Math 493 (Fall 2012). Solutions for Problem Set 1
1.1. (i) Associativity fails because (1 1) 1 = 1 (1 1). The identity axiom fails
because there does not exist an integer e such that e x = x for all x Z. So the third
group axiom cannot even be formulated.
Math 493 (Fall 2012). Solutions for Problem Set 9
9.1. (a) Reexivity and symmetry are clear, so I will only explain the proof of transitivity. Suppose that (a, b) (c, d) and (c, d) (e, f ) where a, b, c, d, e, f A and b, d, f = 0.
Then ad = bc and cf = de
Math 493 (Fall 2012). Solutions for Problem Set 8
8.1. (a) First, the identities that dene multiplication in H immediately imply that G is
closed under quaternion multiplication. The associativity of multiplication on G follows
from the associativity of m
Math 493 (Fall 2012). Solutions for Problem Set 7
7.1. (a) Since K is a p-subgroup of G, the second Sylow theorem implies that there
exists a p-Sylow subgroup P of G such that K P . Then K H P H and |H P |
divides the order of P by Lagranges theorem, so H
Math 493 (Fall 2012). Solutions for Problem Set 6
6.1. In this solution I will use the older convention that gh denotes g h for two permutations g, h of the set cfw_1, 2, . . . , n. Because we wont have to work with explicit cycle
decompositions, this con
Math 493 (Fall 2012). Solutions for Problem Set 5
5.1. (a) Write X as the disjoint union of G-orbits with respect to the given action. The
elements of X G correspond to those G-orbits that consist of a single point. For any other
orbit, we know that its s
Math 493 (Fall 2012). Solutions for Problem Set 4
4.1. (a) The assumption of part (a) means that the map : H K G given by
(h, k ) = hk is surjective. We need to show that it is a group homomorphism and that it
has trivial kernel. It will then follow that
Math 493 (Fall 2012). Solutions for Problem Set 3
3.1. I will use the notation cfw_e, r, r1 , s1 , s2 , s3 for the elements of S3 introduced in my
solution to homework problem 1.5(c); this notation is consistent with the notation of the
current problem.
Math 493 (Fall 2012). Solutions for Problem Set 2
2.1. Consider any element g G such that g = e. The subgroup g of G generated by
g is not cfw_e, so it must be all of G. We deduce that G is cyclic and every non-identity
element of G is a generator of G. W
Math 493 (Fall 2012). Solutions for Problem Set 10
10.1. The cases when a is 0 or a unit are trivial, so assume that this is not the case and consider
a prime factorization a = p1 p2 pn . We induct on n. The case n = 1 is the denition of what
it means for