Mathematics 352 – Abstract Algebra II Midterm Exam 1 Solutions – February 27, 2008

I. Terminology.

A) (5) Let G be a group and let x in G. What is the conjugacy class of x in G?

Solution: The conjugacy class of x is

C_{x} = {g^{−1}xg ∈ G : g ∈ G}.

(In other words, this is the equivalence class of x for the conjugacy relation on G.)

B) (5) Let K be an extension field of the field F and let u ∈ K. What does it mean to say that u is transcendental over F?

Solution: The element u ∈ K is transcendental over F if there is no nonzero polynomial f(x) ∈ F [x] such that f(u) = 0 (or, equivalently, that u is not algebraic over F.)

1. C) (5) Let K be an extension field of the field F. What does it mean to say that K is normal over F?

Solution: K is a normal extension of F if K is algebraic over F and if p(x) is an irreducible polynomial with one root in K, then p(x) splits completely in K[x] (that is, in K[x],

p(x) = c(x − u_{1}) … (x − u_{n}),

so all the roots of p(x) are in K.)

II.

A) (10) State the Structure Theorem for finite abelian groups.

Solution: Every finite abelian group is a direct sum of cyclic subgroups of prime power order.

B) (10) Let G = ℤ_{42} ⊕ ℤ_{36} ⊕ ℤ_{84}. What are the elementary divisors and invariant factors of G?

Solution: From the factorizations 42 = 2 · 3 · 7, 36 = 2^{2} · 3^{2}, and 84 = 2^{2} · 3 · 7, we see that the elementary divisors of G are 2, 4, 4, 3, 3, 3^{2}, 7, 7, so

G ≃ ℤ_{2} ⊕ ℤ_{4} ⊕ ℤ_{4} ⊕ ℤ_{3} ⊕ ℤ_{3} ⊕ ℤ_{9} ⊕ ℤ_{7} ⊕ ℤ_{7}.

Regrouping these as usual, we see the invariant factors of G are 6, 84, 252, so that

G ≃ ℤ_{6} ⊕ ℤ_{84} ⊕ ℤ_{252}.

C) (5) What is the Sylow 7-subgroup of G from part B? Is this a cyclic group?

Solution: The Sylow 7-subgroup is isomorphic to ℤ_{7}⊕ℤ_{7} from the decompositions above. It is not cyclic. If we take the original direct sum decomposition G = ℤ_{42}⊕ℤ_{36}⊕ℤ_{84} = ⟨a⟩⊕⟨b⟩⊕⟨c⟩ then the Sylow 7-subgroup is generated by (a^{6}, 0, 0) and (0, 0, c^{12}) (two elements of order 7).

III.

1. A) (20) State and prove the First Sylow Theorem for finite groups.

Solution: The First Sylow Theorem states that if G is a finite group and p is a prime number such that p^{n}||G|, then G contains a subgroup of order p^{n}.

The proof is by induction on |G|. When |G| = 1, there is nothing to prove. So now for the induction step, assume that the statement of the theorem is true for all groups of order < m and consider G of order m. By the Class Equation, we have

|G| = |Z(G)| + ∑_{i=1}^{t}[G : C(g_{i})],

where the g_{i} are representatives of the distinct conjugacy classes of size > 1 and C(g_{i}) is the centralizer of g_{i}.

Case 1: Assume that there is some i such that [G : C(g_{i})] is not divisible by p. Then p^{n}||C(g_{i})|. We must have |C(g_{i})| < |G| since g_{i} ∉ Z(G). Therefore, by the induction hypothesis, C(g_{i}) has a subgroup of order p^{n}. This subgroup is also a subgroup of G and we are done in this case.

Case 2: Now assume that p|[G : C(g_{i})] for all i = 1, …, t. Then since p||G|, we also have p||Z(G)|. Now Z(G) is a finite abelian group, so we know by a consequence of the Structure Theorem that Z(G) contains an element x of order p. Let C = ⟨x⟩ be the subgroup generated by this x. Since C ⊆ Z(G), C is normal in G and we can form G/C which has order divisible by p^{n−1}. By the induction hypothesis applied to G/C, G/C has a subgroup H of order p^{n−1}. But then the inverse image of H under the quotient map G ⟶ G/C is a subgroup of G of order p^{n}.

B) (10) Use the Sylow Theorems to show that every group of order 33 is cyclic.

Solution: By the Third Sylow Theorem, the number of Sylow 3-subgroups is congruent to 1 mod 3 and divides 33. The only possibility is 1, so let H be the unique Sylow 3-subgroup. Similarly, there is exactly one Sylow 11-subgroup K. This implies that H and K are normal subgroups. Moreover H ∩ K = {e}, since if not any x ≠ e would simultaneously have order 3 and order 11, which is impossible. Therefore, G ≃ H × K ≃ ℤ_{3} × ℤ_{11} ≃ Z_{33}, where the last isomorphism follows because (3, 11) = 1.

A) (5) Let K be an extension field of F, and let u ∈ K. Show that if u^{2} is algebraic over F, then u is also algebraic over F.

