Math 4107: Abstract Algebra I Fall 2017 Assigmet 1 Solutios 1. Webwork Assigmet1-Groups 5 parts/problems) Solutios are o Webwork. 2. Webwork Assigmet1-Subgroups 5 parts/problems) Solutios are o Webwork. 3. Problem 2.1 # 22. Note that you must verify that is, you ca ot assume) the two relatios i Example 10 give i the setece Oe ca verify.... Proof. Let > 2 be a iteger. Cosider S {x, y) x, y R} ad f, h AS) where fx, y) x, y) ad hx, y) is rotatio of the plae about the origi through a agle of / i the couterclockwise directio. I matrix form, we have that fx, y) 1 0 ) ) x cos ad hx, y) y si si cos ) ) x y Let i deote the idetity mappig i AS). We see that f 2 x, y) ffx, y)) f x, y) x, y) so f 2 i. Likewise, h is rotatio of the plae about the origi through a agle of /) so h i. I the matrix form, we ca also cofirm that f 2 1 0 ) ) 1 0 1 0 ad that h i. To see the secod relatio, we give a quick proof by iductio. Let P k) be the statemet that h k cos k si k si k ) cos k for all k N. We kow that h 1 is true. For our iductio hypothesis, we assume that P i) is true for some i 1. We cosider h i+1 h i h. Sice h i cos i h cos si i si cos i si si i cos ) si i cos cos i si si i si cos i cos ad siu ± v) si u cos v ± cosu si v, cosu ± v) cos u cos v si u si v, we have that h i+1 cosi + 1) sii + 1) ) sii + 1) cosi + 1). Sice our assumptio that P i) is true for some i 1 implies that P i + 1) is true, by the priciple of iductio we kow that P k) is true for all k N. I particular, ) h cos si si cos ) ) 1 0. Now cosider cos fh)fh) si si cos ) 2 ) 1 0
sice cos 2 + si2 1. Because fh)fh) i, we have that fh) fh) 1 h 1 f 1 h 1 f by the defiitio of iverses ad our previous idetities. Let G {f k h j k 0, 1; j 0, 1,..., 1}. Cosider f i h j, f s h t G with 0 i, s 1 ad 0 j, t 1. If s 0, the f i h j )f s h t ) f i h t+j f i h t+j) mod. If s 0, the s 1 ad f i h j )f s h t ) f i+s h t j f i+s) mod 2 h t j) mod. Note that this is true eve whe j 0. I either case, we have that f i h j )f s h t ) f a h b for some a {0, 1}, b {0,..., 1}. Hece, G is closed uder the product of compositio of fuctios. We kow that compositio of fuctios is associative i AS). Hece, sice G AS), the product i G is associative. We have the idetity mappig i f 2 h i G. Let f s h t G. If s 0, the h t h t mod h t mod h t h 0 i. If s 1, the fh t )fh t ) f 2 h t t) mod i. Hece, every elemet of G has a iverse i G. Sice G satisfies the four coditios of closure, associativity, existece of a idetity, ad existece of iverses, the G is a group. We see that G is oabelia sice si si ad cos fh si si cos ) cos si si ) hf. cos By defiitio, G has order at most 2. Sice h is a rotatio by a agle of, we kow that h j i for 1 i 1. Hece the elemets h i for 0 i i are all distict. Likewise, the elemets fh i for 0 i i are all distict. If ot, the there would be some fh i fh t for 0 i < t 1. However, cacelig by f o the left ad h o the right we have that h t i i for 1 i t < 1, which cotradicts our previous assertio. Thus, there are 2 distict elemets i G. Hece, G is a oabelia group of order 2. 4. Problem 2.1 # 24. Proof. Let G be the dihedral group of order 2 where > 2 as above. a) Let be odd. Suppose there is a G such that ab ba for all b G. We claim that a e where e is the idetity i G. Let a f i h j for 0 i 1 ad 0 j 1. Suppose that i 1 ad cosider b h 2 G sice 3. We have that ab fh j )h 2 ) fh 2+j) mod ad ba h 2 )fh j ) fh j 2) mod. Sice ab ba, we have fh 2+j) mod fh j 2) mod. By cacellatios, we arrive at the equatio 2 + j) j 2) mod. But this is true if ad oly if 4 0 mod. Sice is odd, though, it s ot possible that 4. Hece i 0. We have that a h j for some 0 j 1. Now let b f. By assumptio ab h j f fh j ba. Sice h j f fh j mod we have that j j mod. But the 2j 0 mod. Sice is odd, it must be that j. But sice j <, this ca be true oly if j 0. Hece a h 0 e ad our claim is proved.
