Math 401: Solutions to HW #3
(Ch. 5 # 8) What is the maximum order of any element in A10 ?
The order of a permutation is the least common multiple of the cycle lengths in its disjoint
cycle form. We consider all of the possible disjoint cycle structures,
Math 401: Solutions to HW #7
(Ch. 9 # 34) In Z , let H = 5 and K = 7 . Prove that Z = HK . Does Z = H K ?
Since 5 and 7 are coprime, then 1 = 5s + 7t for some integers s and t. Then since Z = 1 ,
each n Z can be written as n = 5(ns) + 7(nt) H + K , so Z =
Math 401: Solutions to HW #6
(Ch. 8 # 2) Show that Z2 Z2 Z2 has seven subgroups of order 2.
Each element of order 2 determines a unique subgroup of order 2, hence we only need to
count the elements of order 2. Since |(a, b, c)| = lcm(|a|, |b|, |c|), and t
Math 401: Solutions to HW #5
(Ch. 7 # 6) Let n be a positive integer. Let H = cfw_n, 2n, 3n, .. Find all left cosets
of H in Z. How many are there?
There are n left cosets: H , 1 + H , 2 + H , ., n 1 + H . Notice theses are the residue
classes modulo n.
Math 401: Solutions to HW #4
(Ch. 6 # 8) Show that the mapping a log10 a is an isomorphism from R+ under
multiplication to R under addition.
First, suppose that log10 a = log10 b. Then 10log10 a = 10log10 b , hence a = b, so the mapping
is one-to-one. For
Math 401: Solutions to HW #2
(Ch. 3 # 4) Prove that in any group, an element and its inverse have the same order.
Proof. First, suppose a G has innite order. If (a1 )n = e for some integer n > 0, then
an = e, and multiplying by an on both sides yields e =
Math 401: Solutions to HW #1
(Ch. 1 # 4) Describe the elements of D5 .
D5 consists of ve rotations counterclockwise of 0, 72, 144, 216, and 288 degrees, and ve
reections across the lines which connect each vertex to the midpoint of the opposite edge.
Math 401 Exam #2 Solutions
1. Suppose G is a group of order 27 (not necessarily abelian!). Prove that G has an element
of order 3.
The non-identity elements of G have order 3, 9 or 27. Let x G be such an element. If
x has order 3, we are done. If x has or
Exam #1 Solutions
1. Let G be the group of permutations on a set X . Let a X and dene
stab(a) = cfw_ G|(a) = a.
Prove that stab(a) is a subgroup of G.
Many of you were confused and thought the set X was the group G but remember that G
is the set