Download MATH 103A Homework 5 - Solutions Due February 15, 2013

MATH 103A Homework 5 - Solutions
Due February 15, 2013
Version February 4, 2013
Assigned reading: Chapters 4, 6,7 of Gallian.
Recommended practice questions: Chapter 4 of Gallian, exercises
Chapter 6 of Gallian, exercises
2, 6, 11, 12, 15, 16, 18
Chapter 7 of Gallian, exercises
3, 4, 5, 6, 7, 10, 15, 17, 26, 33
Assigned questions to hand in:
(1) (Gallian Chapter 6 # 10) Let G be a group. Prove that the mapping α g g 1 for all
g G is an automorphism if and only if G is Abelian.
Solutions: Suppose α is an automorphism. Let x, y G. We want to show that xy
Applying the property of automorphism to xy 1 :
α xy 1 definition of α
xy 1 1
yx.
xy,
but also
autom.
α xy 1 α y 1x1 α y 1α x1 y 1 1 x1 1
yx.
Thus, xy yx. Conversely, suppose G is abelian. To prove that α is an automorphism,
we need two facts:
(1) WTS α is a bijection. It is sufficient to exhibit an inverse for α. In fact, we will show
that α is its own inverse. Let x G, then
α α x α x1 x1 1
x.
Thus, α α identity and α has an inverse so is a bijection.
(2) WTS α preserves the operation. Let x, y G. Then
α xy xy 1
y 1x1
abelian
x1 y 1
α xα y ,
as required.
(2) (Gallian Chapter 6 # 22) Let φ be an automorphism of a group G. Prove that H
G : φ x x is a subgroup of G.
x Solution: We use the one-step subgroup test.
Nonempty? Consider e G. By Theorem 6.2.1, φ e e so e H. Thus, H .
Closed? Let a, b H. WTS that ab1 H. By definition of H, φ a a and
φ b b.
6.2.2
φ ab1 φ aφ b1 φ aφ b1 ab1 .
Thus, ab1 satisfies the condition for membership in H.
(3) (Gallian Chapter 6 # 27) Let r U n. Prove that the mapping α : Zn
α s sr mod n for all s Zn is an automorphism of Zn .
Zn defined by
Solution: Fix n and r. By question 3 on homework 1 (Ch 0 # 11), there is x such that
rx mod n 1; in fact, x mod n r 1 in U n.
(1) Bijection? We will prove that the function β s sr 1 mod n (for r 1 U n) is
the inverse of α. Note that the domain and codomain of α and β is Zn , as required.
Let s Zn .
α β s α sr 1
mod n sr 1
mod n s r 1 r mod n s,
mod nr
by associativity of multiplication mod n.
β α s β sr
mod n sr
mod nr 1
mod n s rr 1
mod n s,
by associativity of multiplication mod n. Thus, α and β are inverse functions and
are bijections.
(2) Preserves operation? Let s, t Zn . Then
α s t s tr
mod n sr
mod n tr
mod n α s α t,
by distributivity of multiplication mod n over addition mod n.
(4) (Gallian Chapter 6 # 28) The group
group? What if Z is replaced by R?
1 a
0 1
:aZ
is isomorphic to what familiar
Solution: It is isomorphic
to Z or R under addition (respectively). To prove this, consider
1 a
the function φ
a.
0 1
1 a
works.
(1) Bijection? The inverse ψ a 0 1 1 a
1 b
and
be given. Then
(2) Preserves operation? Let
0 1
0 1
φ
1 a
0 1
1 b
0 1
Matrix Mult
φ
1 ba
0
1
baabφ
1 a
0 1
φ
1 b
0 1
.
(5) (Gallian Chapter 6 # 34) Prove or disprove that U 20 and U 24 are isomorphic.
Solution: These groups are not isomorphic. We prove this by demonstrating that every
element in U 24 has order 1 or 2 whereas there is an element of U 20 with order 4.
U 24 1, 5, 7, 11, 13, 17, 19, 23.
