Fall 1994 Algebra Prelim Solutions

1.
(i)
By definition (x) = {gxg-1 | g G} . Since x H G, we see that gxg-1 H for all g G and we deduce that (x) H. Let CG(x) denote the centralizer of x in G. Then |(x)| = [G : CG(x)]. This last quantity is a power of p and it cannot be 1 because x is not in the center of G. We deduce that |(x)| is a nontrivial power of p and (i) follows.
(ii)
It follows from (i) that H Z is a union of conjugacy classes of the form (x) where x H Z. Since different conjugacy classes intersect trivially, we see from (i) that p divides | H Z|. Also | H| divides | G| and so | H| is a nontrivial power of p. We deduce that p divides | Z H| and (ii) is proven.
(iii)
Let H be the centralizer of A in G. Then H is a normal subgroup of G containing A and it will be sufficient to show that H = A. Let X/A be the center of G/A and suppose H > A. Then we see from (ii) that X/A H/A > 1 because H/A G/A, and we deduce that there is a nontrivial cyclic subgroup Y/A contained in X/A H/A. Since Y/A X/A we see that Y/A G and we deduce that Y G. Also A is contained in the center of Y and Y/A is cyclic, hence Y is abelian. This contradicts the fact that A is a maximal normal abelian subgroup of G and the result follows

2.
(i)
The number of Sylow 5-subgroups of G divides 36 and is congruent to 1 modulo 5, which means that this number must be 1,6 or 36. It cannot be 1, for then G would have a normal Sylow 5-subgroup, which contradicts the fact that G is simple. Nor can G have 6 Sylow 5-subgroups, for then G would be isomorphic to a subgroup of A6 because G is simple. This would mean that A6 has a subgroup of index 2, and since subgroups of index 2 are always normal, this would contradict the fact that A6 is simple. We conclude that G has 36 Sylow 5-subgroups.
(ii)
Let N be the normalizer of a Sylow 3-subgroup. Then the number of Sylow 3-subgroups is [G : N]. Also the number of Sylow 3-subgroups divides 20 and is congruent to 1 modulo 3, so this number is 1,4 or 10. It cannot be 1 because that would mean G has a normal Sylow 3-subgroup, which would contradict the fact that G is simple. Nor can it be 4, for then G would be isomorphic to a subgroup of A4 because G is simple, which is clearly impossible. Therefore the number of Sylow 3-subgroups is 10. We deduce that [G : N] = 10 and hence N has order 18.

(iii)
A Sylow 3-subgroup of a group of order 18 has order 9. This means the subgroup has index 2, hence the subgroup is normal because in any group, a subgroup of index 2 is normal.

(d)
Let C be the centralizer in G of A B and suppose A B is not 1. Then C contains A, B and A B is a normal subgroup in C, so C cannot be the whole of G. Therefore the order of C is a multiple of 9 and divides 180, and is neither 9 nor 180. Let d be the index of C in G, so | C| = | G|/d. Then G is isomorphic to a subgroup of Ad because G is simple. Since the order of Ad is less than 180 if d5, we see that d6 and consequently C has order 18. From part (iii), the Sylow 3-subgroup of C is normal in C and therefore C has exactly one subgroup of order 9. We now have a contradiction because C has two subgroups of order 9, namely A and B. We conclude that A B = 1.

(e)
We count the elements in G. Since two Sylow 5-subgroups intersect in 1 and there are 36 Sylow 5-subgroups by (i), we see that G has 36*4 = 144 elements of order 5. Also two Sylow 3-subgroups intersect in 1 by (iv) and G has 10 Sylow 3-subgroups by (ii). Therefore G has 8*10 = 80 elements of order 3 or 9. We conclude that G has at least 144 + 80 = 224 elements, which contradicts the fact that G has only 180 elements. It follows that no such G can exist and thus there is no simple group of order 180.

