Mathematics 352 – Abstract Algebra II Solutions for Problem Set 4 February 11, 2008

Section 8.5

17. Let G be a group of order 21. We have 21 = 3 · 7. Since 7 ≡ 1 mod 3, Corollary 8.18 does not apply. However, by Sylow 3, there is one Sylow 7-subgroup H, which is normal by Sylow 2. Moreover, there are either 1 or 7 Sylow 3-subgroups.

In the case that the Sylow 3-subgroup K is unique the reasoning of the proof of Corollary 8.18 applies. We have H ∩ {e} and HK = G since HK contains 21 distinct elements of G. Therefore by Theorem 8.3,

G ≃ H × K ≃ ℤ_{7} × ℤ_{3} ≃ ℤ_{21},

and G is cyclic.

We now consider the case where there are 7 Sylow 3-subgroups K_{1}, …, K_{7}. We claim that there is only one possible structure for G, up to isomorphism: We will show that G is generated by two elements a, b with a^{7} = e, b^{3} = e and b^{−1}ab = a^{2}. Let H = ⟨a⟩ so |a| = 7. Similarly, let K_{1} = ⟨b⟩, so |b| = 3. Note that since H ∩ K = {e}, we have G = {b^{j}a^{i} : 0 ≤ k ≤ 6, 0 ≤ j ≤ 2}. Since H is normal, the conjugate b^{−1}ab = a^{k} for some k ∈ {1, 2, 3, 4, 5, 6}. Moreover,

b^{−2}ab^{2} = b^{−1}(b^{−1}ab)b = b^{−1}a^{k}b = (b^{−1}ab)^{k} = (a^{k})^{k} = a^{k^{2}}
a = b^{−3}ab^{3} = a^{k^{3}},

so

k^{3} ≡ 1 mod 7.  (1)

As in the dihedral group case, knowing k in (1), or the equivalent form ab = ba^{k} lets us determine the products

(b^{j}a^{i})(b^{j′}a^{i′}) = b^{j+j′}a^{ik^{j′}+i′}.

This gives closure under the product operation. Associativity follows from this also since

((b^{j}a^{i})(b^{j′}a^{i′})) (b^{j′′}a^{i′′}) = (b^{j+j′}a^{ik^{j′}+i′})(b^{j′′}a^{i′′}) = b^{(j+j′)+j′′}a^{(ik^{j′}+i′)k^{j′′}+i′′}
(b^{j}a^{i}) ((b^{j′}a^{i′})(b^{j′′}a^{i′′})) = (b^{j}a^{i})(b^{j′+j′′}a^{i′k^{j′′}+i′′}) = b^{j+(j′+j′′)}a^{ik^{j′+j′′}+i′k^{j′′}+i′′},

which are the same. The identity element is e = b^{0}a^{0}. The inverse of b^{j}a^{i} is b^{j′}a^{i′} where j′ + j ≡ 0 mod 3 and ik^{j′} + i′ ≡ 0 mod 7. Note that ik^{j′} + i′ ≡ 0 mod 7 implies that i′k^{j} + i ≡ 0 mod 7 also when j′ + j ≡ 0 mod 3 because of (1), so this is a well-defined two-sided inverse. So we get a group of order 21 whenever the condition (1) holds.

• If k = 1, then G is abelian and we are back in the first case.

• We cannot have k = 3, 5, 6 since then 3^{3} = 27 ≡ 6 mod 7, 5^{3} = 125 ≡ 6 mod 7, and 6^{3} = 216 ≡ 6 mod 7.

• This leaves k = 2, 4. We claim these two groups are actually isomorphic. This follows since if b^{−1}ab = a^{2}, then b^{−2}ab^{2} = a^{4}. The element b^{2} also generates the Sylow 3-subgroup K_{1}, so in fact this is the same group up to isomorphism.

Section 10.2

17. (a) We have x = \sqrt{1 + \sqrt{5}} so x^{2} = 1 + \sqrt{5} and (x^{2} − 1)^{2} = 5, so x^{4} − 2x^{2} − 4 = 0. It is not hard to check that this polynomial is irreducible over ℚ, so this is the minimal polynomial.

(b) For x = i\sqrt{3} + \sqrt{2}, x^{2} = −1 + 2i\sqrt{6}, x^{3} = −3i\sqrt{3} − 7\sqrt{2} and x^{4} = −23 − 4i\sqrt{6}. Hence x^{4} + 2x + 25 = 0. It is not hard to check that this polynomial is irreducible over ℚ, so this is the minimal polynomial.

19. By Theorem 10.4, we have [F (u) : F] = [F (u) : E][E : F]. However, by Theorem 10.7, [F (u) : F] = p, which is prime. Therefore either [F (u) : E] = p and [E : F] = 1, in which case E = F, or [F (u) : E] = 1 and [E : F] = p, in which case E = F (u).

20. We have F ⊆ F (u^{2}) ⊆ F (u) since u^{2} ∈ F (u). Therefore, [F (u) : F] = [F (u) : F (u^{2})][F (u^{2}) : F] by Theorem 10.4. By Theorem 10.7, [F (u) : F (u^{2})] is either 1 or 2 since the polynomial f(x) = x^{2} − u^{2} with coefficients in F (u^{2}) has x = u as a root. But the minimal polynomial of u over F has odd degree so [F (u) : F] is odd by Theorem 10.7. This implies that [F (u) : F (u^{2})] = 1. Hence F (u) = F (u^{2}).

