January 1998 Algebra Prelim Solutions

1. Let a be a root of f in K. Then [F(a) : F]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 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 pn - 1 and since n2, 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 , w and w2 and it follows easily that K is the splitting field for X3 - 2. Therefore K/ is a Galois extension of . Also [() : ] = 3 and ()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 : ] = 6 and the Galois group of K over is isomorphic to S3.

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

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

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 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 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 H, we see that [G : H] divides [G : N] and the result follows.

1. A5 X /59.

2. /2 X /2 X /885.

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

 G a /piai H b /pibi

for certain integers a, b, ai, bi, n, and the pi are distinct primes. Since G G H H, we see that

2a /pi2ai 2b /pi2bi

Using the uniqueness statement in the structure theorem for finitely generated abelian groups, we see that 2a = 2b and 2ai = 2bi for all i. Therefore a = b and ai = bi for all i, which proves that G H.

Peter Linnell
1999-06-16