Algebra Homework Set 5 Hung Tran. 6.3.4 Suppose W Is

Algebra Homework Set 5 Hung Tran. 6.3.4 Suppose w is the word of finite order h. Since w can be written in reduced form as x1 ...xn , w can also be written in the form aba−1 for b is in reduced form and furthermore, the last term in b is not the inverse of the first term (a can be chosen to be the identity). Then 1 = wh = abh a−1 so bh = 1. But bh is a reduced word by the condition on its first and last term. So bh = 1 iff b = 1 or w is the identity elemement. Thus every nonidentity element of a free group is of infinite order. QED 6.3.12 Formulate the notion: A free nilpotent group on S of nilpotence class c is described by the presentation: NS,c =< S | R > for R contains all the word of the form: [[...[x1 , x2 ], ...], xc+1 ] with xi ∈ S. Another equivalent notion is to define NS,c = F (S)/Fc (S) with Fc the c-th term in the lower central series. Universal mapping property: Let G be a nilpotent group of class less then or equal c, and ϕ : S → G a set map and ι : S → NS,c the inclusion map. Then there is a unique group homomorphism φ : NS,c → G such that φι = ϕ Proof of the UMP: If G is a nilpotent class of nilpotent class less then or equal c then Gc = 1. Thus the identity [[...[x1 , x2 ], ...], xc+1 ] = 1 holds with xi ∈ G. By the UMP for presentation, there is a unique homomorphism Φ : F (S) → G that extends the set map ϕ (R ≤ Ker(Φ) so the normal closure of R in F(S) is also contained in the kernel of Φ as the kernel is itself a normal group. QED 6.3.13 Suppose there is such a nilpotent group N generated by two elements, then denote its nilpotence class c. Since every nilpotent group K which is generated by 2 elements is a homomorphic image, the homomorphism from N to K is surjective. Therefore, any commutator in K is the image of a commutator in N; and hence, any term in the lower central series of K is the image of a corresponding term in the lower central series of N. Since N has nilpotence class c, the lower central series terminates (goes to the identity) at the c-th term. Therefore, the lower central series of K must also terminate before or at the c-th term. Thus K has nilpotence class less than or equal c. (1) Now we show that a D2n is a nilpotent group of class n-1 by induction. It certainly trues for n = 1. Suppose it is true for D2k . Consider D2k+1 . Using the k−1 notation as in section 1.2, it is easy to see that Z(D2k+1 ) = (1, r2 ) and the list k−1 k−1 (ri , r2 +i ), (sri , sr2 +i ) for 0 ≤ i < 2k−1 exhausts all coset with respect to the center. So it is straightforward calculation to show that D2k+1 /Z(D2k+1 ) ∼ = D2k . Then we can apply the induction hypothesis to conclude the claim. Certainly a finite dihedral group is generated by 2 elements and the above claim shows that the nilpotence class can be as large as we want. That contradicts (1) and leads the priori assumption to absurdity. So there does not exist such a group N. QED Additional problem 1. 1

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a. (1) UMP: If S=(a,b) the generating set of a free vector space FV(S) over a field k, then, for given V, a vector space over a field k, and v1 , v2 two arbitrary vectors in V, there exists a linear map φ : F V → V such that φ(a) = v1 and φ(b) = v1 (with a and b the identification (canonical inclusion) of genetators in FV) (2) and (3): Let F V (S) = {(λ1 a + λ2 b)} for λ1 , λ2 in k, then FV(S) is certainly a vector space over k. The map: φ(λ1 a + λ2 b) = λ1 v1 + λ2 v2 is certainly linear. On the other hand, if φ∗ is any linear map that extends the set map that sends a to v1 and b to v2 must satisfy:φ∗ (λ1 a + λ2 b) = λ1 φ∗ (a) + λ2 φ∗ (b) = λ1 v1 + λ2 v2 . So φ is unique. QED b. (1) UMP: If S=(a,b) the generating set of a free abelian FA(S), then, for given A, an abelian group and α, β ∈ A, there exists a homomorphism φ : F A(S) → A such that φ(a) = α and