August 1999 Algebra Prelim Solutions

  1. We first factor 480 as 32*3*5. Note that since G is simple, it cannot have a nontrivial subgroup of index $ \le$7, because that would mean that G is isomorphic to a subgroup of A7, which is not possible by Lagrange's theorem.

    1. Let A = P $ \cap$ Q and suppose A > 1. Since P, Q $ \leqslant$ CG(A), we see that P < CG(A). Using Lagrange's theorem, we deduce that | CG(A)| = 96, 160 or 480. We cannot have 480 because then A would be a central and hence a normal subgroup of G, which would contradict the hypothesis that G is simple. Also we cannot have | CG(A)| = 96 or 160, because that would mean that G has a subgroup of index 5 or 3. We now have a contradiction, and we conclude that A = 1.

    2. The number of Sylow 2-subgroups is congruent to 1 modulo 2 and divides 15. It cannot be 1 because that would mean that G has a normal Sylow 2-subgroup. Nor can it be 3 or 5, because then G would have a subgroup of index 3 or 5. Therefore G has 15 Sylow 2-subgroups. Next the number of Sylow 3-subgroups is congruent to 1 modulo 3 and divides 96. This number cannot be 1, because that would mean that the Sylow 3-subgroup is normal. Nor can it be 4, because that would yield a subgroup of index 4. Therefore G has at least 10 Sylow 3-subgroups. Finally the number of Sylow 7-subgroups is congruent to 1 modulo 7 and divides 160. This number cannot be 1, because that would mean that the Sylow 7-subgroup is normal. Therefore G has at least 8 Sylow 7-subgroups.

      We now count elements. Since by (a) any two Sylow 2-subgroups intersect trivially, there are 15*31 nontrivial elements whose order is a power of 2. Next any two Sylow 3-subgroups intersect trivially, because a Sylow 3-subgroup has prime order 3, and we see that there are at least 10*2 elements of order 3. Finally any two Sylow 5-subgroups intersect trivially, because a Sylow 5-subgroup has prime order 5, and we deduce that G has at least 6*4 elements of order 5. We now count elements: we find that G has at least 15*31 + 10*2 + 6*4 = 509 nontrivial elements. Since G has only 480 elements altogether, we have now arrived at a contradiction. We conclude that there is no such group G.

  2. Since R is a domain and 0$ \notin$S, we see that S-1R is a domain. Also for a, b $ \in$ R and s, t $ \in$ S, we have a/s = b/t if and only if at = bs. Suppose p/1 divides (a/s)(b/t) in S-1R. This means that there exists c/u $ \in$ S-1R such that (p/1)(c/u) = (a/s)(b/t), which means that pstc = abu. Since p is prime, we see that p divides at least one of a, b, u. If p divides u, then p/1 is a unit in S-1R because u/1 is a unit in S-1R. Therefore we may assume that p does not divide u; without loss of generality, we may assume that p divides a, say pq = a. Then (p/1)(q/s) = a/s and we see that p/1 divides a/s. Therefore if p/1 is not a unit, it is prime and the result follows.

  3. Let P be a finitely generated projective k[X]/(X3 + X)-module. Then there is a k[X]/(X3 + X)-module Q and an integer e such that P $ \oplus$ Q $ \cong$ (k[X]/(X3 + X))e. Note that X3 + X = X(X + 1)2 and k[X]/(X3 + X) $ \cong$ k[X]/(X) $ \oplus$ k[X]/(X + 1)2, so P $ \oplus$ Q $ \cong$ (k[X]/(X))e $ \oplus$ (k[X]/(X + 1)2)e. We may view this as an isomorphism of finitely generated k[X]-modules. We use repeatedly without comment that a map between k[X]/(X3 + X)-modules is an isomorphism as k[X]/(X3 + X)-modules if and only if it is an isomorphism as k[X]-modules. Since k is a field, k[X] is a PID, so the structure theorem for finitely generated modules over a PID tells us that

    P $\displaystyle \cong$ $\displaystyle \bigoplus_{i}^{}$k[X]/(fi)    and    Q $\displaystyle \cong$ $\displaystyle \bigoplus_{i}^{}$k[X]/(gi)

    where the fi, gi are either 0 or positive powers of monic irreducible polynomials. Then we have

    $\displaystyle \bigoplus_{i}^{}$k[X]/(fi) $\displaystyle \oplus$ $\displaystyle \bigoplus_{i}^{}$k[X]/(gi) $\displaystyle \cong$ (k[X]/(X))e $\displaystyle \oplus$ (k[X]/(X + 1)2)e.

    The uniqueness part of the structure theorem for finitely generated modules over a PID now tells us that fi = X or (X + 1)2 for all i. It follows that a finitely generated k[X]/(X3 + X)-module is isomorphic to a finite direct sum of modules of the form k[X]/(X) or k[X]/(X2 + 1).

  4. Suppose there is a positive integer n such that MJn = MJn + 1$ \ne$ 0. Then MJn is finitely generated because M is Noetherian, and (MJn)J = MJn + 1 = MJn. By Nakayama's lemma we deduce that MJn = 0, as required.

  5. Since R is a right Artinian ring with no nonzero nilpotent ideals, the Wedderburn structure theorem tells us that R $ \cong$ R1 $ \oplus$ ... $ \oplus$ Rn, where n is a positive integer, and the Ri are matrix rings over division rings. If n > 1, then (1, 0,..., 0) is a nontrivial idempotent, so n = 1 which means that R is a matrix ring over a division ring. If this matrix ring has degree > 1, then the matrix with 1 in the (1, 1) position and zeros elsewhere is a nontrivial idempotent. Therefore R is isomorphic to a matrix ring over a division ring, and the result follows.

  6. The character table for S4 is given below; the irreducible characters are c1,...,c5. c1 is the principal character, c2 is the character coming from the sign of the permutation, r is the permutation character (not irreducible), c4 = r - c1, and c5 = c2c4. The remaining row, the character of c3, can easily be filled in using the orthogonality relations.

    Class Size 1 6 8 6 3
    Class Rep (1) (12) (123) (1234) (12)(34)
    c1 1 1 1 1 1
    c2 1 -1 1 -1 1
    c3 2 0 -1 0 2
    c4 3 -1 0 1 -1
    c5 3 1 0 -1 -1
    r 4 2 1 0 0

  7. Since splitting fields are determined up to isomorphism, we may as well assume that K $ \subseteq$ $ \mathbb {C}$. Since the roots of X4 - 2 are ±$ \sqrt[4]{2}$, ±i$ \sqrt[4]{2}$, we see that K = $ \mathbb {Q}$([$ \sqrt[4]{2}$], i). Now X4 - 2 is irreducible over $ \mathbb {Q}$ by Eisenstein's criterion for the prime 2, so [$ \mathbb {Q}$($ \sqrt[4]{2}$) : $ \mathbb {Q}$] = 4. Also i$ \notin$$ \mathbb {Q}$($ \sqrt[4]{2}$) and i satisfies X2 + 1 = 0, consequently [K : $ \mathbb {Q}$($ \sqrt[4]{2}$)] = 2. We deduce that [K : $ \mathbb {Q}$] = 8 and therefore | Gal(K/$ \mathbb {Q}$)| = 8. Now a group of order 8 has a normal subgroup of order 2 (a p-group has normal subgroups of any order dividing the order of the group); let H be a normal subgroup of order 2 in G, and let L be the fixed field of H. Then H has index 4 in G, so by the fundamental theorem of Galois theory, we see that L is a normal extension of degree 4 over $ \mathbb {Q}$, as required.

Peter Linnell