Mathematics 351 – Abstract Algebra 1 Solutions for Problem Set 8 November 1, 2007

Section 7.1

9.

(a) The order of U_{n} is the number of k, 0 ≤ k ≤ n − 1 that are relatively prime to n. So U_{10} = {1, 3, 7, 9} has order 4, U_{12} = {1, 5, 7, 11} has order 4 and U_{24} = {1, 5, 7, 11, 13, 17, 19, 23} has order 8. Comment: The numbers of elements here are values of a famous function called “Euler’s ϕ-function.” It can be shown that the number of k as above relatively prime to n is given by the following formula if the prime factorization of n = p_{1}^{e_{1}} ⋯ p_{s}^{e_{s}}:

ϕ(n) = p_{1}^{e_{1}−1}(p_{1} − 1) ⋯ p_{s}^{e_{s}−1}(p_{s} − 1).

For instance 24 = 2^{3} · 3, so ϕ(24) = 2^{2}(2 − 1) · 3^{0}(3 − 1) = 8.

(b) U_{20} = {1, 3, 7, 9, 11, 13, 17, 19} (order 8). The orders of the elements are

|1| = 1 (it’s the identity element)
|9| = |11| = |19| = 2 since for instance 9^{2} = 81 ≡ 1 mod 20
|3| = |7| = |13| = |17| = 4 since 3^{2} ≡ 7^{2} ≡ 13^{2} ≡ 17^{2} ≡ 9 mod 20

10.

a. In ℤ_{4}, |0| = 1, |1| = 4, |2| = 2, |3| = 4.

b. In ℤ_{4} × ℤ_{2}, |(0, 0)| = 1, |(1, 0)| = |(1, 1)| = |(3, 0)| = |(3, 1)| = 4, and |(0, 1)| = |(2, 0)| = |(2, 1)| = 2.

c. In S_{3}, |()| = 1, |(123)| = |(132)| = 3, and |(12)| = |(13)| = |(23)| = 2.

d. (Using Hungerford’s notation), |r_{0}| = 1, |r_{1}| = |r_{3}| = 4, and |r_{2}| = |d|, |t|, |h|, |v| = 2.

e. In ℤ, 0 has order 1, but every other element has infinite order.

16. We have x = a_{1}a_{2} ⋯ a_{n}, so x^{2} = (a_{1}a_{2} ⋯ a_{n})(a_{1}a_{2} ⋯ a_{n}). G is a group, so each of the a_{i} has an inverse in G. This inverse must be one of the factors appearing in the second product. Since G is abelian, we can rearrange the factors to write x^{2} = (a_{1}a_{1}^{−1})(a_{2}a_{2}^{−1}) ⋯ (a_{n}a_{n}^{−1}) = e, which is what we wanted to show.

22. If (ab)^{2} = a^{2}b^{2}, then we have abab = aabb in G. Since G is a group each element has an inverse in G, so we can multiply the given equation on both sides by a^{−1} on the left and b^{−1} on the right: a^{−1}(abab)b^{−1} = a^{−1}(aabb)b^{−1}. After simplifying, we get ba = ab. Since this is true for all a, b ∈ G, G is abelian.

25. We assume that every a ≠ e in G has order 2, so a^{2} = a, and hence a = a^{−1}. It follows that every element of G satisfies a = a^{−1}, since this is true of e as well. If a, b are any two elements of G, then by the “reverse order” rule for the inverse of a product: ab = a^{−1}b^{−1} = (ba)^{−1} = ba. Hence G is abelian.

30. Consider the mapping ι : G → G where ι(g) = g^{−1}. Then ι(e) = e, and if g ∈ G, then ι(ι(g)) = (g^{−1})^{−1} = g. If g is an element of order > 2 then g and ι(g) are distinct, and are interchanged by ι. Hence, “pairing up” elements with their inverses via ι, since |G| is even, there must be at least one other element g ∈ G other than e with ι(g) = g. Since g^{−1} = g, g^{2} = e but g ≠ e. This shows that G must contain an element of order 2.