φ(b) = β (2) and (3): Let F A(S) = {(λ1 a + λ2 b)} λ1 , λ2 in Z, and equip FA(S) with component-wise addition then FA(S) is an abelian group with respect to that operation. Now if the group law in G is written additively then the map: φ(λ1 a+λ2 b) = λ1 α + λ2 β is certainly a group homomorphism (note that the (+) symbol on LHS and RHS are two different operations in different groups). On the other hand, if φ∗ is any homomorphism that extends the set map that sends a to α and b to β must satisfy:φ∗ (λ1 a + λ2 b) = λ1 φ∗ (a) + λ2 φ∗ (b) = λ1 α + λ2 β. So φ is unique. QED c. (1) UMP: If S=(a,b) the generating set of a free G-set FG(S), then, for given G-set A and α, β ∈ A, there exists a G-map: φ : F G(S) → A such that φ(a) = α and φ(b) = β (2) and (3): Let F G(S) = {(g, x)} for g ∈ G, x ∈ S and identify x with (e, x) for e the identity of G. G acts on FG(S) as h • (g, x) = (hg, x) so FG(S) is a G-set. The map: φ((g, x)) = g • λ(1, x) with λ the set map that sends (1,a) to α and (1,b) to β is a G-map because h • φ((g, x)) = hg • λ(x) = φ((hg, x)) = φ(h • (g, x)). On the other hand, if φ∗ is any G-map that extends λ must satisfy:φ∗ ((g, x)) = g • λ(1, x). So φ is unique. QED d. (1) UMP: If S=(a,b) the generating set of a free communitive ring with identity FR(S), then, for given R, a commutative ring with identity and α, β ∈ R, there exists a ring homomorphism φ : F R(S) → R such that φ(a) = α and φ(b) = β (2) and (3): Let FR(S) ≡ all finite linear combination with coefficients in Z of terms of the form an bm for n,m non-negative integers. Also we require 1 ≡ a0 ≡ b0 as the multiplicative identity. Now the addition and multiplication are defined component-wise (for example an bm as bt = an+s bm+t ) and require the operations to follow the distributive law. Thus, we make FR(S) a polynomial ring. Then it is straightforward to verity FR(S) is a commutative ring with identity (addition is the group law while multiplication is certainly commutative). For any x in FR(S), x = Σki=1 ci ani bmi and φ any Ring homomorphism, we must have: φ(x) = Σki=1 ci φ(a)ni φ(b)mi = Σki=1 ci αni β mi . So if φ exists it must be unique. But it is straightforward to verity that φ as above is a Ring homomorphism. QED 2 Given a group G, then G =< G >. By UMP for free group, the identity map of G to itself has a unique extension to a homomorphism: φ : F (G) → G. Let R = Ker(φ) then R can be considered as a set of words. Since R as a kernel is normal the normal closure of R (as a set of words in F(G)) is also R itself. Therefore, G has

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the presentation G =< G | R >. QED 3. a. For every set S, F(S) is a free group, i.e., an object in the category of groups. For any set map ϕ : S → T , by the UMP of free groups, there is a unique group homomorphism:φ : F (S) → F (T ) such that it extends the map ϕ. So the functor assigns any ϕ ∈ Homset (S, T ) to φ ∈ Homgroup (F (S), F (T )). Furthermore, it is clear that if gf is a composition of set maps, then F (gf ) = F (g)F (f ) as a composition of homomorphism in the category of groups. Finally, if ϕ is the identity map, then φ is also an identity homomorphism (since the identity homomorphism satisfies the condition and we know there is a unique map). Thus, the free group construction can be viewed as a functor from the category of sets to the category of groups. QED b. Categorical sum in the category of sets is disjoint union while categorical sum in the category of groups is the free product(assumed to exist for the moment). A functor preserves sum if F (ti∈I Si ) ∼ = ∗i∈I F (Si ). To ease notation, we prove it for the case of disjoint union of two sets S, T (it is a matter of notation to generalize for arbitrary sets). Let ιS be the canonical injection S → S tT then by UMP for free group, there is an unique homomorphism φS : F (S) → F (S t T ) that extends the map ιS . Similarly, there is an unique homomorphism φT : F (T ) → F (S t T ) that extends the map ιT , the canonical injection T → S t T . By the UMP for a free product (described below in exercise 4), there exists a unique homomorphism φ : F (S) ∗ F (T ) → F (S t T ) such that φρX = φX for X in {S, T } and ρX is the canonical injection from F(X) to F (S) ∗ F (T ). Let x = x1 ...xn a reduced word in F (S t T ) with xi is in S or T. Then each xi is in the image of the function ιS and ιT ; hence in the image of φS or φT . So for each xi , there must be zi in F (S) ∗ F (T ) such that φ(zi ) = xi . Since φ is a group homomorphism, x = φ(z1 ...zn ). That is, φ is surjective. (1) Now since F (S t T ) is a free group, there is a unique group homomorphism η : F (S t T ) → F (S) ∗ F (T ) that extends the canonical injection: ι : S t T → F (S) ∗ F (T ) (the injection is letter by letter and can be viewed as the combination of ρX ιX for X ∈ {S, T }). Then φη is the identity on the set of generators of F (S t T ); hence, it is the identity automorphism of F (S t T ). Thus, φ is injective. (2) (1) and (2) show that φ is an isomorphism between groups. That is, F (ti∈I Si ) ∼ = ∗i∈I F (Si ). So the functor preserves sum. QED 4. a. UMP: Let A and B be groups and ρA , ρB group homomorphisms of A, B into a fixed group A ∗ B. For everygroup G, and any homomorphism ψA and ψB from A, B into G, there is a unique homomorphism: χ : A ∗ B → G such that χρX = ρX for X in {A, B}. b. Let < S1 | R1 > be a presentation of A and < S2 | R2 > be a presentation of B. Let A ∗ B =< S1 t S2 | R1 t R2 > and ρX the unique group homomorphism that

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extends the canonical inclusion of Si into S1 t S2 for X ∈ {A, B} and i ∈ {1, 2} (that homomorphism exists uniquely since generators in A ∗ B correspondingly satisfy conditions for generators in A and B) Let w = x1 ...xn a word in A ∗ B with xi is in S1 or S2 . Then if χ is a group homomorphism then it must satisfy: χ(w) = χ(x1 )...χ(xn ) = ψ(x1 )...ψ(xn ) for ψ is the combination map of ψA and ψB . So such a map is unique and it is clear that if we define χ as above then it satisfies the conditions. c. We define W as the set of all words x1 ...xn such that each xi ∈ X − {identity} for X ∈ {A, B} and two consecutive terms come from different groups and the empty word. Note that any word of A ∗ B can be reduced to that form after finite steps. It remains to show that the expression in that form is unique. Now we define an operation by nonidentity element g in A as:g • (w) = gx1 ...xn if g and x1 belong to different groups = g 0 x2 ...xn if g and x1 belong to the same group and g 0 = gx1 in that group The identity operates trivially on the set. It is clear that the operation is actually a (left) action. Similarly we define action by B on W. So each element of A and B can be identified with an element of the symmetric group on W. So there is a unique group homomorphism φ : A ∗ B → SW such that for s ∈ S1 t S2 φ(s) =the permutation induced by s seen as a generator in A or B. Consider the map ιW → A ∗ B such that w is sent to [w] the equivalence class that contains w and π : A ∗ B → W as [w] is sent to the [w] • () with () denotes the empty word. Then, πι(w) = (w) so πι is the identity map on W. Hence, ι is 1-1. By the observation at the beginning, ι is also surjective. Therefore, there is 1-1 correspondence; i.e., every element of A ∗ B is uniquely expressible as a word consisting of factors coming alternately from A/{1} and B/{1}. QED d. We prove by induction on the length of the word expressed in alternative form as in part (c). It is obviously true for words of length 0,1. Suppose it is true for words of length less then or equal k. Let w = x1 ...xk+1 is a word of finite order m and of length k+1. Then x1 and xk+1 belong to a same group since otherwise 1 = wm = x1 ...xk+1 x1 ......xk+1 (1) is a word expressed in alternative form of length m(k + 1) and by result of part (c) it is not in the same equivalence class with the identity element (empty word). Furthermore, x1 xk+1 = identity = empty word since otherwise, wn can be expressed in alternative form of length mk by replacing in (1) x1 xk+1 by x∗ the product in the group of x1 and xk . So w = x1 vx−1 1 for v is also expressed in alternative form m m −1 of length k-1.1 = wm = (x1 vx−1 1 ) = x1 v x1 . m Thus v = 1 so by the induction hypothesis, v is conjugate to an element of either A or B. Now it follows immediately that w is also conjugate to an element of A or B. That completes the induction. QED e. Since the free group construction preserves sum, a free group can be considered as a free product of free groups on one generators. By part (d), an elment of finite order must be conjugate to one element of one free group on one generator. But a

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free group on one generator is a cylic group of infinite order, every element has infinite orders. So there does not exist an element of finite order in a free group. QED 5. a. Let G =< r, s | rn = s2 = 1, srs−1 = r−1 >. The elements of G are products of r’s, s’s, r−1 ’s, s−1 ’s. Since r and s have finite order, the elements of G are in fact products of r’s and s’s. Since sr = rn−1 s every product of r’s and s’s can be rewritten so that r’s precede s’s. Furthermore, since rn = s2 , every element of G is a product of fewer then n r’s and fewer than 2 b’s. So G has at most 2n elements. (1) Furthermore, using the notation of 1.2, D2n is generated by two elements and they satisfy the condition given by the presentation. Therefore, according to the remark at the end of example 6.3.1, G ∼ = D2n . QED b. Let a, b are generators for Z and Z2 respectively (both will be writtend multiplicatively with 1 as the identity). Then by the definition of semidirect product, D∞ = (an , bt ) with the group law as: (an , bt )(am , bl ) = (an bt • am , bt+l ) Then it is obvious that D∞ =< (a, 1), (1, b) > and (1, b)2 = 1; (1, b)(a, 1)(1, b) = (a−1 , 1). By the UMP for presentation there is a unique homomorphismφ : G → D∞ for G =< r, s; s2 = 1, srs−1 = r−1 > that sends r to (a, 1) and s to (1, b). Since sr = r−1 s any word in G can be written uniquely in the form rn sm for m ≤ 1. Suppose 1 = φ(rn sm ) = φ(r)n φ(s)m = (am , bn ) then m = n = 0, so φ is injective. And the equation in the previous sentence also shows that φ is surjective. So φ is an isomorphism between groups and D∞ ∼ = G. Thus D∞ admits that presentation. QED c. We use the presentation proved in part (b). (sr)2 = srsr = r−1 r = 1 so sr has order 2. Furthermore, s(sr) = r and {s, r} generates D∞ , so does {s, sr}. Thus the infinite dihedral group is generated by two elements of order 2. QED d. Let A = {1, a} and B = {1, b} be two groups of order 2 (that is a2 = b2 = 1). From problem 4, the free product admits the presentation: A ∗ B =< a, b; a2 = b2 = 1 >. So by UMP for free product, there exist a unique homomorphism χ : A ∗ B → D∞ that sends a, the identification of a, to s and b, the identification of b, to sr (the homomorphism extends the obvious homomorphism from A, B to D∞ ). Let h = ab then it is obvious to check aha−1 = h−1 . By the UMP for presentation applied for D∞ , there is a unique homomorphism ζ : D∞ → A ∗ B that sends s to a and r to h. Let τ = ζχ then τ is an endomorphism on A ∗ B. Since τ (a) = a, τ (b) = ζ(sr) = ah = aab = b, τ is the identity map on the set of generators of A ∗ B. Hence, τ is the group identity map. Similarly, χζ is the group identity on D∞ . Therefore, each χ and ζ is an isomorphism between groups implying: G∼ = D∞ . QED e. So an element of D∞ is of the form (x1 ....xn ) with each xi is a nontrivial element in a group of order 2, A or B, and two consecutive terms belong to different groups. QED