Math 210A: Algebra, Homework 6


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1 Math 210A: Algebra, Homework 6 Ian Coley November 13, 2013 Problem 1 For every two nonzero integers n and m construct an exact sequence For which n and m is the sequence split? 0 Z/nZ Z/mnZ Z/mZ 0 Let α : Z/nZ Z/mnZ be defined by α(1 = m Since 1 = Z/nZ, this determines the entire homomorphism Let x ker α Then mx 0, which implies mn mx so n x But since 0 x < n, we must have x = 0 Therefore the kernel of α is trivial, so α is an injection Since m mn, by the fundamental theorem on cyclic groups Z/mnZ has a subgroup of order m isomorphic to Z/mZ Let β : Z/mnZ Z/mZ be defined by projection onto that subgroup, which is a surjective homomorphism Further, let x Z/nZ Them α(x = mx and β(mx = 0, so im α ker β Since im α = ker β = n, im α = ker β Thus this is an exact sequence Recall that a sequence is split if and only if the middle term is the direct sum of the outer terms In our case, we need Z/mnZ = Z/mZ Z/nZ By the Chinese Remainder Theorem, this is true if and only if (m, n = 1 Problem 2 Let N be the normal subgroup in G H generated by G G H Prove that (G H/N = H Consider the projection π : G H H given by the following: for a word w = g 1 h 1 g n h n, let π(w = h 1 h n We have π(ε = π(e H = e H, where ε is properly the empty word but representable by (reducible e H (and by earlier considerations any group homomorphism is consistent on any representative for a word Now let w, w G H where w is as above and w = g 1h 1 g nh n Then π(ww = h 1 h n h 1 h n = π(wπ(w so indeed π is a group homomorphism π is clearly surjective since π(h = h for any h H We claim that ker π = N If so, then (G H/N = (G H/ ker π = im π = H, so we would be done We prove double inclusion 1
2 If w ker π, then removing all g G from w must yield ε Therefore for every instance of h in w, there must be h 1 in the form hgh 1 Therefore w is of the form w = h 1 g 1 h 1 1 h n g n h 1 n where some of the h i may be e H By Problem 8, we know the normal subgroup generated by G is generated by elements of the form hgh 1, so w N Therefore ker π N Now suppose w is a generator of N, so of the form w = vgv 1 for some word v G H and g G Then π(vgv 1 = π(vπ(gπ(v 1 = e H since π(g = e H Therefore since every generator of N is contained in ker π, N ker π as well Therefore N = ker π and we are done Problem 3 Let D = Z Z/2Z with respect to the (unique isomorphism Z/2Z Aut Z Prove that D = Z/2Z Z/2Z The nonidentity automorphism of Z is given by f(x = x element of Z/2Z Then let a = (1, τ and b = (0, τ Let τ be the nonidentity Then an element n(a + b = (n, 0 and n(a + b + b = (n, τ Further, a + a = (0, 0 and b+b = (0, 0 Therefore we let ϕ : Z/2Z Z/2Z Z Z/2Z be defined by letting α (1, τ and β (0, τ, where α and β are the nonidentity letters of the free product Then by the above reasoning, ϕ is surjective Further, since we have established ϕ(αβ = (1, 0 (0, 0 and ϕ(βα = ( 1, 0 (0, 0 (and ϕ(α, ϕ(β 0, the only word that maps to (0, 0 is the empty word Thus ϕ is an isomorphism, and we are done Problem 4 Show that there exists a surjective homomorphism Z/nZ Z/2Z S n Recall that S n is generated by an σ = (1 2 n and τ = (1 2 Therefore any ρ S n can be written in the form ρ = σ k 1 τ σ km τ where perhaps this last τ is omitted or another τ added at the beginning This form is sufficient without loss of generality Let {0, 1,, n 1} = Z/nZ and {0, a} = Z/2Z Define a map ϕ : Z/nZ Z/2Z S n by 1 σ and a τ Then ϕ(k 1 a k m a = ϕ(k 1 ϕ(a ϕ(k m ϕ(a = ϕ(1 k 1 τ ϕ(1 km τ = σ k 1 τ σ km τ It is clear by construction that ϕ is a group homomorphism, and we have just shown it is surjective This competes the proof 2
