Fall 1995 Algebra Prelim Solutions

1.
The order of G is 53 73. The number of Sylow 5-subgroups of G divides 73 and is congruent to 1 modulo 5; the only possibility is 1. Therefore G has a normal Sylow 5-subgroup A. The number of Sylow 7-subgroups of G divides 53 and is congruent to 1 modulo 7; the only possibility is 1. Therefore G has a normal Sylow 7-subgroup B. Since (| A|,| B|) = 1, we see that A B = 1. We next show that every element of A commutes with every element of B. Suppose a A and b B. Then aba-1b-1 = a(ba-1b-1) and since A G, we see that ba-1b-1 A and consequently aba-1b-1 A. Similarly aba-1b-1 B and we deduce that aba-1b-1 A B = 1. Therefore aba-1b-1 = 1 and we conclude that ab = ba, in other words every element of A commutes with every element of B.

Since a group of prime power order has normal subgroups of order m for all m dividing the order of the group, we see that A has a normal subgroup H of order 25. From the previous paragraph, B centralizes H and so certainly normalizes H. Thus A and B normalize H, hence | A| and | B| divide the order of the normalizer of H in G and we conclude that H G. This completes the solution.

2.
First we write G as a direct product of cyclic groups of prime power order: G /4/9/9/5 . Any subgroup of G is isomorphic to a product of subgroups, where one subgroup is taken from each factor. Thus H /9/3 or /3/9 . These last two groups are isomorphic, so we conclude that H /3/9 .

3.
(a)
Since D is a conjugacy class in f-1(C), we may write D = {bdb-1 | b f-1(C)} for some fixed d D. Then

f (D) = {f (bdb-1) = f (b)f (d )f (b)-1 | b f-1(C)}.

Therefore f (D) = {cf (d )c-1 | c C} , and (a) follows.
(b)
Let D(g) denote the conjugacy class of g in f-1(C). Since f (g) is centralized by C, the conjugacy class containing f (g) is precisely {f (g)} . Since f (D(g)) is by (a) the conjugacy class containing f (g), we see that | f (D(g))| = 1. Therefore all elements of D(g) are in the same coset of ker f and we conclude that | D(g)|| ker f| as required.

(c)
Let K denote the centralizer of g in f-1(C). Then the order of the centralizer of g in G is at least | K|. Now the order of the conjugacy class of g in f-1(C) is [f-1(C) : K], and by (b) this order is at most | ker f|. Therefore [f-1(C) : K]| ker f|, consequently

| K|| f-1(C)/ker f| = | C|

because f-1(C)/ker f C. The result follows.

4.
(a)
If R has no prime elements, then R is a field and so certainly a PID. Therefore we may suppose that R has exactly one prime p (up to a multiple of a unit), and we need to prove that R is a PID. Let I be a nonzero ideal of R. Then each nonzero element of I can be written in the form upn for some nonnegative integer n and some unit u, because p is the only prime (up to a multiple of a unit) of R. Let N be the smallest nonnegative integer such that upN I for some unit u. We now show that I = pNR; clearly pNR I. If x I \ 0, then we may write x = vpn for some unit v and some integer nN. Thus x = pNvpn - N which shows that x pNR, and the result follows.
(b)
Suppose every maximal ideal of R is principal. Then each maximal ideal of R is of the form pR where p is a prime of R. Suppose by way of contradiction that I is a nonprincipal ideal of R. Clearly 0IR. Choose a nonzero element x I, and write x = p1d1... pndn, where the pi are nonassociate primes and the di are positive integers.

For each prime p, let e(p) denote the largest integer such that pe(p)R I. If p is not an associate of one of the pi, then e(p) = 0. Set y = p1e1... pnen. We claim that I = yR.

First we show that I yR. If z I, then by unique factorization we may write z = qp1f1... pnfn, where the fi are nonnegative integers and q is a product of primes which are not associate to any of the pi. Again using unique factorization, we must have fiei for all i and we deduce that z yR.

Finally we show that yR I. Set J = {r R | yr I} (so J = y-1I). Clearly J is an ideal of R and yJ = I. If JR, then by Zorn's lemma J is contained in a maximal ideal of R, which we may assume is of the form pR where p is a prime of R. It would follow that ypR I, which contradicts the maximality of the e(p). Therefore J = R and we deduce that yR I. Thus I = yR and the proof is complete.

5.
Let : N XN denote the projection onto N, so (n, x) = n for all n N and x X, and let denote the identity map on N. Then (f )i = = . This shows that i has a left inverse, consequently the sequence

0 N M M/N 0

splits (the map MM/N above is of course the natural epimorphism). This shows that M N M/N as required.

6.
Let G denote the Galois group of E over F. Since E is a Galois extension of F with [E : F] = pn, we see that | G| = pn. Since G is a p-group, it has a series G = G0 G1 ... Gn - 1 Gn = 1, with Gi G and | Gi/Gi - 1| = p for all i. Set Ki = {e E | ge = e for all g Gi} . Then by the Galois correspondence, Ki is normal over F and [Ki : Ki - 1] = [Gi : Gi - 1] = p for all i, as required.

7.
Suppose K is a finite field which is algebraically closed. Let n be a positive integer which is prime to the characteristic of K, and consider the polynomial Xn - 1. The derivative of Xn - 1 is nXn - 1 which is prime to Xn - 1 in K[X], because n is prime to the characteristic of K. This tells us that the roots of Xn - 1 in a splitting field for K are distinct. If K is algebraically closed, then all these roots would be in K, and we would deduce that | K|n. Since n can be arbitrarily large, this would contradict the assumption that K is finite, and the result is proven.

Peter Linnell
1997-09-28