23. If u ∈ K but u ∉ ℚ, then have ℚ ⊆ ℚ(u) ⊆ K By Theorem 10.4 and the hypothesis [K : ℚ] = 2, this implies [ℚ(u) : ℚ] = 2 and [K : ℚ(u)] = 1, so K = ℚ(u). By Theorem 10.7, the minimal polynomial of u is of the form x^{2} + bx + c for some b, c ∈ ℚ. Applying the quadratic formula,

u = \frac{−b ± \sqrt{b^{2} − 4c}}{2} = \frac{−b}{2} ± \frac{\sqrt{b^{2} − 4c}}{2}.

First, we claim that ℚ(u) = ℚ(\sqrt{b^{2} − 4c}). The ⊆ inclusion here is clear since u = \frac{−b}{2} ± \frac{1}{2} \sqrt{b^{2} − 4c} ∈ ℚ(\sqrt{b^{2} − 4c}). Hence ℚ(u) ⊆ ℚ(\sqrt{b^{2} − 4c}). On the other hand \sqrt{b^{2} − 4c} = ∓(2u + b) which implies \sqrt{b^{2} − 4c} ⊂ ℚ(u) and hence ℚ(\sqrt{b^{2} − 4c}) ⊆ ℚ(u).

Next, write b^{2} − 4c = \frac{m^{2}}{n^{2}} · \frac{e}{f} where m, n, e, f ∈ ℤ are integers and e, f are not divisible by the

square of any prime, and assume b^{2} − 4c = \frac{m^{2}e}{n^{2}f} is in lowest terms. Then \sqrt{b^{2} − 4c} = \frac{m}{n} \sqrt{\frac{e}{f}} and reasoning similarly to the above, we have

ℚ(u) = ℚ(\sqrt{b^{2} − 4c}) = ℚ (\sqrt{\frac{e}{f}} ) .

Finally, note that \sqrt{\frac{e}{f}} = \frac{\sqrt{ef}}{f}, where f ∈ ℤ, and d = ef is not divisible by the square of any prime in ℤ. Then by similar reasoning to the above, we have

ℚ(u) = ℚ (\sqrt{\frac{e}{f}} ) = ℚ(\sqrt{d}),

which is what we wanted to show.

25. Aiming for a contradiction, suppose that u is transcendental over F, but some element of F (u) not in F is algebraic over F. Then we have some

v = \frac{a_{0} + a_{1}u + ⋯ + a_{n}u^{n}}{b_{0} + b_{1}u + ⋯ + b_{m}u^{m}} ∈ F (u)

(but not in F, so v ≠ 0) which is a root of some nonzero polynomial f(x) = c_{0} + c_{1}x + ⋯ + c_{k}x^{k} ∈ F [x]. In the expressions for v and for f(x), we may assume that a_{n} ≠ 0, b_{m} ≠ 0, and c_{0}, c_{k} ≠ 0 in F. (The reason we can assume c_{0} ≠ 0 is that v ≠ 0. So if v is a root of any polynomial in F [x], then v is a root of some polynomial in F [x] with nonzero constant term.) We may also assume that f(x) = 0 has no roots in F. If there are any such roots a ∈ F, then we know f(x) will factor in F [x] as f(x) = (x − a)q(x) for some q(x) ∈ F [x]. Then f(v) = 0 implies q(v) = 0 since v ≠ a ∈ F. So we can remove any factors producing roots in F and we will be left with f(x) having no roots in F.

Substitute x = v ≠ 0 in f(x) and put the terms over a common denominator and collect like powers of u on the top. The result is an equation

0 = c_{0} + c_{1}v + ⋯ + c_{k}v^{k} = c_{0} + ⋯ + c_{k} (\frac{a_{0} + a_{1}u + ⋯ + a_{n}u^{n}}{b_{0} + b_{1}u + ⋯ + b_{m}u^{m}} )^{k} = \frac{A_{0} + A_{1}u + ⋯ + A_{N}u^{N}}{(b_{0} + b_{1}u + ⋯ + b_{m}u^{m})^{k}} ,

where A_{i} ∈ F for all i. This implies

0 = A_{0} + A_{1}u + ⋯ + A_{N}u^{N}.

If n > m then N = nk and A_{N} = c_{k}a_{n}^{k} ≠ 0 so the polynomial g(x) = A_{0} + A_{1}x + ⋯ + A_{N}x^{N} is not zero in F [x]. Similarly, if m > n, then N = mk and A_{N} = c_{0}b_{m}^{k} ≠ 0, so g(x) is not zero in F [x]. Finally, if m = n, then N = mk = nk and the coefficient A_{N} is

A_{N} = c_{0}b_{n}^{k} + c_{1}b_{n}^{k−1}a_{n} + ⋯ + c_{k−1}b_{n}a_{n}^{k−1} + c_{k}a_{n}^{k}.

If A_{N} = 0, then dividing through by b_{n}^{k} we get

0 = c_{0} + c_{1} (\frac{a_{n}}{b_{n}} ) + ⋯ + c_{k} (\frac{a_{n}}{b_{n}} )^{k} .

But this implies that f(x) = 0 has x = \frac{a_{n}}{b_{n}} ∈ F as a root. This contradicts what we said above about f(x). Hence A_{N} ≠ 0 in this case too. It follows that u is a root of a nonzero polynomial in F [x], which is a contradiction to the hypothesis that u is transcendental over F.