ex 6 solutions

7/31/2019 Ex 6 Solutions

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CM121A, Abstract Algebra Solution Sheet 6

1. Suppose that G is a group and g G. Prove that g and g1

have thesame order.Solution: Suppose first that g has finite order n. So n is the least positiveinteger n such that gn = e. We have to show that n is also the least positiveinteger such that (g1)n = e.We will show that if k is any integer, then gk = e if and only if (g1)k = e.Since (g1)k = gk = (gk)1, and (gk)1 = e if and only ifgk = e1 = e,it follows that gk = e if and only if (g1)k = e. It follows that if g hasorder n, then so does g1. Moreover if g has infinite order, then there isno positive integer k such that gk = e, so there is no positive integer ksuch that (g1)k = e, and therefore g1 also has infinite order.

2. Find the order of every element of each of the following groups:

(a) Z8 under addition;Solution:

[0] has order 1 since 1[0] = [0] (just writing [a] for [a]8).

[1] has order 8 (n[1] = [n] = [0] if 0 < n < 8, but 8[1] = [0]).Similarly,

[2] has order 4;

[3] has order 8;

[4] has order 2;

[5] has order 8 (note that for n = 1, 2, . . . , 8, the multiples n[5] =[5n] are [5], [2], [7], [4], [1], [6], [3], [0]);

[6] has order 4;

[7] has order 8.

(b) Z15 under multiplication;Solution: Recall that Z15 = {[1], [2], [4], [7], [8], [11], [13], [14]}.

[1] has order 1 (now writing [a] for [a]15).

[2] has order 4 (for n = 1, 2, . . . , 4, the powers [2]n = [2n] are[2], [4], [8], [1]). Similarly,

[4] has order 2;

[7] has order 4;

[8] has order 4;

[11] has order 2;

[13] has order 4;

[14] has order 2.

(c) D4.Solution: Recall that D4 has 4 rotation elements (counting the iden-tity e) and 4 reflections. Writing for clockwise rotation by 90, wehave

e has order 1;

has order 4;

2 has order 2;

3 has order 4;

every reflection has order 2.


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(g) The subset

a b0 1

a R, b R

of GL2(R).

Solution: Yes. 1) IfA =

a b0 1

and A =

a b

0 1

are in the

subset, then so is AA

= aa

ab + b

0 1

.

2) The identity I =

1 00 1

is in the subset.

3) IfA =

a b0 1

is in the subset, then so is A1 =

a1 ba1

0 1

.

5. Recall that if z = x + iy is a complex number, then the modulus (or

absolute value) of z is |z| =

x2 + y2.

(a) Prove that if z1, z2 C, then |z1z2| = |z1||z2|.Solution: Ifz1 = x1+iy1 and z2 = x2+iy2, then |z1z2| is the squareroot of

(x1x2 y1y2)2 + (x1y1 + x2y2)

2 = x21x22 + y

21y

22 + x

21y

21 + x

22y

22

= (x2

1 + y2

1)(x2

2 + y2

2)

.

Therefore |z1z2| = (x21 + y21)

1/2(x22 + y22)

1/2 = |z1||z2|.

(b) Prove that { z C | |z| = 1 } is a subgroup ofC.Solution: 1) If z1 and z2 are in the subset, then |z1| = |z2| = 1,so |z1z2| = |z1||z2| = 1 (by part (a)). It follows that z1z2 is in thesubset. 2) The identity 1 = 1 + i 0 is in the subset since |1| = 1. 3)If |z| = 1, then |z||z1| = |zz1| = |1| by (a). Therefore |z1| = 1and z1 is in the subset.

6. Suppose that H and K are subgroups of a group G.

(a) Show that H K is also a subgroup of G.Solution: 1) Suppose that g, g HK. Then g, g H gg H

(since H is a subgroup), and g, g

H gg

K (since K is asubgroup), so gg H K.2) e H and e K since H and K are subgroups, so e H K.3) Suppose that g H K. Then g H g1 H and g Kg1 K (since H and K are subgroups), so g1 H K.Therefore H K is a subgroup.

(b) Given an example to show that H K need not be.Solution: Let G = D4, H = {e, 2}, K = {e, 4} (in the notationof Exercise 1f) above). Then H and K are subgroups of G, butH K = {e, 2, 4} is not.


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7. Suppose that H is a subgroup of a group G, and g is an element of G.Prove that the following subset of G

{ ghg1 |h H}

(denoted gH g1

) is a subgroup of G.Solution: 1) Suppose that k and k are elements of gH g1. This meansthat k = ghg1 and k = ghg1 for some h, h H. Therefore

kk = (ghg1)(ghg1) = gh(g1g)hg1 = ghhg1,

which is in gH g1 since hh H (as H is a subgroup of G).2) Since e H, we have e = gg1 = geg1 gH g1.3) Let k = ghg1 be an element of gH g1 (with h H, so h1 H).Then

k1 = (ghg1)1 = (g1)1h1g1 = gh1g1 gH g1.

Therefore gH g1 is a subgroup of G.

8. Let H be a non-empty subset of a group G.

(a) Suppose that H has the property that if g, h H, then gh1 H.Prove that H is a subgroup of G.Solution: Let g be any element of H. Then applying the propertywith g = h gives e = gg1 H. So we can apply the property as wellwith e in place of g, giving g1 = eg1 H. Finally if g and h areany elements ofH, then we now know that h1 H, so we can applythe property with h1 in place ofh, giving gh = g(h1)1 H.

(b) Suppose that G is finite, and H has the property that if g, h H,then gh H. Prove that H is a subgroup of G.Solution: Let h be any element of H. Then hn H for everypositive integer n. (This is clear by induction on n. For n = 1 it isobvious. If n > 1 and hn1 H, then hn = hhn1 H as well.) ByCor. 4.6.5, h has finite order, so hd = e for some positive integer d.Therefore e H. Furthermore, h1 = hd1 H. Therefore H is asubgroup of G.

ex 6 solutions

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