3.
Since M is a cyclic R-module, we know that M R/Rs for some s R. By the uniqueness part of the fundamental structure theorem for finitely generated modules over a PID, we cannot write R = A B where A, B are nonzero R-modules. Therefore s 0. Since sM = 0, we may take r = s.

Since rM = 0, we see that rR sR and hence s divides r. Suppose there do not exist distinct primes p, q dividing r. Then the same is true for s because s divides r, and we deduce that s is a prime power, say pe for some prime p. From the uniqueness statement in the fundamental structure theorem for finitely generated modules over a PID, we cannot write R/Rpe = A B where A, B are nonzero R-modules and we have a contradiction. This finishes the proof.

4.
(i)
Choose integers r, s such that qr + 2s = 1. Then in [[X]]/(X - 2) we have qr = 1 - sX mod (X - 2). Since 1 - sX is invertible in [[X]] (with inverse 1 + sX + s2X2 + ...), it follows that q is invertible in [[X]]/(X - 2).
(ii)
The general element of R is aiXi mod (X - 2), where ai for all i. Now each ai is of the form p/q where p, q and q is odd. Using (i), we may now write ai = bi mod (X - 2) where bi [[X]], and then we may write the general element of R in the form biXi mod (X - 2). This proves that is surjective. We now determine the kernel of . Obviously (X - 2) ker . Conversely suppose f ker . Then we may write f = (X - 2)g where g [[X]]. We want to show that g [[X]]. Write g = giXi where gi . Then the coefficient of Xn in g(X - 2) is gn - 1 - 2gn for n > 0, and the constant coefficient is -2g0. By induction on n we see that 2gn and since gn , we conclude that gn for all n. This proves that ker = (X - 2) and it now follows from the fundamental isomorphism theorem that R [[X]]/(X - 2).

(iii)
By considering the homomorphism [X] determined by sending X to 2, we see that [X]/(X - 2) . Since 3 is not invertible in , we see that 3 is not invertible in [X]/(X - 2). But 3 is invertible in R and the result follows.

5.
(i)
Obviously K() K(). Also is separable over K() and satisfies the polynomial Xp - . Since is the only root of Xp - , it follows that K() and hence K() = K(). Now we consider the minimum polynomial of over K. This has degree p because [K() : K] = p, and must be a polynomial in Xp because is not separable over K. Thus the minimum polynomial must be of the form Xp - b for some b K and it follows that = b K.

(ii)
Since we are in characteristic p, we have ( + )p = + . But K by (i), hence K( + ). Therefore K() K( + ) and it now follows from (i) that K() K( + ) as required.

(iii)
Since K( + ) K() by (ii), we have [K( + ) : K] = [K( + ) : K()][K() : K] and since [K() : K] = d, it remains to prove that [K( + ) : K()] = p. Now ( + )p = + K() which shows that [K( + ) : K()] = p or 1, because we are in characteristic p. It remains to prove that [K( + ) : K()] 1, or equivalently that K(). But is separable over K, hence every element of K() is separable over K which shows that K() because is not separable over K. This completes the proof.

6.
(i)
It will be sufficient to prove that Xp - t is irreducible in [t, X], or equivalently that t - Xp is irreducible in [X, t]. Let R = [X] and let F = (X), the field of fractions of R. Then t - Xp is a monic polynomial in R[t] and is irreducible in F[t], hence it is irreducible in R[t] = [X, t] and the result follows.
(ii)
Let y be one of the roots of Xp - t in L. Since Xp - t is irreducible in K[X], we see that [K(y) : K] = p. Also the roots of Xp - t are e2ni/py where n = 0,..., p - 1, and since e2ni/p for all n, we deduce that all the roots of Xp - t are in K(y). It follows that K(y) = L and we conclude that [L : K] = p. Therefore the Galois group of L over K has order p (note that L/K is a separable extension because we are in characteristic zero). Since groups of order p are cyclic, we conclude that the Galois group of L over K is cyclic of order p and hence isomorphic to /p .

Peter Linnell
1998-06-21