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MATH 103A Homework 4 - Solutions
Due February 4, 2013
Version February 5, 2013
Assigned reading: Chapters 3-4 of Gallian.
Recommended practice questions: Chapter 3 of Gallian, exercises
11, 12, 13, 14, 15, 16, 17, 20, 22, 23, 28, 31, 33, 34, 37, 50, 53, 78
Chapter 4 of Gallian, exercises
4, 5, 7, 13, 24, 30, 63, 82
Assigned questions to hand in:
(1) (Gallian Chapter 3 # 10) How many subgroups of order 4 does D4 have?
Solution: Recall that
e, R90 , R180 , R270 , F, F R90, F R180 , F R270.
There are three subgroups of order 4, one cyclic and two not:
R90 R270 e, R90 , R180 , R270 e, F, R180 , F R180 e, R180 , F R90, F R270 .
(2) (Gallian Chapter 3 # 26) Prove that a group with two elements of order 2 that commute
must have a subgroup of order 4.
Solution: Let G be such a group and let a, b G be elements of order 2 with ab
Consider the subgroup of G containing a and b. Then it must contain
e, a, a2 , a3 , . . . , a1 , . . . , b, b2 , b3 , . . . , b1 , . . . , ab, ba, abb, baba, . . .
But, since a b 2, a1 a and b1 b. Moreover, all powers of a are either a or e
and all powers of b are either b or e. Thus, we’ve reduced the subgroup to
e, a, b,
Since ab
and products containing both a, b.
i1 j1 i2 j2
a b a b
in jn
in j1
a, or
depending on the parity of i1 . . . in and j1 jn . Thus, the subgroup of G containing
a, b is e, a, b, ab. It remains to argue that ab a and ab b. Suppose, towards a
contradiction, that ab a. By cancellation (left multiplication by a1 ), this gives b e
so b 1, contradicting b 2. The symmetric argument shows that ab b.
(3) (Gallian Chapter 3 # 45) Let G be an Abelian group with identity e and let n be some
fixed integer. Prove that the set of all elements of G that satisfy the equation xn e is a
subgroup of G. Give an example of a group G in which the set of all elements of G that
satisfy the equation x2 e does not form a subgroup of G.
x G : xn e. To show this is a
Solution: Let G be an abelian group and H
subgroup, we use the one-step subgroup test:
Is H nonempty? Take x e. Then em e for all m so in particular en e and
e G.
For a, b H, is ab1 H? Let a, b H. Then an e and bn e. We check if
ab1 H by considering the criterion for membership in H:
ab1 n abelian anbn
an bn 1
Thus, ab1 H.
For the example, consider the group D4 . Then
x D4 : x2
e, R180 , F, F R90, F R180 , F R270.
But, this set is not a group because it is not closed under multiplication. For example,
F R90 F R180
F F R270 R180
R90 ,
which is not in the set.
(4) (Gallian Chapter 3 # 69) Let H be a subgroup of R under addition. Let K
Prove that K is a subgroup of R under multiplication.
2a : a H .
Solution: We use the one-step subgroup test:
Is K nonempty? Since H is nonempty, let a H and then 2a K.
Let x, y K. Is xy1 K? By definition, there are a, b H such that x 2a and
y 2b . Then y 1 2b and xy 1 2a 2b 2ab . But, H is a subgroup of R under
addition so, since a, b H, a b H as well. Thus, 2ab K and xy 1 K.
(5) (Gallian Chapter 4 # 28) Let a be a group element that has infinite order. Prove that
ai aj if and only if i j.
Solution: Suppose i j. Then either ai aj or ai is the inverse of aj . In either case,
they generate the same group. For the converse, suppose ai aj . Then ai aj and
aj ai . By definition, there are m, n Z such that ai amj and aj ani . Therefore,
aimj e and aj ni e. But, since a , the only way this can happen is if i mj 0
and j ni 0. In other words, ij and j i. This implies that i j.
(6) (Gallian Chapter 4 # 38) Consider the set 4, 8, 12, 16. Show that this set is a group
under multiplication modulo 20 by constructing its Cayley (multiplication) table. What
is the identity element? Is the group cyclic? If so, find all of its generators.
4 8 12 16
4 16 12 8 4
8 12 4 16 8
12 8 16 4 12
16 4 8 12 16
The Cayley table shows closure under the operation (which is associative) and
This group is cyclic.
8 80 , 81, 82, 83, . . . 16, 8, 4, 12
12 120, 121, 122, 123, . . . 16, 12, 4, 8.
Note that 16 16 and 4 16, 4 so the only generators of the group are 8 and 12.
(7) (Gallian Chapter 4 # 64) Let a and b belong to a group. If a and b are relatively
prime, show that a b e.
Solution: Recall that the intersection of two subgroups is itself a subgroup, and is a
subgroup of each group. That is,
a b a
a b b.
By the Fundamental Theorem of Cyclic Groups (Theorem 4.3), every subgroup of a cyclic
group is cyclic. Therefore, a b is cyclic. Moreover, the same theorem says that the
order of this group divides both a and b. But, by Corollary 1 of Theorem 4.1 (page
79), a a and b b. We assume that these two numbers are relatively prime.
Therefore, the only common divisor is 1. Thus, 1 a b. The only group of order
1 is the trivial group, so a b e.
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