ag)(g^{−1}bg)(g^{−1}a^{−1}g)(g^{−1}b^{−1}g) = (g^{−1}ag)(g^{−1}bg)(g^{−1}ag)^{−1}(g^{−1}bg)^{−1} ∈ G′.

If we have any element c of G′, then c = c_{1}c_{2} ⋯ c_{n} is a product of elements c_{i} = a_{i}b_{i}a_{i}^{−1}b_{i}^{−1} of the form above for i = 1, …, n. Hence g^{−1}cg = (g^{−1}c_{1}g)(g^{−1}c_{2}g) ⋯ (g^{−1}c_{n}g) is also a product of commutators, hence an element of G′. This shows g^{−1}G′g ⊆ G′ for all g ∈ G. Hence G′ is normal by part (2) of Theorem 7.34.

(b) In the quotient group G/G′, consider any two cosets G′a and G′b. By the definition of the coset product, we have

(G′a)(G′b)(G′a)^{−1}(G′b)^{−1} = G′(aba^{−1}b^{−1}) = G′ = G′e,

since aba^{−1}b^{−1} ∈ G′. This implies (G′a)(G′b) = (G′b)(G′a) for all a, b. Hence G/G′ is an abelian group.

Section 7.8

6. (a) Every subgroup of ℤ_{12}/H is K/H for a subgroup K ⊂ ℤ_{12} containing H (Theorem 7.44, part (3)). These subgroups are K_{0} = H itself, K_{1} = {0, 3, 6, 9}, K_{2} = {0, 2, 4, 6, 8, 10}, and K_{3} = ℤ_{12}. The corresponding subgroups of ℤ_{12}/H are K_{0}/H = {0 + H}, K_{1}/H = {0 + H, 3 + H}, K_{2}/H = {0 + H, 2 + H, 4 + H}, and K_{3}/H = ℤ_{12}/H = {0 + H, 1 + H, 2 + H, 3 + H, 4 + H, 5 + H}.

(b) Similar to part (a). This time there are only three such subgroups, L_{i}/K where L_{0} = K, L_{1} = {0, 2, 4, 6, 8, 10, 12, 14, 16, 18}, and L_{2} = ℤ_{20}.

10. (a) The set K is the set of all elements of order 2 or less. Since e ∈ G has order 1, K is nonempty. Moreover, if a, b ∈ K, then using the abelian hypothesis on G, (ab)^{2} = (ab)(ab) = a^{2}b^{2} = e. Hence ab has order at most 2 and is an element in K. Finally, if a ∈ G, then the order of a^{−1} is the same as the order of a, so a^{−1} ∈ K as well. Thus K is a subgroup of G.

(b) The set H is the set of squares of elements in G. This is nonempty since G is nonempty. If a = x^{2} and b = y^{2} are in H, then ab = x^{2}y^{2} = (xy)(xy) = (xy)^{2} is also in H since x and y commute. Finally, if a = x^{2} is in H, then a^{−1} = (x^{2})^{−1} = (x^{−1})^{2} ∈ H. Hence H is a subgroup of G.

(c) Let φ : G → H be the mapping φ(x) = x^{2}. This is a group homomorphism under the hypothesis that G is abelian: φ(xy) = (xy)^{2} = (xy)(xy) = x^{2}y^{2} = φ(x)φ(y). By definition φ is onto H. The kernel of φ is precisely the subgroup K from part (a). Hence by the First Isomorphism Theorem, H ≃ G/K.

15. The group ℝ^{∗} is the multiplicative group of nonzero reals. The mapping

det : GL(2, ℝ) → ℝ^{∗}
A ↦ det(A)

is a group homomorphism by Exercise 22 of section 7.6. The kernel of det is the subgroup SL(2, ℝ) (this was one way to show that SL(2, ℝ) is a normal subgroup of GL(2, ℝ) – see Exercise 23 of section 7.6). Hence by the First Isomorphism Theorem, GL(2, ℝ)/SL(2, ℝ) ≃ ℝ^{∗}.

16. (a) By the distributive law in R, f(x+y) = r(x+y) = rx+ry = f(x)+f(y). Hence f : R → R is a group homomorphism for the additive group structure on (R, +).

(b) Let R = ℤ and r = 4 ∈ ℤ. Then f(1 · 2) = 4 · (1 · 2) = 8 but f(1) · f(2) = (4 · 1) · (4 · 2) = 32. Hence f(1 · 2) ≠ f(1) · f(2). So f is not a ring homomorphism from ℤ to itself.