To prove that x 2 for x U 24, it is sufficient to show that x2
12
72
13
2
19
2
49
1,
52
25 mod 24 1,
112
mod 24 1,
169 mod 24 1,
361 mod 24 1,
2
121 mod 24 1,
17
2
289 mod 24 1,
23
2
529 mod 24 1.
1 for each x U 24.
On the other hand,
U 20 1, 3, 7, 9, 11, 13, 17, 19
and
3 1, 3, 9, 7
so
3 4 2.
By Theorem 6.2.7, two finite isomorphic groups have exactly the same number of elements
of every order. Since this condition fails for U 20 and U 24, they cannot be isomorphic.
(6) (Gallian Chapter 6 # 48) Let φ be an isomorphism from a group G to a group Ḡ and let
a G. Prove that φ C a C φ a.
Solution: Fix G and a. We prove subset inclusion in two directions.
Let y φ C a. Then there is x C a such that y φ x. WTS that y
namely that y commutes with φ a:
Because x
yφ a φ xφ a φ xa
C a
C φ a,
φ ax φ aφ x φ ay,
as required.
Let y C φ a. Then yφ a φ ay. WTS that y φ C a, namely that there
is x C a with y φ a. Since φ is an isomorphism, it has an inverse and φ1 y exists. We claim that φ1 y C a. Recall that by Theorem 6.3.1, φ1 is an
isomorphism. So:
C φa 1
φ φ ay φ1 φ aφ1 y aφ1 y ,
as required. Thus, φ1 y C a and φ φ1 y y so y φ C a.
φ1 y a φ1 y φ1 φ a φ1 yφ a
Because y
(7) (Gallian Chapter 7 # 8) Suppose that a has order 15. Find all the left cosets of a5 in a.
Solution: The subgroup
H
a e, a
5
5
, a5 2 , a5 3 , . . . e, a5 , a10 .
The left cosets of H are of the form xH for x G, namely
ak a5 since each x G is ak for some k
eH
a2 H
e, a
5
a
2
Z. In particular,
,a
10
,
aH
a, a
6
, a11 , a7 , a12 ,
a3 H a3 , a8 , a13 a4 H a4 , a8 , a14 but
a5 H a5 , a10 , a15 a5 , a10 , e eH.
Thus, there are 5 distinct left cosets of this subgroup.
(8) (Gallian Chapter 7 # 24) Suppose that H and K are subgroups of G and there are elements a, b G such that aH bK. Prove that H K.
Solution: Let a, b G such that aH bK. In particular, since H is a subgroup of G, e G
and a ae bK so there is k K such that a bk. Note that a1 bk 1 k 1 b1
and that k 1 K (since K is a subgroup of G).
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Suppose x H. We will show that x K. By definition ax
ax bK. Namely, there is y K such that ax by.
x a1 ax a1 ax a1 by k 1 b1 by
k 1 y
aH so by assumption,
K
since k 1 , y K and K is a subgroup of G so is closed under the operation.
(9) (Gallian Chapter 7 # 26) Suppose that G is a group with more than one element and
G has no proper, nontrivial subgroups. Prove that G is prime. (Do not assume at the
outset that G is finite.) Note that by Corollary 3 to Lagrange’s theorem, this exercise
proves that G is cyclic.
Solution: G contains at least the elements e, x for x some non-identity element. We have
proved that x G. Since x e, x e. By assumption that G contains no proper
nontrivial subgroups, it must be the case that x G. We consider two cases.
Case 1: x . By Theorem 4.1, for each i, j Z, xi xj iff i j. In particular,
e x2n : n Z xn : n Z. But, x2n : n Z x2 is a subgroup of
G. In this case, we have a proper nontrivial subgroup of G, which contradicts the
assumption on G.
Case 2: x . By the Fundamental Theorem of Cyclic Groups (Theorem 4.3),
each positive divisor of x corresponds to a subgroup of G x of that order.
To guarantee that G has no proper nontrivial subgroups, x can have no positive
divisors other than 1 and x. That is, x is prime. But, x x G so we have
shown that G is prime.
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