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1 Prolem 2.3: Which of the following collections of 2 2 mtrices with rel entries form groups under [ mtrix ] multipliction? i) Those of the form for which c d 2 Answer: The set of such mtrices is not closed under mtrix multipliction, so it does not form group. To see tht it is not closed under mtrix multipliction, it is enough to consider the following exmple: [ ][ ] [ ] = [ ] [ ] 2 5 where the mtrix is not of the form. 3 7 d [ ] ii) Those of the form such tht c 2 c [ ] [ ] Answer: Agin, consider the two mtrices nd which re [ ] 7 6 of this form. Their product is, which is not in the correct form. So this 2 3 set is not closed under mtrix [ multipliction, ] nd does not form group. iii) Those of the form where c is not zero. 0 c Answer: These do form group. To show this, we need to check the following things. [ ] [ ] (1) The set is closed under multipliction: Suppose nd 0 c 0 c stisfy c, c 0. Then [ ][ ] [ ] 0 c 0 c = + c 0 c c where cc 0 ecuse c, c 0. Thus the product of two mtrices in this set is gin in the set, so it is closed under multipliction. [ ] 1 0 (2) The identity is in this set: This is true ecuse the mtrix is of 0 1 the correct form. [ ] (3) We hve inverses: Suppose the mtrix A = is in our set. Then 0 c consider the mtrix B = [ 1 c 1 0 c ]. Since c 0, the term c mkes sense, so B is well-defined. It is simple to check tht AB = BA = I, where I is the identity mtrix. Thus every mtrix in our set hs n inverse. (4) Associtivity: We hve this ecuse mtrix multipliction is ssocitive in generl. iv) The set of mtrices with non-zero determinnt nd integer entries. Answer: This set does not form [ group ecuse ] inverses my not e in the 2 0 set. For exmple, consider the mtrix. This mtrix is in our set. But its 0 2 inverse is the mtrix [ 1/ /2 ], which is not in our set. 1

2 2 Prolem 2.6 Show tht the collection of ll rottions of the plne out fixed point P forms group under composition of functions. Is the sme true of the set of ll reflections in lines which pss through P? Wht hppens if we tke ll the rottions nd ll the reflections? Answer: Fix point P in the plne. Let G ro e the set of ll rottions out P, let G re e the set of ll reflections out lines through P nd let G e the set of ll rottions nd reflections together. First, note tht ll of these sets stisfy the ssocitivity property. This is ecuse the composition of functions is lwys ssocitive. Suppose we hve three functions f,g nd h. We need to see tht for ech point x in their domin, f (g h)(x) = (f g) h(x). To show this tkes just it of mnipultion. Since (g h)(x) = g(h(x)), we hve f (g h)(x) = f(g(h(x))), nd likewise, (f g) h(x) = (f g)(h(x)), which is gin just f(g(h(x))). Therefore, composition of functions is lwys ssocitive, so composition of rottions nd/ or reflections is lso ssocitive. Now we show tht G ro is group. Let r θ G ro e the clockwise rottion y ngle θ out P. So if θ is negtive, we men counter-clockwise rottion y θ. Then given two ngles θ nd φ, we clerly hve r θ r φ = r θ+φ. Thus the composition of two rottions is gin rottion, so G ro is closed under composition of functions. Now we hve to check the 3 group properties. (1) Associtivity: Composition of functions is ssocitive. (2) Identity: Clerly the identity is r 0, the rottion y ngle 0, since for ny ngle θ, r θ r 0 = r θ = r 0 r θ. (3) Inverses: Fix n ngle θ. Then the inverse of r θ is r θ since r θ r θ = r 0 = r θ r θ. Thus the set of rottions is group under function composition. Forthenextprtsoftheprolem,weusethefollowingfctoutthecomposition of two reflections. Property 1. Let R,R G re e reflections out lines nd through the point P. Suppose the ngle tht line sweeps out s it moves in clockwise direction to line is θ. Then R R = r θ/2, where R R mens we do R first nd then R. Proof. Choose some point Q in the plne, where Q P. Then s in the digrm, mesure the signed cute ngle etween the line segment QP nd line. We sy this ngle is positive if when you use QP to sweep out the ngle you go in clockwise direction. Otherwise, the ngle is negtive. Cll this signed ngle θ 1. In the digrm, the ngle is positive. Then the signed cute ngle etween the line segment R (Q),P nd is lso θ 1. Now let θ 2 e the signed cute ngle etween R (Q),P nd the line. Agin, the signed cute ngle etween R R (Q),P nd the line is lso θ 2. Note tht the distnce etween Q nd P is the sme s the distnce etween R R (Q) nd P. So if you rotte Q y n ngle of 2θ 1 +2θ 2 out P, we get to R R (Q). But θ 1 +θ 2 is just θ in the cse when t lest one of these two ngles is positive, nd θ 1 + θ 2 = θ π when oth of these ngles re negtive. So rottion y 2θ 1 +2θ 2 is either rottion y 2θ or rottion y 2(θ π). But the ltter is, in fct, the sme s rottion y θ since rotting y 2π is the sme s not rotting t ll. So we re done.