Solution: (Method 1) If u^{2} is algebraic over F, then there is some nonzero polynomial p(x) ∈ F [x] such that p(u^{2}) = 0. This implies that the polynomial q(x) = p(x^{2}) has u as a root. Therefore u is also algebraic over F. (Note: If p(x) = a_{n}x^{n}+⋯+a_{1}+a_{0}, then the polynomial q(x) is just q(x) = a_{n}x^{2n} + ⋯ + a_{1}x^{2} + a_{0}.)

Solution: (Method 2) Since u^{2} is algebraic over F, we know that [F (u^{2}) : F] = n, the degree of the minimal polynomial of u^{2} in F [x]. Now consider the polynomial x^{2} − u^{2} ∈ F (u^{2})[x]. This has u as a root, so u is algebraic over F (u^{2}), and [F (u^{2})(u) : F (u^{2})] = [F (u) : F (u^{2})] ≤ 2. Therefore [F (u) : F] = [F (u) : F (u^{2})][F (u^{2}) : F] ≤ 2n is finite, which implies that u is algebraic over F.

B) (10) Show that ℚ(\sqrt{2}, \sqrt{3}) is a splitting field of f(x) = x^{4} + 2x^{3} − 8x^{2} − 6x − 1 over ℚ. (Hint: Look for quadratic factors of f(x).)

Solution: The factorization of f(x) in ℚ[x] is f(x) = (x^{2} + 4x + 1)(x^{2} − 2x − 1). The roots of the first factor are x = \frac{−4±\sqrt{16−12}}{2} = −2 ± \sqrt{3}. The roots of the second factor are x = \frac{2±\sqrt{4+4}}{2} = 1 ± \sqrt{2}. Hence the splitting field of f(x) over ℚ is, by definition,

K = ℚ(−2 + \sqrt{3}, −2 − \sqrt{3}, 1 + \sqrt{2}, 1 − \sqrt{2}).

We claim that K = ℚ(\sqrt{2}, \sqrt{3}). The inclusion K ⊆ ℚ(\sqrt{2}, \sqrt{3}) is clear since each root of f is contained in ℚ(\sqrt{2}, \sqrt{3}). The other inclusion follows since \sqrt{2} = \frac{1}{2}(1 + \sqrt{2}) + \frac{−1}{2}(1 − \sqrt{2}) and \sqrt{3} = \frac{1}{2}(−2 + \sqrt{3}) + \frac{−1}{2}(−2 − \sqrt{3}) are both in K. So ℚ(\sqrt{2}, \sqrt{3}) ⊆ K.

V. (15) Show that if F ⊆ E ⊆ K are field extensions and [E : F], [K : E] are both finite, then [K : F] = [K : E][E : F].

Solution: Say {u_{1}, …, u_{m}} is a basis for E over F and {v_{1}, …, v_{n}} is a basis for K over F. Then each element a in K can be written as

a = c_{1}v_{1} + ⋯ + c_{n}v_{n}

for some c_{i} ∈ E. Similarly, we have c_{i} = a_{i1}u_{1} + ⋯ + a_{im}u_{m} were a_{ij} ∈ F. Substituting into the last displayed equation,

a = (a_{11}u_{1} + ⋯ + a_{1m}u_{m})v_{1} + ⋯ + (a_{n1}u_{1} + ⋯ + a_{nm}u_{m})v_{n}
= ∑_{i=1}^{n} ∑_{j=1}^{m} a_{ij}v_{i}u_{j}.

This shows that the m · n elements v_{i}u_{j} for i = 1, …, n and j = 1, …, m span K over F. To show that these form a basis of K over F, we need to show that they are linearly independent. Suppose there are a_{ij} ∈ F such that

0 = ∑_{i=1}^{n} ∑_{j=1}^{m} a_{ij}v_{i}u_{j}
= (a_{11}u_{1} + ⋯ + a_{1m}u_{m})v_{1} + ⋯ + (a_{n1}u_{1} + ⋯ + a_{nm}u_{m})v_{n}.

Since the v_{i} are linearly independent over E, this shows that

a_{i1}u_{1} + ⋯ + a_{im}u_{m} = 0

for all i = 1, …, n. But then, the u_{j} are linearly independent over F, so all a_{ij} = 0. This shows the linear independence so {v_{j}u_{i} : 1 ≤ i ≤ n, 1 ≤ j ≤ m} is a basis of K over F and

[K : F] = m · n = [E : F][K : E].

Extra Credit (10) Let p(x) and q(x) be irreducible polynomials in F [x] such that deg p(x) and deg q(x) are relatively prime integers. Show that if u is a root of p(x) in some extension field of F, then q(x) is also irreducible in F (u)[x].

Solution: If u is a root of p(x) in some extension field of F, then by Theorem 10.7, [F (u) : F] = deg p(x) = m. Similarly if v is a root of q(x), we have [F (v) : F] = deg q(x) = n. Since m = deg p(x) and n = deg q(x) are relatively prime, it follows as in Exercise 11 from Section 10.3 of Hungerford that [F (u, v) : F] = mn. From the tower F ⊆ F (u) ⊆ F (u, v), we get [F (u, v) : F] = mn = [F (u, v) : F (u)][F (u) : F], so [F (u, v) : F (u)] = n. This says that the minimal polynomial of v over F (u) must have degree n, and it must be a divisor of q(x). Since q(x) has degree n itself, that minimal polynomial must be q(x), and hence q(x) is still irreducible in F (u)[x].