(c) Suppose that R is a commutative ring and r is an idempotent element (that is r^{2} = r). Then f(xy) = r(xy) = r^{2}(xy) = (rx)(ry) = f(x)f(y). So in this case f is a ring homomorphism. (Comment: Another correct answer would be to say that r = 0, 1 always work! Any ring R contains idempotents r = 0, 1. Some rings have others too. For instance, R = ℤ × ℤ under the component-wise sum and product operations also has idempotents r = (1, 0) and r = (0, 1). )

17. (a) G is clearly nonempty since a, b, c ∈ ℚ are arbitrary. Let

A = (\begin{matrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{matrix}) , B = (\begin{matrix}1 & a′ & b′ \\ 0 & 1 & c′ \\ 0 & 0 & 1 \end{matrix}) .

Then

AB = (\begin{matrix}1 & a + a′ & b + ac′ + b′ \\ 0 & 1 & c + c′ \\ 0 & 0 & 1 \end{matrix}) .  (1)

Since this matrix is also in G, G is closed under matrix multiplication. The computation above also shows that AB = I, the 3 × 3 identity matrix, if a′ = −a, c′ = −c, and b′ = −b − ac′ = −b + ac. The matrix

B = A^{−1} = (\begin{matrix}1 & −a & −b + ac \\ 0 & 1 & −c \\ 0 & 0 & 1 \end{matrix})

is also in G. Hence G is a subgroup of GL(3, ℚ), hence a group.

(b) From the equation for AB above and the reverse product

BA = (\begin{matrix}1 & a′ + a & b′ + a′c + b \\ 0 & 1 & c′ + c \\ 0 & 0 & 1 \end{matrix}) ,

we see that AB = BA if and only if a′c = ac′. Hence A is in the center of G if and only if a′c = ac′ for all choices of a′, c′ ∈ ℚ. For a′ = 1, c′ = 0 we get c = 0 and then since c′ ≠ 0 is also possible, a = 0 as well. The center of G is

Z(G) = \{A ∈ G : A = (\begin{matrix}1 & 0 & b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) , b ∈ ℚ\} .

To show the last statement in this part, consider the mapping

φ : Z(G) → ℚ

defined by

φ ((\begin{matrix}1 & 0 & b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix})) = b.

This is clearly one-to-one and onto, and moreover by the product formula (1) above in the case a = c = a′ = c′ = 0, we see if A and B are as in part A, φ(AB) = b + b′ = φ(A) + φ(B). This shows that Z(G) and (ℚ, +) are isomorphic groups.

(c) Now define ψ : G → ℚ × ℚ (a group under component-wise sums) by

ψ ((\begin{matrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{matrix})) = (a, c).

By the product formula (1), ψ is a group homomorphism: ψ(AB) = (a+a′, c+c′) = (a, c)+(a′, c′) = ψ(A) + ψ(B). Then we note that ker(ψ) is the set of matrices A in G with a = c = 0. This is precisely Z(G) by part (b). Hence by the First Isomorphism Theorem, ℚ × ℚ ≃ G/Z(G).

Section 7.9

3. (a) This is (12)(45)(679). (b) (13)(254)(789). (c) (13)(254)(69)(78). (d) (1573)(24). (e) (123)(456)(78).

4. (a) One way (there are many other correct ones too – see for instance the different factorization from Corollary 7.48!): (12)(45)(79)(69). (b) (13)(54)(24)(89)(69). (c) (13)(54)(24)(69)(78). (d) (73)(53)(13)(24). (e) (23)(13)(56)(46)(78).

5. (a) (2468) = (68)(48)(28) is odd. (b) even. (c) even.

23. Let τ = (ij). Then we claim that for any σ ∈ S_{n}, ρ = στσ^{−1} = (σ(i)σ(j)). To see this note that ρ(σ(i)) = σ(τ(i)) = σ(j) and ρ(σ(j)) = σ(τ(j)) = σ(i). Moreover if k is any element of {1, 2, …, n} besides σ(i), σ(j) then we can write k = σ(ℓ) for some ℓ ≠ i, j. Then ρ(k) = σ(τ(ℓ)) = σ(ℓ) = k. Hence ρ fixes every other element of {1, 2, …, n} besides i, j. Therefore ρ = στσ^{−1} = (σ(i)σ(j)) is a transposition.

