January 1998 Algebra Prelim Solutions

1. Let a be a root of f in K. Then [F(a) : F]$\ne$1 because f has no roots in F, is less than 6 because f has degree 6, and divides 21. It follows that [F(a) : F] = 3. Let g be the minimum polynomial of a over F. Then g is an irreducible polynomial of degree 3 which divides f in F[X], so we may write f = gh in F[X] where h has degree 3. Since h has no root in F, we see that h is irreducible in F[X], so f = gh is the factorization of f into irreducible polynomials in F[X].

   Since f has at most two roots in K, we see that g and h have at most one root in K. It follows that we may write g = g₁g₂ and h = h₁h₂ where g₁, g₂, h₁, h₂ are irreducible in K[X], g₁ and h₁ have degree 1, and g₂ and h₂ have degree 2. Then f = g₁g₂h₁h₂ is the factorization of f into irreducible polynomials in K[X].

2. The number of Sylow 59-subgroups divides 33 and is congruent to 1 modulo 59. Therefore there is only one Sylow 59-subgroup which means that G has a normal subgroup H of order 59.

   Now G/H is a group of order 33 and so the number of Sylow 3-subgroups of G/H is congruent to 1 modulo 3 and divides 11. Therefore G/H has a normal Sylow 3-subgroup, which we may write as A/H where A is a normal subgroup of G. Then A is a group of order 3*59 and the number of Sylow 3-subgroups of A is congruent to 1 modulo 3 and divides 59. Therefore A has a normal subgroup K of order 3. Observe that if g $\in$ G, then gKg^-1 is a subgroup of order 3 contained in gAg^-1. Since A is normal, gAg^-1 = A, so gKg^-1 is a subgroup of order 3 in A and hence gKg^-1 = K, because A has exactly one subgroup of order 3. Therefore K is a normal subgroup of order 3 in G.

   Using exactly the same argument as above with the primes 3 and 11 interchanged, we see that G has a normal subgroup L of order 11. We have now proved that all the Sylow subgroups of G are normal, so G is isomorphic to a direct product of its Sylow subgroups. Also each nontrivial Sylow subgroup has prime order and is therefore cyclic. It follows that G abelian, and then by using the structure theorem for finitely generated abelian groups, we conclude that G is cyclic.

3. Since G is a nontrivial p-group, its center is nontrivial and therefore it has a central subgroup Z of order p. Then G/Z is a group of order p^{n - 1} and since n$\ge$2, we see that G/Z is nontrivial p-group and hence it has a central subgroup of order p. We may write this subgroup as A/Z where A is a normal subgroup of G. Then A has order p² and since groups of order p² are abelian, it follows that A is a normal abelian subgroup of order p² as required.

4. The roots of X³ - 2 are $\sqrt[3]{2}$, $\sqrt[3]{2}$w and $\sqrt[3]{2}$w² and it follows easily that K is the splitting field for X³ - 2. Therefore K/$\mathbb {Q}$ is a Galois extension of $\mathbb {Q}$. Also [$\mathbb {Q}$($\sqrt[3]{2}$) : $\mathbb {Q}$] = 3 and $\mathbb {Q}$($\sqrt[3]{2}$)$\ne$K. Since the splitting field of a polynomial of degree 3 has degree dividing 6 and the Galois group is isomorphic to a subgroup of S₃, we conclude that [K : $\mathbb {Q}$] = 6 and the Galois group of K over $\mathbb {Q}$ is isomorphic to S₃.

5. 1. Obviously 0 $\in$ A $\cap$ R. Let a, b $\in$ A $\cap$ R. Then a - b $\in$ A and a - b $\in$ R, so a - b $\in$ A $\cap$ R. Finally let r $\in$ R. Then ar $\in$ A because A is an ideal of S, and ar $\in$ R. Thus ar $\in$ A $\cap$ R and we have proved that A $\cap$ R is an ideal of R.

   2. Let x $\in$ A. Since F is the field of fractions of the PID R, we may write x = ab^-1 with a, b $\in$ R and (a, b) = 1. Then there exist p, q $\in$ R such that ap + bq = 1, so px + q = b^-1. Since p, q, x $\in$ S, we see that b^-1 $\in$ S. Now A is an ideal of S, so xb = a $\in$ A $\cap$ R = Rd, so there exists r $\in$ R such that xb = rd. Then we have x = b^-1rd $\in$ Sd and the result follows.

6. 1. We have [G/M : PM/M] = [G : PM] and [G : P] = [G : PM][PM : P], so [G/M : PM/M] divides [G : P]. Since P is a Sylow p-subgroup of G, we see that [G : P] is prime to p and hence [G/M : PM/M] is prime to p. This shows that PM/M is a Sylow p-subgroup of G.

   2. Let n $\in$ N. Then nPn^-1 = P and nMn^-1 = M, hence nPMn^-1 = PM. This shows that Mn is in the normalizer of PM/M in G/M and we conclude that n $\in$ H. The result follows.

   3. The number of Sylow p-subgroups of G/M is [G/M : H/M] = [G : H], and the number of Sylow p-subgroups of G is [G : N]. Since N $\subseteq$ H, we see that [G : H] divides [G : N] and the result follows.

7. 1. A₅ X $\mathbb {Z}$/59$\mathbb {Z}$.

   2. $\mathbb {Z}$/2$\mathbb {Z}$ X $\mathbb {Z}$/2$\mathbb {Z}$ X $\mathbb {Z}$/885$\mathbb {Z}$.

8. We shall use the structure theorem for finitely generated abelian groups. We may write

   $$\cong \mathbb {Z} \oplus \bigoplus_{i=1}^{n} \mathbb {Z} \mathbb {Z}$$ G

   $$\cong \mathbb {Z} \oplus \bigoplus_{i=1}^{n} \mathbb {Z} \mathbb {Z}$$ H

   
   
   for certain integers a, b, a_i, b_i, n, and the p_i are distinct primes. Since G $\oplus$ G $\cong$ H $\oplus$ H, we see that
   $$\mathbb {Z}$$^2a $$\oplus$$ $$\bigoplus_{i=1}^{n}$$$$\mathbb {Z}$$/p_i^2a_i$$\mathbb {Z}$$ $$\cong$$ $$\mathbb {Z}$$^2b $$\oplus$$ $$\bigoplus_{i=1}^{n}$$$$\mathbb {Z}$$/p_i^2b_i$$\mathbb {Z}$$
   Using the uniqueness statement in the structure theorem for finitely generated abelian groups, we see that 2a = 2b and 2a_i = 2b_i for all i. Therefore a = b and a_i = b_i for all i, which proves that G $\cong$ H.