b) Let be eve ad let a G where a h /2. We claim that ab ba for all b G. Let b f i h j G for 0 i 1, 0 j 1. Suppose i 0. The ab h j+/2 h /2+j ba. Suppose i 1. The ab)a h /2 fh j ))h /2 fh j /2) mod )h /2 fh j b. Sice aba b ad a 2 h e, we have that ba aba)a ab)e ab so our claim is true. c) Let be eve. Let a G such that ab ba for all b G. Let a f i h j for 0 i 1, 0 j 1. ad let b f s h t for 0 s 1, 0 t 1. Suppose i 0. The ab h j f s h t ) ad ba f s h t )h j. If s 0, the ab h j+t ba for j. Suppose s 1. The ab fh t j) mod ad ab ba implies that fh t j) mod fh t+j) mod. This is true if ad oly if t+j t j) mod which holds exactly whe 2j 0 mod, that is whe j 0, /2. Suppose that i 1. The ab fh j )f s h t ) ad ba f s h t )fh j ). If s 0, the ab fh j+t) mod ad ba fh j t) mod. Assumig that ab ba implies that j+t) j t) mod ad that 2t 0 mod. Let t 1. Sice > 2, we have that 2 ad 2t 0 mod. Hece for b h, we have that ab fh j+1 fh j 1 ba. Thus, there is o a f i h j G with i 1. 5. Problem 2.1 # 28. Proof. Let G be a set with a operatio such that: a) G is closed uder. b) is associative. c) e G such that e x x x G. d) x G y G such that y x e. We claim that G is a group. We kow that G satisfies the first two requiremets of a group, closure ad associativity. We must show that the left idetity of G is also a right idetity, ad that left iverses are also right iverses. For all x, z G, we write x z as xz. Let e G such that ex x for all x G. Let x G ad y G such that yx e. Cosider xy G. The xy)xy) xyx)y) xey)) xy by repeated applicatio of associativity ad the fact that e is a left idetity i G ad y is a left iverse of x i G. There exists z G such that zxy) e. Hece zxy)xy)) zxy))xy) exy) xy. We also have that zxy)xy)) zxy)) e. Thus xy e ad y is also a right iverse of x i G. Now cosider xe. The xe xyx) xy)x ex x sice we have show that xy e. Thus, e is also a right idetity for ay a G. Sice G has a elemet which is both a left ad right idetity ad for every x G there exists some y G which is both a left ad a right iverse, we kow that G is a group. 6. Problem 2.2 # 2.
Proof. We are give G as a set, closed uder a associative operatio, such that, for all a, x, y, u, w G, ax ay implies x y ad ua wa implies u w. Assume the operatio is called. By defiitio, G is closed uder, ad is associative. Hece, we eed oly prove two thigs: the existece of a idetity elemet, ad the existece of iverse elemets. Suppose G. To fid the idetity elemet, fix a G ad defie the map f : G G defied by fx) ax for all x G. For all x, y G, ax ay implies x y. Therefore, f is a ijective map. Because f is ijective, the elemets i the domai G map to distict elemets i the rage G. Hece, by the pigeohole priciple, f is also surjective. This implies that, for all z G, there exists x G such that z ax. Similarly, if we defie g : G G such that gu) ua for all u G, for all u, w G, ua wa implies u w. Thus, g is ijective, ad by the same pigeohole argumet above, g must also be bijective. Hece, for all z G, there exists u G such that gu) ua z. Therefore, there exists elemet b G such that fb) ab a ad elemet c G such that gc) ca a. Applyig the elemets b ad c, for all z G, zb uab ua z ad cz cax ax z. Hece, c is a left-idetity elemet, ad b is a right-idetity elemet. Sice the above equatios apply to all z G, by applyig c z i the first equatio, ad b z i the secod, we have cb c ad cb b. Hece b c. Reamig this elemet as e, it satisfies ze ez z for all z G. Therefore, G has a idetity elemet uder. For each elemet z G, we ca use the defiitios of f ad g above as templates to defie surjective maps f z x) xz ad g z u) zu. Let z be fixed. Because f z ad g z are bijective, there exist elemets s, t G such that e f z s) zs ad e g z t) tz. The s es tz)s tzs) te t. This implies s t, ad thus zs sz e. Thus, s is the iverse of z, z 1. Sice this argumet holds for all z, every elemet of G has a iverse. We are give that G is closed uder the operatio, with associative. As we have show, G has a idetity elemet ad each elemet has a iverse. Therefore, G is a group. Additioally, give a example to that this result is false if G is ifiite. Proof. Let G be the group of atural umbers, G N {x Z x > 0}, ad let the operatio be stadard additio, +. The set of atural umbers is a ifiite set. We already kow that additio is associative o Z: for all x, y, z Z, x + y) + z x + y + z). Therefore, + is associative o G Z. If x, y G, the x, y Z, ad sice Z is closed, x + y Z. Further, sice x > 0 ad y > 0, x + y > 0. Hece, x + y G. Therefore, G is also closed uder +. Now let a, x, y, u, w G, such that a + x a + y ad u + a w + a. By applyig the properties of additio, we kow x y ad u w. Therefore, G is a ifiite set with the operatio + satisfyig all the other properties listed i Problem 2.2 # 2.
However, G does ot cotai a idetity elemet. The equatio a + e a is valid if ad oly if e 0, but 0 / G. Similarly, G does ot cotai ay of the iverses of its elemets. For all a G, a + b e is oly valid if b a, but a < 0, which meas b a / G. Therefore, G N with the operatio + is ot a group.