Section 7.3

6. The definition of H is equivalent to saying that H = {a ∈ G : a^{k} = e} for the fixed integer k. We use Theorem 7.10 to show that H is a subgroup of G when G is abelian. Notice that H is nonempty since e^{k} = e implies e ∈ H. Next, if a, b ∈ H, then a^{k} = b^{k} = e. Consider ab ∈ G. Since G is assumed abelian we can reorder the factors to rewrite (ab)^{k} = (ab)(ab) ⋯ (ab) = (aa ⋯ a)(bb ⋯ b) = a^{k}b^{k}. Since a, b ∈ H, this shows (ab)^{k} = e, so ab ∈ H. Finally, let a ∈ H, so a^{k} = e. This implies e = (a^{k})^{−1} = (a^{−1})^{k}. Hence a^{−1} ∈ H as well. It follows that H is a subgroup of G.

7. We reason as in problem 6 above, but now note that if a ∈ T, then the power k > 0 such that a^{k} = e will vary with which element a we are considering. Notice that T is nonempty since e^{k} = e implies e ∈ H. Next, if a, b ∈ T, then a^{k} = b^{ℓ} = e for some integers k, ℓ. Consider ab ∈ G, and let m = kℓ. Since G is assumed abelian we can reorder the factors to rewrite (ab)^{m} = (ab)(ab) ⋯ (ab) = (aa ⋯ a)(bb ⋯ b) = a^{m}b^{m}. Then a^{m} = (a^{k})^{ℓ} = e and b^{m} = (b^{ℓ})^{k} = e. This shows (ab)^{m} = e, so ab ∈ T. Finally, let a ∈ T, so a^{k} = e for some k > 0. This implies e = (a^{k})^{−1} = (a^{−1})^{k}. Hence a^{−1} ∈ T as well. It follows that T is a subgroup of G.

15. This is false. The group S_{3} gives a counterexample since every proper subgroup of S_{3} is cyclic of order 2 or 3, but S_{3} is not cyclic itself. There are no elements of order 6 in S_{3}.

16. We must show that ℚ^{∗∗} is not equal to ⟨r⟩ for any r ∈ ℚ^{∗∗}. Certainly r = 1 does not work since {1^{k} : k ∈ ℤ} = {1} is the trivial subgroup. So consider r ≠ 1 and the subgroup ⟨r⟩. Without loss of generality, we may assume r > 1, since if r < 1, then r^{−1} = \frac{1}{r} > 1, and ⟨r⟩ = ⟨r^{−1}⟩. If r > 1, then the elements of ⟨r⟩ are ordered as follows in ℝ:

0 < ⋯ r^{−2} < r^{−1} < 1 < r < r^{2} < r^{3} < ⋯

Recall from Principles of Analysis (MATH 242) that the set of rational numbers is dense in ℝ. In other words, in every interval of positive length in ℝ, there are rational numbers. In particular, there must be rational numbers s in the open interval (1, r). Any such rational number s is in ℚ^{∗∗}, since s > 0. But s ≠ r^{k} for any k because of the way the powers of r are ordered as above. This shows that ⟨r⟩ is not all of ℚ^{∗∗}. Since this is true for all r ≠ 1, ℚ^{∗∗} is not a cyclic group.

20. First note that if (a, b) ∈ ℤ×ℤ, then (a, b) = a(1, 0)+b(0, 1). So the set {(1, 0), (0, 1)} generates ℤ × ℤ. On the other hand,

(1, 0) = (3, 1) + (−2, −1) and (0, 1) = 2(−2, −1) + (4, 3).

Hence for any (a, b) ∈ ℤ × ℤ, we have

(a, b) = a(1, 0) + b(0, 1)
= a((3, 1) + (−2, −1)) + b(2(−2, −1) + (4, 3))
= a(3, 1) + (a + 2b)(−2, −1) + b(4, 3).

This shows ℤ × ℤ is contained in the subgroup generated by {(3, 1), (−2, −1), (4, 3)}. The opposite inclusion is true by definition. Hence {(3, 1), (−2, −1), (4, 3)} generates ℤ × ℤ.

21. (a) Check that the element (1, 1) ∈ ℤ_{2} × ℤ_{3} has order 6:

(0, 0), (1, 1), 2 · (1, 1) = (0, 2), 3 · (1, 1) = (1, 0), 4 · (1, 1) = (0, 1), 5 · (1, 1) = (1, 2).

Since we have all 6 elements of ℤ_{2} ×ℤ_{3}, this group is the same as the cyclic subgroup ⟨(1, 1)⟩, hence cyclic.