3 3 Q R (Q) θ 1 θ 1 P θ θ2θ 2 R (R (Q)) Figure 1. The point Q reflected first through nd then through By the ove property, we see tht the set of reflections is not closed under composition of functions, s two reflections give rottion. So G re is not group. On the other hnd, we clim tht G is group. Clim 1. The set of ll rottions nd reflections forms group. Proof.. It isn t hrd to see y the sme methods s ove tht if r θ is rottion nd R is reflection out line then r θ R = R r12 θ () tht is, reflecting out nd then rottion y ngle θ is the sme s reflecting out the line rotted y ngle 1 2θ. Likewise, R r θ = R r 1 2 θ () So the set G of rottions nd reflections is closed under composition of functions. The identity is in G, ecuse the identity is rottion y ngle 0. Inverses re in G ecuse ech reflection is its own inverse, nd the inverse of rottion y ngle θ is rottion y ngle θ. The set is ssocitive ecuse, gin, composition of functions is ssocitive. Therefore, G is group. Prolem 2.8 If x nd y re elements of group, prove tht (xy) 1 = y 1 x 1. Proof. The element (xy) 1 is the element s.t. (xy) 1 (xy) = 1 = (xy)(xy) 1. We show tht the element y 1 x 1 stisfies this property. (y 1 x 1 )(xy) = y 1 (x 1 (xy)) y ssocitivity = y 1 ((x 1 x)y) y ssocitivity = y 1 y y prop. of inverses = 1 y prop. of inverses so the first prt of tht eqution is stisfied. Now for the second prt. (xy)(y 1 x 1 ) = ((xy)y 1 )x 1 y ssocitivity So we hve shown tht (xy) 1 = y 1 x 1. = (x(yy 1 ))x 1 y ssocitivity = x 1 x y prop. of inverses = 1 y prop. of inverses

4 4 Prolem 3.2 Write Q( 2) for the set descried in Exercise 3.1 (iii). Given non-zero element + 2 express 1/(+ 2) in the form c+d 2 where c,d Q. Prove tht multipliction mkes Q( 2)\{0} into group. Proof. The set Q( 2) consists of ll rel numers of the form + 2 where nd re in Q. The numer 1/(+ 2) is the unique rel numer with the property tht (+ 2) 1/(+ 2) = 1. Let c = 2 2 nd let d = Then we 2 clim tht 1/(+ 2) = c+d 2. To see this, we clculte: Then nd (+ 2) c+d 2 = c+2d+(d+c) 2 2 c+2d = = 1 d+c = = 0. Thus, for the ove choices of c nd d, (+ 2) c+d 2 = 1, s promised. Now we show tht the set G = Q( 2)\{0} is group. The formul (+ 2) c+d 2 = c+2d+(d+c) 2 shows tht this set is closed under multipliction. Since G is suset of rel numers, it is ssocitive. The identity is in G since 1 = Finlly, the ove clcultion mens tht every non-zero element of Q( 2) hs multiplictive inverse lso in Q( 2). Therefore, Q( 2)\{0} is group. Prolem 3.3 Let n e positive integer nd let G consist of ll those complex numers z which stisfy z n = 1. Show tht G forms group under multipliction of complex numers. Proof. First we show tht G is closed under multipliction. Let z,z G. We need to show tht (zz ) n = 1. We hve tht (zz ) n = zz zz }{{} n times But multipliction in C is commuttive, so in fct, (zz ) n = z n z n. Since z,z G, this just mens tht (zz ) = 1. Thus G is closed under multipliction. Since C is group, nd G is suset of C, we know tht G is ssocitive. The identity is in G ecuse 1 n = 1. So the lst thing we need to show is tht G hs inverses. Let z G. Then z 1 is some complex numer (since ll complex numers hve inverses which re complex numers). We need to show tht (z 1 ) n = 1. But we know tht z n = 1. So multiply oth sides of this eqution y (z 1 ) n. This gives us tht (z 1 ) n z n = (z 1 ) n. But (z 1 ) n z n = 1 y the fct tht multipliction in C is commuttive, so this mens tht 1 = (z 1 ) n, s required. Thus G is group under multipliction. Prolem 3.5 Let n e positive integer. Prove tht (x ny) nz = x n(y nz).

