Homework 2 Model Solution - Han-Bom Moon

[Pages:5]MATH 3005 Homework Solution

Han-Bom Moon

Homework 2 Model Solution

Chapter 2.

5. In each case, find the inverse of the element under the given operation.

(a) 13 in Z20 (b) 13 in U (14).

13 + 7 mod 20 = 0 -13 = 7 in Z20

132 mod 14 = 169 mod 14 = 12 ? 14 + 1 mod 14 = 1 13-1 = 13 in U (14)

(c) n - 1 in U (n) (n > 2)

(n - 1)2 mod n = n2 - 2n + 1 mod n = (n - 2)n + 1 mod n = 1

(n - 1)-1 = n - 1 in U (n)

(d) 3 - 2i in C, the group of nonzero complex numbers under multiplication.

(3 - 2i)-1 = 1 =

3 + 2i

3 + 2i 3 2

=

= +i

3 - 2i (3 - 2i)(3 + 2i) 13 13 13

26 11. Find the inverse of the element 3 5 in GL(2, Z11).

-1

26

1

5 -6

1 5 -6

=

=

35

2 ? 5 - 6 ? 3 -3 2

-8 -3 2

In Z11, -8 = 3, -6 = 5, and -3 = 8.

1

5

-6

1 =

55

= 3-1

55

-8 -3 2

3 82

82

Finally, 3 ? 4 mod 11 = 1 so 3-1 = 4 in Z11. Therefore

3-1 5 5 = 4 5 5 = 20 20 = 9 9 .

82

82

32 8

10 8

You may check the axiom of the inverse directly.

13. Translate each of the following multiplicative expressions into its additive counterpart. Assume that the operation is commutative.

(a) a2b3

2a + 3b

1

MATH 3005 Homework Solution

Han-Bom Moon

(b) a-2(b-1c)2

-2a + 2(-b + c)

(c) (ab2)-3c2 = e

-3(a + 2b) + 2c = 0

14. For group elements a, b, and c, express (ab)3 and (ab-2c)-2 without parentheses.

(ab)3 = ababab

(ab-2c)-2 = ((ab-2c)-1)2 = (((ab-2)c)-1)2 = (c-1(ab-2)-1)2 = (c-1(b-2)-1a-1)2 = (c-1b2a-1)2 = c-1b2a-1c-1b2a-1

15. Let G be a group and let H = {x-1 | x G}. Show that G = H as sets.

Here is a standard technique to show the equality between sets. To show the fact that H = G as a set, we need to show that H G and G H. Also to show that H G, we have to prove that for any element a H, a G as well.

Step 1. H G

If a H, then a = x-1 for some x G. Because G is a group, x-1 G. Thus a G and H G.

Step 2. G H

Let a G. Then a = (a-1)-1, because a ? a-1 = e = a-1a so a satisfies the axiom for the inverse of a-1. Therefore a is an inverse of some element of G. Thus a H and G H.

19. Prove that the set of all 2 ? 2 matrices with entries from R and determinant +1 is a group under matrix multiplication.

Let G be the set of all 2 ? 2 matrices with entries from R and determinant +1. Step 1. Closedness of the binary operation First of all, we need to show that matrix multiplication is a well-defined binary operation on G, i.e., for any two A, B G, AB G as well. Because

det(AB) = det A ? det B = 1 ? 1 = 1,

AB G.

Step 2. Associativity

It comes from the associativity of matrix multiplication. (G is a subset of all 2 ? 2 matrices. For all matrices the associativity holds so for a subset G it automatically holds.)

Step 3. Existence of the identity

On the set of 2 ? 2 matrices, the multiplicative identity is I := so I G.

10 . det I = 1

01

2

MATH 3005 Homework Solution

Han-Bom Moon

Step 4. Existence of inverses Let A G. Because its determinant is nonzero, there is a multiplicative inverse A-1 in the set of 2 ? 2 matrices. We need to show that A-1 G. From AA-1 = I,

1 = det I = det(AA-1) = det A ? det A-1 = det A-1

so A-1 G. Therefore G is a group. (We discussed this group G = SL(2, R) in our class already.)

22. Let G be a group with the property that for any x, y, z in the group, xy = zx implies y = z. Prove that G is Abelian. ("Left-right cancellation" implies commutativity.)

We need to show that for any two elements a, b G, ab = ba. Let's take x = b, y = ab, and z = ba. Then xy = bab = zx. From the condition above, y = z so ab = ba.

25. Prove that a group G is Abelian if and only if (ab)-1 = a-1b-1 for all a and b in G. Step 1. G is Abelian (ab)-1 = a-1b-1 for all a, b G. Note that (ab)-1 = b-1a-1 in general. Because G is Abelian, b-1a-1 = a-1b-1. Step 2. (ab)-1 = a-1b-1 for all a, b G G is Abelian. Take a, b G. Note that a-1b-1 = (ba)-1. Therefore (ab)-1 = (ba)-1. Thus ab = ((ab)-1)-1 = ((ba)-1)-1 = ba and G is Abelian.

