Homework #4 Solutions (due 10/3/06) Chapter 2 Groups

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1 UNIVERSITY OF PENNSYLVANIA DEPARTMENT OF MATHEMATICS Math 370 Algeba Fall Semeste 2006 Pof. Gestenhabe, T.A. Ashe Auel Homewok #4 Solutions (due 10/3/06) Chapte 2 Goups Recall: Let G be a goup. Fo x G let #x denote the ode of x in G. The cental manta of odes (poved in the pevious solution set) is: x n = e #x n and the ode #x of x is the smallest such positive intege n. Definitions/Facts: About gcd and lcm. Fo positive integes n and m define thei geatest common diviso to be the positive intege gcd(n, m) chaacteized by the following equivalent conditions: i) any common diviso of n and m is a diviso of gcd(n, m), i.e. a n and a m a gcd(n, m), ii) gcd(n, m) is the smallest positive intege that can be witten in the fom kn+lm fo k, l Z, iii) witing n = p e 1 1 pe and m = p f 1 1 pf as a poduct of powes of distinct pime numbes p 1,..., p with nonnegative exponents e 1,..., e, f 1,..., f 0, then we have that gcd(n, m) = p g 1 1 pg whee g i = min{e i, f i } fo i = 1,...,. Fo positive integes n and m define thei least common multiple to be the positive intege lcm(n, m) chaacteized by the following equivalent conditions: i) any common multiple of n and m is a multiple of lcm(n, m), i.e. n b and m b lcm(n, m) b, ii) lcm(n, m) is the smallest positive intege that can be witten simultaneously in the fom kn and lm fo k, l 1, note that in this case l k is the educed faction of n m, iii) witing n = p e 1 1 pe and m = p f 1 1 pf as a poduct of powes of distinct pime numbes p 1,..., p with nonnegative exponents e 1,..., e, f 1,..., f 0, then we have that gcd(n, m) = p g 1 1 pg whee g i = max{e i, f i } fo i = 1,...,. The gcd and lcm have the following useful popeties: gcd(n, m) lcm(n, m) = n m, n and m ae elatively pime gcd(n, m) = 1 lcm(n, m) = nm, n m gcd(n, m) = n lcm(n, m) = m 2.10 Let G be a goup. a) Claim: If #x = s fo some, s 1 then #x = s. Poof. Fist note that (x ) s = x s = e since #x = s so #x s. Futhemoe, fo 0 < k < s we have that 0 < k < s, so that (x ) k = x k e. So #x eally is s. b) Claim: If #x = n then fo any 1. #x = Poof. Fo l 1 we have that n gcd(n, ) lcm(n, ) =. (x ) l = x l = e n l nk = l fo some k 1, and if l = #x, i.e. the least possible such l, then nk = l = lcm(n, m) is then the least common multiple of n and m. But then #x = l = nk = lcm(n, m) = n gcd(n, m), whee the final equality comes fom the fomula elating gcd and lcm. 1

2 2.11 Let a, b G be elements of a goup, and suppose ab is of finite ode n. Then (ab) n = e a 1 (ab) n a = a 1 a = e (a 1 aba) n = e (ba) n = e, whee the second equivalence is execise 3.4. Thus ba has finite ode and #ba n. Now similaly, fo 0 < k < n we have (ab) k e a 1 (ab) k a a 1 a = e (a 1 aba) k e (ba) k e, and so indeed the ode of ba is n. This also poves that if ab has infinite ode, then so does ba Let G be a cyclic goup of ode n. Then an element x G geneates G if and only if #x = n. Now fixing a geneato x G, we have G = {e, x, x 2,..., x n 1 }, and so in view of the fomula fom execise 2.10b, we see that x also geneates G #x n = n = n gcd(n, ) = 1 gcd(n, ) is elatively pime to n. Thus in asking the question how many of its elements geneate G? we ae foced to deal with the following numbe ϕ(n) = { : 0 < < n and gcd(n, ) = 1} = the numbe of numbes fom 1, 2,..., n 1 that ae elatively pime to n, usually called the Eule phi-function of n. a) Fo n = 6, we see that of the numbes 1, 2, 3, 4, 5, only 1, 5 ae elatively pime to 6, so ϕ(6) = 2. Fo completeness I ll compute the cyclic subgoups geneated by evey element: < e > = {e} < x > = {e, x, x 2, x 3, x 4, x 5 } < x 2 > = {e, x 2, x 4 } < x 3 > = {e, x 3 } < x 4 > = {e, x 4, x 2 } < x 5 > = {e, x 5, x 4, x 3, x 2, x} and we see that only x and x 5 ae geneatos. b) Why don t we make a little table fo n = 2,..., 12: n numbes 1,..., n 1 elatively pime to n ϕ(n) , 2 1, , 2, 3 1, , 2, 3, 4 1, 2, 3, , 2, 3, 4, 5 1, , 2, 3, 4, 5, 6 1, 2, 3, 4, 5, , 2, 3, 4, 5, 6, 7 1, 3, 5, , 2, 3, 4, 5, 6, 7, 8 1, 2, 4, 5, 7, , 2, 3, 4, 5, 6, 7, 8, 9 1, 3, 7, , 2, 3, 4, 5, 6, 7, 8, 9, 10 1, 2, 3, 4, 5, 6, 7, 8, 9, , 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 1, 5, 7, 11 4 c) As aleady noted, the numbe of elements that geneate a cyclic goup of ode n is ϕ(n). 2.20a Claim: Let x, y G be commuting elements of a goup and let #x = n and #y = m. Then all we can say is that #xy lcm(n, m).