3 Problem 5 Prove that the group SL 2 (Z is generated by two matrices ( ( and Elements of SL 2 (Z are of the form ( a b c d such that ad bc = 1 Let α and β be the matrices above, respectively We can see that α 2 = I Further, ( ( n α 3 β = and (α 3 β n = where n Z makes sense since α 1, β 1 SL 2 (Z Further, ( 1 n (α 3 β n α 2 = 0 1 so we can construct all matrices of the form ( ±1 n 0 ±1 for n Z Now, taking matrices of the form (α 3 β n α and (α 3 β n α 3 we can construct matrices of the form ( n ±1 1 0 Taking inverses of the above constructions yields ( ( 0 ±1 ±1 0 and 1 n n ±1 Problem 6 Let H and K be two subgroups of G Assume that G acts on a set X and there are two subsets A, B X and an element x X \ (A B such that h(a {x} B for every e h H and k(b {x} A for every e k K Prove that the natural homomorphism H K G is injective Let w = h 1 k 1 h n k n a word in H K and let ϕ be the natural homomorphism from H K to G Suppose that ϕ(w = e G Note that for arbitrary h H, k K, we have (hk x = h (k x B and (kh x = k (h x A, where the action of H K on X is induced by the action of G In particular, w x B since the first letter of w is in H (the case is identical if the first letter is in K But e G x = x / B, so w cannot have a first letter, ie w is the empty word Therefore if ϕ(w = e G, then w = ε the identity in H K Therefore ϕ is injective 3
4 Problem 7 Let G and H be two nontrivial groups Show that G H is an infinite group with trivial centre Let g G and h H be two nonidentity elements Then each of the words gh }{{} gh n times is distinct Therefore G H is infinite Further, suppose that w = g 1 h 1 g n h n is in the centre of G H (without loss of generality; the case where h 1 comes first is identical Then for nontrivial h H, hw = hg 1 h 1 g n h n g 1 h 1 g n h n h = wh since, in particular, hw has n + 1 letters in H and wh has only n Even if w = g, we have hw = hg gh = wh (the case similar for h Therefore the centre of G H is trivial Problem 8 Let X be a subset in a group G Prove that X = Y where Y = g G gxg 1 We prove double inclusion Let N = X First, let gxg 1 Y Then since x N and N is closed under conjugation, we must have gxg 1 N as well Therefore since Y N, Y N We claim now that Y is normal by proving its generators are closed under conjugation Indeed, if gxg 1 Y, then hgxg 1 h 1 = (hgx(hg 1 Y as well Since N is the smallest normal subgroup containing X and Y is one such subgroup, we have N Y Therefore N = Y, so we are done Problem 9 Let G = a, b : w 3 = e, w a word in a, b Show that G is finite and find G We prove this in two steps First we claim that [a, b] Z(G We see a 2 ba 2 b 2 = a 2 (ab 2 ab 2 ab 2 b 2 aba = b 2 ab 2 ababa = b 2 ab = b 2 ab 2 a 2 ba 2 ba 2 = a 2 b(b 2 ab 2 ab 2 aaba 2 ba 2 = ab 2 aba 2 = ab 2 (bababaaba 2 = aba 2 (abababb 2 a = aba 2 b 2 a so that [a, b] commutes with a, the case for b being analogous Since it commutes with the generators of G, it commutes with every w G We additionally have ab = [a, b]ba, so ab = ba[a, b] We proceed by a combinatorial argument Through these insertions and commuting ab with ba, we can reduce any word to the form [a, b] i a j b k for some i, j, k N Then since w 3 = e, we have 0 i, j, k 2 Therefore there are 27 elements of G, and we are done 4
5 Problem 10 Prove that if the free groups F (X and F (Y for finite sets X and Y are isomorphic, then X = Y Suppose that X Y, so we may assume X < Y Let ϕ : F (X F (Y be a group homomorphism, and suppose it is an isomorphism Then there exists a letter y Y such that ϕ 1 (y = w has at least two letters Suppose that w = ab, a, b X (the case of w with n letters being similar Then y = ϕ(ab = ϕ(aϕ(b = a b where a, b Y must be proper letters (or words since ϕ is an isomorphism Since a b has at least two letters, y = a b is impossible Therefore ϕ is not an isomorphism Therefore we have proved the contrapositive, and we are done 5
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