26. Prove that in a group, (a-1)-1 = a for all a.

This exercise should appear earlier... From the definition of the inverse of a, aa-1 = e. On the other hand, from the definition of the inverse of a-1, (a-1)-1a-1 = e. So we have aa-1 = (a-1)-1a-1. By the right cancellation, we have a = (a-1)-1.

32. Construct a Cayley table for U (12).

As a set, U (12) = {1, 5, 7, 11}.

1 5 7 11 1 1 5 7 11 5 5 1 11 7 7 7 11 1 5 11 11 7 5 1

35. Let a, b, and c be elements of a group. Solve the equation axb = c for x. Solve a-1xa = c for x.

axb = c xb = exb = a-1axb = a-1c 3

MATH 3005 Homework Solution

Han-Bom Moon

xb = a-1c x = xe = xbb-1 = a-1cb-1

Therefore x = a-1cb-1.

a-1xa = c xa = exa = aa-1xa = ac

xa = ac x = xe = xaa-1 = aca-1

So x = aca-1.

Note that there is a unique solution for given `linear' equation.

37. Let G be a finite group. Show that the number of elements x of G such that x3 = e is odd. Show that the number of elements x of G such that x2 = e is even.

A simple technique to show that the number of elements of a set S is even is to show that we can pair all elements of S. (More formally, we can think a pairing on a set S as a bijection map p : S S such that p(x) = x and p(p(x)) = x. Then we can make pairs {x, p(x)} for all elements x S.) Of course, to show that the number of elements of S is odd, take one element x out from S and show that we can pair all elements of S - {x}.

Let S = {x G | x3 = e} We want to show that the number of elements of S is odd. Note that e S. Let S = S - {e}. We will show that the number of elements in S is even.

Take x S and consider x-1. First of all, because (x-1)3 = (x3)-1 = e-1 = e, x-1 S. Furthermore, if x-1 = e, then x = e. Because x = e, x-1 = e and x-1 S .

We claim that collecting {x, x-1} for all x S forms pairs for all x S . First of all, this is indeed a pair, because if x = x-1, then x2 = xx-1 = e and x = xe = xx2 = x3 = e. Secondly, obviously each element in S is in one of such pairs. Next, suppose that {x, x-1} {y, y-1} = . If x = y, then {x, x-1} = {y, y-1}. If x = y-1, then because x-1 = (y-1)-1 = y, {x, x-1} = {y, y-1} again. In any cases, such two pairs {x, x-1} and {y, y-1} are disjoint or equal. Therefore we can divide the set S into pairs. Therefore the number of elements of S is even and that of S is odd.

For next, now consider the set T = {x G | x2 = e}. First of all, if (x-1)2 = e, then x2 = (x-2)-1 = ((x-1)2)-1 = e-1 = e so x / T . So for x T , x-1 T . For x T , make a pair {x, x-1} of elements of T .

We claim that this is indeed a `pair' of elements of T . That is, x = x-1. If x = x-1, then x2 = xx-1 = e so x / T . Thus x = x-1 and {x, x-1} is a set of two elements of T .

Obviously, each element of T is in one of such pairs. Moreover, suppose that {x, x-1} {y, y-1} = . If x = y, then {x, x-1} = {y, y-1}. If x = y-1, because x-1 = (y-1)-1 = y, {x, x-1} = {y, y-1} again. Therefore again, such pairs {x, x-1} and {y, y-1} are disjoint or equal. Hence we are able to divide the set T into pairs and the number of elements of T is even.

4

MATH 3005 Homework Solution

Han-Bom Moon

41. Suppose F1 and F2 are distinct reflections in a dihedral group Dn. Prove that F1F2 = R0.

(Simple proof) Note that for any reflection F , F 2 = R0, the identity. In particular, F1F1 = R0. If F1F2 = R0 = F1F1, then by left cancellation, F1 = F2. (Geometric proof) Suppose that F1 is a reflection across a line 1, and F2 is a reflection across a line 2. Because F1 = F2, 1 = 2. Pick a point P on 2 - 1. Then since F2 fixes all points on 2, F1F2(P ) = F1(P ) = P . On the other hand, R0(P ) = P , because it is the identity map. Therefore F1F2 = R0.

47. Prove that if G is a group with the property that the square of every element is the identity, then G is Abelian.

Let a, b G. From the assumption, we know that a2 = e and b2 = e. Moreover, abab = (ab)2 = e. Then

ab = aeb = a(abab)b = a2bab2 = ebae = ba.

So G is Abelian.

51. List the six elements of GL(2, Z2). Show that this group is non-Abelian by finding two elements that do not commute. Note that because Z2 = {0, 1}, there are only two possible distinct entries.

10

10

01

01

11

11

,

,

,

,

,

01

11

10

11

01

10

It is not Abelian, because for example,

10

01

01

?

=

,

11

10

11

01

10

11

?

=

.

10

11

10

5

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