3 Poof. Fist, note that since x and y commute, (xy) l = x l y l fo all l Z. Now let l = lcm(n, m). Then since n l and m l, i.e. thee exist a, b 1 such that l = an = bm, we know that thus #xy lcm(n, m). (xy) l = x l y l = (x n ) a (y m ) b = e a e b = e, Note: The ode #xy is difficult to elate exactly to the individual odes #x and #y. Fo example, let G =< a > be a cyclic goup of ode 6, then the following table displays the ange of possible behavio: x y xy #x #y #xy lcm(#x, #y) =? a a a no a a 2 a no a a 3 a no a a 4 a yes a a 5 e no a 2 a 2 a yes a 2 a 3 a yes a 2 a 4 e no a 2 a 5 a yes a 3 a 3 e no a 3 a 4 a yes a 3 a 5 a no a 4 a 4 a yes a 4 a 5 a no a 5 a 5 a no 3.11 Claim: Let G be a goup. Then the set Aut(G) of goup automophisms of G foms a goup unde composition. Poof. We need to veify the goup axioms fo the set Aut(G) unde the opeation of composition. Fist, we show that Aut(G) is closed unde composition. We ll need the following: Lemma: Let ϕ, ψ : G G be maps. Then i) if ϕ and ψ ae injective then so is ϕ ψ, ii) if ϕ and ψ ae sujective then so is ϕ ψ, iii) if ϕ and ψ ae bijective then so is ϕ ψ, iv) if ϕ and ψ ae goup homomophisms then so is ϕ ψ, v) if ϕ and ψ ae goup isomophisms then so is ϕ ψ. Poof. To i), let x, y G, then (ϕ ψ)(x) = (ϕ ψ)(y) ϕ(ψ(x)) = ϕ(ψ(y)) ψ(x) = ψ(y) x = y, whee the second and thid implications follow if ϕ and ψ ae injective, espectively. Thus ϕ φ is injective. To ii), let x G, then since ψ is sujective, thee exists x G such that ψ(x ) = x. Since ϕ is sujective, thee exists x G such that ϕ(x ) = x. But then so we see that ϕ ψ is sujective. To iii), combine i) and ii). To iv), let x, y G, then (ϕ ψ)(x ) = ϕ(ψ(x )) = ϕ(x ) = x, (ϕ ψ)(xy) = ϕ(ψ(xy)) = ϕ(ψ(x)ψ(y)) = ϕ(ψ(x)) ϕ(ψ(y)) = (ϕ ψ)(x) (ϕ ψ)(y), if both ϕ and ψ ae homomophisms. So we indeed see that ϕ ψ is a homomophism. To v), combine iii) and iv).

