January 1998 Algebra Prelim Solutions
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
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.
f = g1g2h1h2 is the factorization of f into
irreducible polynomials in K[X].
- 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
gAg-1 = A, so gKg-1 is a subgroup of order 3 in A
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
- 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.
- The roots of X3 - 2 are
w2 and it follows easily that K is the
splitting field for X3 - 2. Therefore
K/ is a Galois
[() : ] = 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.
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.
ar A R and we have proved that A R is an ideal
- 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.
- We have
[G/M : PM/M] = [G : PM] and
[G : P] = [G : PM][PM : P],
[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.
- 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.
- 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].
N H, we see that [G : H] divides [G : N] and the
- A5 X
- We shall use the structure theorem for finitely generated
abelian groups. We may write
for certain integers
a, b, ai, bi, n, and the pi are distinct
G G H H, we see that
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.