5 5 Proof. By definition, (x n y) = xy kn where k Z is the numer s.t. xy kn [0,n 1]. Thus, (x n y) n z = (xy kn) n z = (xy kn)z k n = xyz k n where k is the numer s.t. (xy kn)z k n [0,n 1] nd so k = kz+k is the numer s.t. xyz k n [0,n 1]. With similr steps we get x n (y n z) = x n (yz ln) = x(yz ln) l n = xyz l n where gin l is the numer s.t. xyz l n [0,n 1]. We wnt to show tht xyz k n nd xyz l n re the sme numer. Note tht xyz k n (xyz l n) = (l k )n. So the difference etween these two numers is n integer multiple of n. On the other hnd, oth xyz k n nd xyz l n re etween 0 nd n 1. So (n 1) xyz k n (xyz l n) n 1. The only multiples of n inside these ounds is 0. Thus l k = 0 nd so xyz k n = xyz l n. Therefore, (x n y) n z = x n (y n z). Prolem 3.7 Which of the following sets form group under multipliction modulo 14? Answer: Note tht 5 5 = 11 mod 14. The sets {1,3,5} nd {1,3,5,7} contin 5 nd not 11, so they cnnot form groups. The set {1,7,13} cnnot form group mod 14 since 7 nd 14 shre common fctor, so 7 cnnot hve n inverse mod 14. In fct, suppose 7 14 x = 1. Then n = 2. But = 0, so 0 = 2 mod 14, which is impossile. Finlly, = 5, so the set {1,9,11,13} does not form group. So none of these sets form groups mod 14. Prolem 3.8 Show tht if suset of {1,2,...,21} contins n even numer, or contins the numer 11, then it cnnot form group under multipliction. Proof. Suppose n is n even integer etween 1 nd 21, inclusive. Then we cn write n = 2m for some other integer m. We clim tht n cnnot hve n inverse mod 22. Suppose not. Then n = 2m hs n inverse with the property tht 2m 22 n 1 = 1. Thus, m 22 n 1 = 11. By ssocitivity, I cn multiply 11 nd 2 first. Note tht = 0. Thus I get the sttement tht 0 22 n 1 = 11, i.e. 0 = 11. But this is not true mod 22. Therefore, n hs no inverse. Similrly, suppose n = 11. We clim tht 11 cnnot hve n inverse mod 22. Suppose not. Then n = 11 hs n inverse with the property tht n 1 = 1. Thus, n 1 = 2. By ssocitivity, I cn multiply 11 nd 2 first. Note tht = 0. Thus I get the sttement tht 0 22 n 1 = 2, i.e. 0 = 2. But this is not true mod 22. Therefore, 11 hs no inverse. So if we were given suset of {1,2,...,21} contining the numer 11 or ny even numer, then tht suset couldn t possily hve inverses for ll of its elements. Therefore it would not form group.

6 6 Prolem 4.7 Let G e the collection of ll rtionl numers x which stisfy 0 x < 1. Show tht the opertion { x+y if 0 x+y < 1 x+ G y = x+y 1 if 1 x+y < 2 mkes G into n infinite elin group ll of whose elements hve finite order. Proof. We will first show tht G is group. Agin, we do this in four steps. (1) G needs to e closed under the group opertion. Let x,y e in G. Then 0 x,y < 1 so 0 x + y < 2. Thus 0 x + G y < 1, so this is ok. Furthermore, oth x+y nd x+y 1 re rtionl numers. So x+ G y is rtionl numer etween 0 nd 1, s required. (2) Identity. The identity is 0, which is in the group. To see tht 0 is the identity, note tht if 0 x < 1, then 0 x+0 < 1, so x+ G 0 = x. (3) Inverses. If 0 x < 1 then 0 1 x < 1. Note tht x+(1 x) = 1, so x+ G (1 x) = 0. So for ny x G, 1 x is the inverse of x. Therefore the inverse of x is in G. (4) Associtivity. Let x,y nd z e in G. Ech time we dd two numers in G we sutrct 1 enough times to mke the result t lest 0, nd strictly less thn 1. Since 0 x+y +z < 3, we see tht { x+(y +G z) if 0 x+(y + x+ G (y + G z) = G z) < 1 x+(y + G z) 1 if 1 x+(y + G z) < 2 x+y + G z if 0 x+y +z < 1 = x+y +z 1 if 1 x+y +z < 2 x+y +z 2 if 2 x+y +z < 3 since y + G z is either y +z or y +z 1. Similrly, we get tht x+y + G z if 0 x+y +z < 1 (x+ G y)+ G z = x+y +z 1 if 1 x+y +z < 2 x+y +z 2 if 2 x+y +z < 3 Thus the group opertion is ssocitive. Therefore, G is group under this group opertion. There re infinitely mny rtionl numers etween 0 nd 1, so G is infinite. G is elin ecuse ddition of rtionl numers is elin. Finlly, we need to show tht elements of G hve finite order. Let e n element of G. Then define n G = + G + G }{{ } n times Since every time we dd on nother we sutrct either 0 or 1, we hve tht n G = n n m where m is n integer s.t. 0 m n nd 0 m < 1. Thus for n =, G = m where m is the integer s.t. 0 m < 1. But is lso n integer. So we must hve = m, nd so G = 0. But tht mens exctly tht hs order t most, since dding to itself times got us ck to the identity. So every element of G hs finite order.

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