4 Thus we see that fo automophisms ϕ, ψ Aut(G) the composition ϕ ψ Aut(G) is again an automophism, so Aut(G) is closed unde composition. Next we quickly veify that composition is associative. Fo ϕ, ψ, λ Aut(G) and fo x G we have ((ϕ ψ) λ)(x) = (ϕ ψ)(λ(x)) = ϕ(ψ(λ(x))) = ϕ((ψ λ)(x)) = (ϕ (ψ λ))(x), so that indeed (ϕ ψ) λ = ϕ (ψ λ), so composition is associative. Next, we find an identity. Let id : G G be the identity function, which is clealy an automophism. Fo ϕ Aut(G) and fo x G note that (ϕ id)(x) = ϕ(id(x)) = ϕ(x), and (id ϕ)(x) = id(ϕ(x)) = ϕ(x), so that indeed ϕ id = ϕ and id ϕ = ϕ. Thus id Aut(G) is indeed an identity. Finally, we check that inveses exist, but we aleady did this in execise 3.5. Fo an isomophism ϕ : G G, we peviously showed that the invese function ϕ 1 : G G is again an isomophism, and by definition satisfies ϕ ϕ 1 = id and ϕ 1 ϕ = id, so ϕ 1 is an invese of ϕ fo composition. So indeed, Aut(G) has inveses. We ve finished showing that Aut(G) is a goup unde composition Detemining some automophism goups. a) We e aleady show that Aut(Z) = {±id} in execise 4.4. b) Since Z/10Z is a cyclic goup geneated by 1, any homomophism ϕ : Z/10Z Z/10Z is completely defined by the image of 1. Now we also know by execise 3.6a that if ϕ is an isomophism, then it peseves odes of elements, i.e. #ϕ(x) = #x fo all x Z/10Z. In paticula, a geneato must be sent to a geneato. Now in execise 2.16b, we aleady know that the only elements in Z/10Z that geneate ae 1, 3, 7, 9. It s also easy to see that each of the fou choices of whee to send 1 gives an automophism of Z/10Z, so we ll label them accodingly: Aut(Z/10Z) = {ϕ 1, ϕ 3, ϕ 7, ϕ 9 }. Note that ϕ 1 = id. Now we compute the goup stuctue on Aut(Z/10Z). Fo example, fo x Z/10Z, we have (ϕ 3 ϕ 7 )(x) = ϕ 3 (ϕ 7 (x)) = ϕ 3 (7x) = 3(7x) = 21x = x, so we find that ϕ 3 ϕ 7 = id = ϕ 1. Continuing like this we can calculate the multiplication table fo Aut(Z/10Z): ϕ 1 ϕ 3 ϕ 7 ϕ 9 ϕ 1 ϕ 1 ϕ 3 ϕ 7 ϕ 9 ϕ 3 ϕ 3 ϕ 9 ϕ 1 ϕ 7 ϕ 7 ϕ 7 ϕ 1 ϕ 9 ϕ 3 ϕ 9 ϕ 9 ϕ 7 ϕ 3 ϕ 1 Notice that we have a nice goup isomophism (Z/10Z) Aut(Z/10Z) a ϕ a We also see that both ϕ 3, ϕ 7 Aut(Z/10Z) have ode 4, i.e. they each geneate. This shows that Aut(Z/10Z) is cyclic, and we can constuct two diffeent isomophisms Z/4Z Aut(Z/10Z) Z/4Z Aut(Z/10Z) 0 ϕ 1 0 ϕ 1 1 ϕ 3 1 ϕ 7 2 ϕ 9 2 ϕ 9 3 ϕ 7 3 ϕ 3 neithe of which seems paticulaly appealing, but just illustates the two ways we can foce ouselves to think of Aut(Z/10Z) as a cyclic goup of ode 4.

5 c) Witing S 3 =< s, t : s 2 = t 3 = e, ts = st 2 >, we see that the symmetic goup S 3 is geneated by elements s, t o odes 2, 3, espectively, subject to a futhe elation. Any automophism ϕ : S 3 S 3 is detemined by the images of s, t, and as befoe, must peseve the odes of elements. Now S 3 has thee elements s, st, st 2 of ode 2, and two elements t, t 2 of ode 3. So any automophism must take s to one of s, st, s 2 and t to one of t, t 2. Thee ae only six conceivable ways of doing this: s s s st s st 2 t t t t t t s s s st s st 2 t t 2 t t 2 t t 2 One now checks that each of these in fact does give an automophism of S 3. Thus Aut(S 3 ) just consists of these six elements. We would futhe like to know the stuctue of Aut(S 3 ). One way to do this is to know that thee ae only two isomophism classes of goups of ode six, namely cyclic of ode six and S 3. We then just need to check if two of these automophisms don t commute. In fact Aut(S 3 ) = S 3. Anothe way to see this is to note that the cente Z(S 3 ) is tivial, so that conjugation by each element of S 3 gives a diffeent automophism, since thee ae aleady six of these, these fill up all of Aut(S 3 ). Thus we have the nice isomophism ad : S 3 Aut(S3 ) x ad x : y xyx 1, in the notation fom lab. d) The analysis of Aut(Z/8Z) follows exactly the same way as fo Aut(Z/10Z) in pat b). In the end, we find that Aut(Z/8Z) = {ϕ 1, ϕ 3, ϕ 5, ϕ 7 } and we have the nice isomophism (Z/8Z) Aut(Z/8Z) a ϕ a Incidentally, we check that each element of Aut(Z/8Z) has ode two, so that Aut(Z/8Z) = Z/2Z Z/2Z. e) Is the automophism goup of a cyclic goup necessaily cyclic? Well, no, see pat d). f) Is the automophism goup of an abelian goup necessaily abelian? Well, no eithe. Take fo example the abelian goup Z/2Z Z/2Z Z/2Z. Each pemutation of the enties gives a goup automophism, and as we know, pemutations of thee objects don t usually commute. In paticula, we see that Aut(Z/2Z Z/2Z Z/2Z) has a subgoup isomophic to the pemutation goup S 3. Do you think that is the whole automophism goup? 4.8 Subgoups of goups. a) The subgoups of S 3 =< s, t : s 2 = t 3 = e, ts = st 2 > ae: {e}, {e, s}, {e, st}, {e, st 2 }, {e, t, t 2 }, S 3, and {e}, {e, t, t 2 }, S 3 ae nomal subgoups. b) The subgoups of the quatenion goup Q = {±1, ±i, ±j, ±k} whee i 2 = j 2 = k 2 = 1 and ij = k, jk = i, and ki = j, ae: and evey subgoup is nomal. {1}, {±1}, {±1, ±i}, {±1, ±j}, {±1, ±k}, Q, 4.9b Claim: Let ψ : G G and ϕ : G G be homomophisms of goups. Then Poof. Obvious. ke(ϕ ψ) = ψ 1 (ke(ϕ)) G.