24. We combine Exercise 23 and the factorization of the cycle τ as a product of transpositions as in Corollary 7.48. We have (a_{1}a_{2} ⋯ a_{k}) = (a_{1}a_{k})(a_{1}a_{k−1}) ⋯ (a_{1}a_{2}). Hence

σ(a_{1}a_{2} ⋯ a_{k})σ^{−1} = σ(a_{1}a_{k})(a_{1}a_{k−1}) ⋯ (a_{1}a_{2})σ^{−1}
= (σ(a_{1}a_{k})σ^{−1})(σ(a_{1}a_{k−1})σ^{−1}) ⋯ (σ(a_{1}a_{2})σ^{−1})
= (σ(a_{1})σ(a_{k}))(σ(a_{1})σ(a_{k−1})) ⋯)(σ(a_{1})σ(a_{2}))
= (σ(a_{1})σ(a_{2}) ⋯ σ(a_{k}))

25. Consider H = G ∩ A_{n}, which is a subgroup of G. H consists of all the even permutations that are in H. But now since there is some σ ∈ G that is an odd permutation, we can consider the right coset Hσ. Every element of the coset Hσ is an odd permutation that is in G. Moreover every odd ρ permutation in G has the form ρ = (ρσ^{−1})σ. Since ρ and σ are odd, ρσ^{−1} is even, so it is an element of H. Therefore every odd permutation in G belongs to the coset Hσ. Hence H ∩Hσ = G, since every permutation in G is either even or odd. This shows that H has index 2 in G.

Section 8.1 5. The Klein 4-group ℤ_{2} × ℤ_{2} has order 4, but no elements of order 4 (only orders 1 and 2). Therefore ℤ_{2} × ℤ_{2} is not cyclic even though ℤ_{2} = ⟨1⟩ is a cyclic group.

6. (a) Every element of ℤ_{12} can be written uniquely as a sum of an element of H = {0, 3, 6, 9} and K = {0, 4, 8}. (the cosets of H for the elements in K fill out all of ℤ_{12}). Hence by Theorem 8.1, ℤ_{12} ≃ H × K.

(b) Every element of ℤ_{15} can be written uniquely as a sum of an element of H = {0, 5, 10} and K = {0, 3, 6, 9, 12} (the cosets of H for the elements in K fill out all of ℤ_{15}). Hence by Theorem 8.1, ℤ_{15} ≃ H × K. 7.

(c) Finally, every element of ℤ_{30} can be written uniquely as a sum of an element of H = {0, 5, 10, 15, 20, 25} and K = {0, 6, 12, 18, 24}. Hence by Theorem 8.1, ℤ_{30} ≃ H × K. Similarly, ℤ_{30} ≃ {0, 10, 20} × {0, 3, 6, 9, 12, 15, 18, 21, 24, 27}. and

ℤ_{30} ≃ {0, 15} × {0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28}.

Note that we also have

ℤ_{30} = {0, 15} × {0, 10, 20} × {0, 6, 12, 18, 24}.

7. ⇒: If G_{1} × G_{2} × ⋯ × G_{n} is an abelian group, then for each i, 1 ≤ i ≤ n, and all a, b ∈ G_{i},

(e_{1}, …, e_{i−1}, a, e_{i+1}, …, e_{n})(e_{1}, …, e_{i−1}, b, e_{i+1}, …, e_{n}) = (e_{1}, …, e_{i−1}, ab, e_{i+1}, …, e_{n})

and

(e_{1}, …, e_{i−1}, b, e_{i+1}, …, e_{n})(e_{1}, …, e_{i−1}, a, e_{i+1}, …, e_{n}) = (e_{1}, …, e_{i−1}, ba, e_{i+1}, …, e_{n}).

The left sides of these equations are equal, hence the right sides are equal too, so ab = ba. This shows G_{i} is abelian.

⇐: Suppose each G_{i} is abelian. Then in the direct product for all (a_{1}, …, a_{n}) and (b_{1}, …, b_{n}),

(a_{1}, …, a_{n})(b_{1}, …, b_{n}) = (a_{1}b_{1}, …, a_{n}b_{n}) = (b_{1}a_{1}, …, b_{n}a_{n}) = (b_{1}, …, b_{n})(a_{1}, …, a_{n})

This shows that the direct product is also an abelian group.

9. The answer is No. One reason we can see this is that ℤ_{4} × ℤ_{2} contains only elements of order 1, 2, 4 (none of order 8):

1 = |(0, 0)|
2 = |(2, 0)| = |(2, 1)| = |(0, 1)|
4 = |(1, 0)| = |(1, 1)| = |(3, 0)| = |(3, 1)|.

This shows that ℤ_{4} × ℤ_{2} cannot be isomorphic to the cyclic group ℤ_{8}.