(b) The cyclic subgroups are

⟨(0, 0)⟩ = {(0, 0)}
⟨(1, 0)⟩ = {(0, 0), (1, 0)}
⟨(0, 1)⟩ = {(0, 0), (0, 1), (0, 2), (0, 3)} = ⟨(0, 3)⟩
⟨(1, 1)⟩ = {(0, 0), (1, 1), (0, 2), (1, 3)} = ⟨(1, 3)⟩
⟨(0, 2)⟩ = {(0, 0), (0, 2)}
⟨(1, 2)⟩ = {(0, 0), (1, 2)}.

Since none of these is of order 8, ℤ_{2} × ℤ_{4} is not cyclic. However, ℤ_{2} × ℤ_{4} is generated by (1, 0) and (0, 1).

30. (a) We use Theorem 7.10 to show that HK is a subgroup when G is abelian. First, HK ≠ ∅ since e ∈ H and e ∈ K, so e · e = e ∈ HK. If a, b ∈ HK, then a = h_{1}k_{1} and b = h_{2}k_{2}, where h_{1}, h_{2} ∈ H and k_{1}, k_{2} ∈ K. Since G is abelian,

ab = (h_{1}k_{1})(h_{2}k_{2}) = h_{1}(k_{1}h_{2})k_{2} = h_{1}(h_{2}k_{1})k_{2} = (h_{1}h_{2})(k_{1}k_{2}).

Since H and K are subgroups, h_{1}h_{2} ∈ H and k_{1}k_{2} ∈ K. Therefore ab ∈ HK and HK is closed under products. Finally if a = hk ∈ HK, then since G is abelian again,

a^{−1} = (hk)^{−1} = k^{−1}h^{−1} = h^{−1}k^{−1}.

Then h^{−1} ∈ H and k^{−1} ∈ K since H, K are subgroups of G. Therefore HK is also closed under inverses.

(b) Let G = S_{3}, and let H = {(), (12)}, K = {(), (13)}. Then

HK = {()(), ()(13), (12)(), (12)(13)} = {(), (13), (12), (132)}.

This is not a subgroup of S_{3} since it is not closed under products or inverses. (Comment: Note, it even violates the necessary condition from the statement of Lagrange’s Theorem!)

Section 7.4

3. (a) The group of invertible 2 × 2 matrices with entries in ℤ_{2} is GL(2, ℤ_{2}) =

\{I = (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}) , R = (\begin{matrix}0 & 1 \\ 1 & 1 \end{matrix}) , R^{2} = (\begin{matrix}1 & 1 \\ 1 & 0 \end{matrix}) , S = (\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}) , T = (\begin{matrix}1 & 0 \\ 1 & 1 \end{matrix}) , U = (\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix})\} .

(These are all the 2 × 2 matrices of determinant 1 with ℤ_{2}-entries, a group under matrix multiplication).

(b) Under the mapping I ↦ (), R ↦ (123), R^{2} ↦ (132), S ↦ (13), T ↦ (12), U ↦ (23), we see that the group tables will “match up”:

\begin{matrix}· & I & R & R^{2} & S & T & U \\ I & I & R & R^{2} & S & T & U \\ R & R & R^{2} & I & U & S & T \\ R^{2} & R^{2} & I & R & T & U & S \\ S & S & T & U & I & R & R^{2} \\ T & T & U & S & R^{2} & I & R \\ U & U & S & T & R & R^{2} & I \end{matrix}

and

\begin{matrix}∘ & () & (123) & (132) & (13) & (12) & (23) \\ () & () & (123) & (132) & (13) & (12) & (23) \\ (123) & (123) & (132) & () & (23) & (13) & (12) \\ (132) & (132) & () & (123) & (12) & (23) & (13) \\ (13) & (13) & (12) & (23) & () & (123) & (132) \\ (12) & (12) & (23) & (13) & (132) & () & (123) \\ (23) & (23) & (13) & (12) & (123) & (132) & () \end{matrix}

5. U_{5} is the group {1, 2, 3, 4} under multiplication mod 5, while U_{10} is the group {1, 3, 7, 9} under multiplication mod 10. Note that U_{5} is cyclic with generator 2, since 2^{2} = 4, and 2^{3} = 8 ≡ 3 mod 5. Similarly U_{10} is cyclic since 3^{2} = 9 and 3^{3} = 27 ≡ 7 mod 10. The mapping φ : U_{5} → U_{10} defined by

φ(1) = 1
φ(2) = 3
φ(4) = 9
φ(3) = 7

is an isomorphism since it is injective and surjective, and φ(2^{k}) = 3^{k} for all k, so

φ(2^{k} · 2^{ℓ}) = φ(2^{k+ℓ}) = 3^{k+ℓ} = 3^{k} · 3^{ℓ} = φ(2^{k}) · φ(2^{ℓ}).

