Solutions to the Assignment 1
1. It is not a group. For example associativity fails since
a = a e = a (b c) = (a b) c = e c = c
2. If we let x = a1 , c = b and d = a, then we have b = axb = cxd = b for all a, b G. So the
property of the group G implies th
Solutions to the Midterm Test
1.
F a) Z6 has an element of order 6, but S3 does not. Or, Z6 is abelian, but S3 is not.
T b) By Lagranges theorem, order of any element of G divides |G| which is prime.
T c) By the rst isomorphism theorems, 10 and 15 divide
Solutions to the Assignment 2
1. i) = (12345)(678) = (23847)(56) = (12485736)
ii) = (15)(14)(13)(12)(68)(67) = (27)(24)(28)(23)(56) = (16)(13)(17)(15)(18)(14)(12)
2. Since the order of ak as an arbitrary element in any group is the order of a divided by
g
Solutions to the Assignment 3
1. First we note that H K G is a subgroup of G by theorem 7.4. Now to show H K
G,
enough to show gxg 1 H K for all g G and x H K. But gxg 1 H since x H and
G by assumption. Similarly gxg 1 K since x K and K
H
and so H K
G. Th
Solutions to the Assignment 4
1. By Lagranges theorem |Z(G)| 1, p, p2 , p3 . Since G is a p-group, class equation implies
that |Z(G)| = 1. On the other hand if |Z(G)|
p2 , p3 , G/Z(G) is cyclic and hence G is
abelian. Therefore |Z(G)| = p for G non-abeli
6. Suppose we have an ideal pendulum of length 1 meter and mass 1. From
m
rest we impart our pendulum with a speed of g sec . (Recall two forces act on
such an ideal pendulum, a tension in the direction and the force of gravity
r
m
acting in the directio