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Math 120 Homework 3 Solution
27 April 2006
2.1 Problem 2
(a) (12)(23) = (123) is not a 2-cycle, so the set of 2-cycles is not closed
under products. Hence it’s not a subgroup.
(b) D2n can be viewed as a subgroup of S2n , with generator and relations
given by
{σ = (1 2 . . . 2n), τ = (1 2)(3 4)(5 6) . . . (2n−1 2n)|σ 2n = τ 2 = 1, σ −1 τ = τ σ}
τ and σ −1 τ σ are two reflections, their products = τ σ −1 τ σ = τ 2 σ 2 = σ 2
is not a reflection. That means the set of reflections is not closed under
products, hence it’s not a subgroup.
(c) If n = km for some k > 1, m > 1 (n is assumed to be a composite
integer implies k and m exists), |xk | = m. So xk is not in the set
{ x ∈ G | |x| = n }, therefore the set is not closed under products, so
it’s not a subgroup.
(d) Because 1 + 1 = 2, 2 is an even number, so odd numbers ∪ {0} is not
closed under addition, hence it’s not a subgroup.
(e) Let H
whose √
square is a rational
√
√ number. 1
√ = the set of real numbers
and 2 are in H, but (1 + 2)2 = 3 + 2 2 implies 1 + 2 is not in H.
H is not closed under products so it’s not a subgroup.
2.1 Problem 3
We have r4 = s2 = 1, r−1 s = sr.
1
(a)
r2 s = sr2
(r2 )−1 = r2
(r2 )(sr2 ) = s
s−1 = s
(sr2 )−1
s(sr2 ) = r2
= r2 s = sr2
Therefore the set is closed under product and inverse, so it’s a subgroup
of D8
(b)
r2 sr = sr3
(r2 )−1 = r2
(r2 )(sr3 ) = sr
(sr)−1 = sr3
sr(sr3 ) = r2
(sr3 )−1 = sr
Therefore the set is closed under product and inverse, so it’s a subgroup
of D8
2.1 Problem 4
Let H = {n |n ∈ Z, and n ≥ 0}. H is a infinite subset of the group
(Z, +). H is closed under addition but 1 fails to have an inverse, so H is not
a subgroup of (Z, +)
2.1 Problem 10
(b) Let {Hi }i∈I be some collection of subgroups of G, I is some index set.
If a, b ∈ ∩i∈I Hi , then a, b ∈ Hi for all i. Because Hi are subgroups
of G, 1, ab and a−1 ∈ Hi for all i. It follows 1, ab, a−1 ∈ ∩i∈I Hi .
Therefore ∩i∈I Hi contains 1 and is closed under product and inverse.
Hence ∩i∈I Hi is a subgroup of G.
(a) It follows from (b)
2.3 Problem 1
45 = 32 · 5. All positive divisors of 45 are 1,3,5,9,15,45. By theorem 7 (3)
all cyclic subgroups of Z45 =< x >, < 3x >, < 5x >, < 9x >, < 15x >, < 0 >
2.3 Problem 2
If |G| is finite, then there exists an integer n such that xn = 1. This gives
us x−1 = xn−1 . < x > is a subgroup of G because it contains 1 and is closed
2
under product and inverse. < x > and G has the same number of elements,
so they are the same group.
< 2 > is a subgroups of Z. They have the same size but < 2 > is a proper
subgroup of Z.
2.3 Problem 3
x is a generator of Z/48Z if and only if (48, x) = 1. So all generators are
1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47.
3