10. Let f : G → H be a surjective homomorphism and assume G is abelian. Then for all a, b ∈ H, a = f(x) and b = f(y) for some x, y ∈ G and therefore

ab = f(x)f(y) = f(xy) = f(yx) = f(y)f(x) = ba.

This establishes the desired fact that H is abelian, since the equation holds for all a, b ∈ H.

11. We will prove this for n ≥ 1 by induction on n first. For the base case, we have f(a^{1}) = f(a) = (f(a))^{1}, so there is nothing to prove. Now assume that the statement is true for n = k. Since f is a group homomorphism, using the induction hypothesis at the second to last step:

f(a^{k+1}) = f(a^{k} · a) = f(a^{k}) · f(a) = (f(a))^{k}f(a) = (f(a))^{k+1}.

This establishes the statement for n ≥ 1. The formula is clearly true for n = 0 since f(a^{0}) = f(e) = e = (f(a))^{0}. Finally, we show that the formula is true for n = −m where m ≥ 1 by another induction proof. For the base case here we need to see

f(a^{−1}) = (f(a))^{−1}.

This is true by part 2 of Theorem 7.19. Now assume the formula has been established for m = k. Then

f(a^{−(k+1)}) = f(a^{−k} · a^{−1}) = f(a^{−k}) · f(a^{−1}) = (f(a))^{−k}(f(a))^{−1} = (f(a))^{−(k+1)}.

This establishes the statement for n ≤ −1 as well.

13. (a) We will ignore the hint and prove the parts separately. First we show that a^{−1}Ha is a subgroup of G. First, a^{−1}Ha is not empty since a^{−1}ea = e ∈ a^{−1}Ha. Next, let x = a^{−1}ha and y = a^{−1}ka be elements of a^{−1}Ha. Then

xy = (a^{−1}ha)(a^{−1}ka) = a^{−1}hka.

Since H is a subgroup, hk ∈ H. So xy ∈ a^{−1}Ha. This shows a^{−1}Ha is closed under products. Last, if x = a^{−1}ha ∈ a^{−1}Ha, then x^{−1} = (a^{−1}ha)^{−1} = a^{−1}h^{−1}(a^{−1})^{−1} = a^{−1}h^{−1}a ∈ a^{−1}Ha, since h^{−1} ∈ H (a subgroup). Hence a^{−1}Ha is also a subgroup of G.

Now, we show that a^{−1}Ha is isomorphic to H as a group. Consider the mapping φ : G → G defined by φ(x) = a^{−1}xa. (This is the “inner automorphism” mentioned in the problem.) We check that

φ(xy) = a^{−1}xya = a^{−1}x(aa^{−1})ya = (a^{−1}xa)(a^{−1}ya) = φ(x)φ(y).

This shows that φ is a group homomorphism. It is injective and surjective because φ has an inverse function ψ(x) = axa^{−1}: For all x ∈ G,

φ(ψ(x)) = a^{−1}(axa^{−1})a = x and ψ(φ(x)) = a(a^{−1}xa)a^{−1} = x.

Then φ restricts to a mapping with domain H and range a^{−1}Ha, and similarly ψ restricts to a mapping with domain a^{−1}Ha and range H. The restricted mappings are still inverses. This shows H and a^{−1}Ha are isomorphic via φ.

(b) If H is finite then a^{−1}Ha = φ(H) where φ is as in part a. Since φ is injective and surjective, |H| = |a^{−1}Ha|.

18. Let G = ⟨a⟩ and let f : G → H be a surjective group homomorphism. Then every element b ∈ H is b = f(c) for some c ∈ H. But by hypothesis, c = a^{n} for some integer n. Hence using problem 11 above, b = f(c) = f(a^{n}) = (f(a))^{n}. This shows that every element of H is a power of f(a), so H ⊂ ⟨f(a)⟩. The other inclusion is true by definition, hence H = ⟨f(a)⟩ is cyclic, generated by f(a).