Mathematics 331 Solutions to Some Review Problems for Exam 2 1. Write out all the even permutations in S 3. Solution. The six elements of S 3 are a =, b = 1 3 2 2 1 3 c =, d = 3 2 1 2 3 1 e =, f = 3 1 2 The three transpositions, a, b, c are odd permutations. Since half of the elements of S 3 are even, d, e, f must be even. One can also check the definition to see that d, e, f are in fact even. 2. Write out all the transpositions in S 4. Solution. If σ S 4 is a transposition, then σ interchanges two of the numbers 1, 2, 3, 4 and fixes the other two. So, to write one down one must select two of the four numbers to interchange. Thus, the transpositions are 4, 2 1 3 4 4, 4 2 3 1 4, 1 4 3 2 4 3 2 1 4 4 1 3 2 4 4. 1 2 4 3 3. Let G be a group and let g G. Define a function L g : G G by L g (x) = gx. Prove that L g is 1-1 and onto. Solution. To prove 1-1, suppose that x, y G with L g (x) = L g (y). By definition of L g, this means gx = gy. By cancellation, we get x = y. Thus, L g is 1-1. Next, let z G. To show L g is onto we want to find x G with z = L g (x). That is, we want to find x with z = gx. We can do this by setting x = g 1 z, since then gx = g(g 1 z) = (gg 1 )z = ez = z. Thus, L g is also onto. 1
4. With notation as in the previous problem, if g, h G, show that L g L h = L gh. Conclude that the map ϕ : G Perm(G) given by ϕ(g) = L g is a 1-1 group homomorphism from G to the group of all permutations of the set G. Solution. We proved in the previous problem that each L g is a 1-1, onto function from G to G, and so is an element of the group Perm(G) of all permutations of the set G. Let g, h G. For each x G we then have (L g L h )(x) = L g (L h (x)) = L g (hx) = g(hx) = (gh)x = L gh (x). Since this equation holds for all x G, we conclude that L g L h = L gh. Consequently, the function ϕ : G Perm(G) defined by ϕ(g) = L g is a group homomorphism, since for all g, h G, ϕ(gh) = L gh = L g L h = ϕ(g) ϕ(h). Finally, to see ϕ is 1-1, it is enough to show that ker(ϕ) = {e}. Since e ker(ϕ), it is enough to show that if g ker(ϕ), then g = e. To do this, suppose that ϕ(g) is the identity of Perm(G). Thus, L g is the identity function on G. That means for each x G we have L g (x) = x. Thus, gx = x = ex. By cancellation, g = e. 5. Let σ S n be the permutation that for which σ(i) = i + 1 if 1 i n 1 and σ(n) = 1. Prove that the order of σ is n. Solution. Not that it is needed, but the permutation σ, written in our usual notation, is 1 2 n 1 n. 2 3 n 1 To find the order m of σ, we recall that m is the smallest positive integer satisfying σ m = e, the identity of S n. Now, we have σ(1) = 2, and so σ 2 (1) = σ(σ(1)) = σ(2) = 3. Continuing this pattern, for small enough t, we have σ t (1) = t + 1. This works for n 1, where we ll have σ n 1 (1) = n. Thus, no power, from the first to the n 1st, of σ can be the identity because the identity sends 1 to 1. However, σ n (1) = σ(σ n 1 (1)) = σ(n) = 1. The same reasoning will show that σ n (r) = r for each r {1, 2,..., n}. Therefore, σ n is the identity. Since no smaller positive power of σ is the identity, the order of σ is n. 6. Let σ S n. Prove that A n σ = σa n ; in other words, prove that the right coset of σ is the same as the left coset of σ for the subgroup A n of S n. Solution. We proved this in class. The argument was to show that if σ A n, then σa n = A n σ = A n and, if σ A n, the σ is odd, and in that case σa n = A n σ = {odd permutations}. 2
7. Let G be a group and H a subgroup in which H has exactly two different cosets in G. Recall that H = He is always a coset. Let Ha be the other coset. Show that Hx = H if and only if x H, and Hx = Ha if and only if x / H. Solution. Recall that Hx = Hy iff xy 1 H, since the coset Hx is the equivalence class of the equivalence relation x y if xy 1 H. Now, x e iff xe 1 H iff x H. Thus, x H means x e, and so Hx = He = H. On the other hand, if Ha is a different coset from H, then a / H. Let x / H. Then Hx H from what we did above. Since there are only two cosets and Hx H, we must have Hx = Ha. Conversely, if Hx = Ha, then Hx H. Since Hx = H iff x H, we must have x / H. 8. Let G = Z 2 Z 2 be the direct product of the group (Z 2, +) with itself. Show that G is not cyclic but that each proper subgroup of G is cyclic. Solution. We proved this in class. For a different proof that each proper subgroup of G is cyclic, since G = 4 = 2 2, this part of the problem is a special case of # 10. 9. Let G, H be groups. If a G has order n and b H has order m, prove that (a, b) G H has order lcm(n, m), the least common multiple of n and m. Solution. Let r be the order of (a, b) and let l = lcm(n, m). Recall that r is the smallest positive integer satisfying (a, b) r = (e G, e H ), the identity of G H. Also, recall that l is the smallest positive integer that is a multiple of both n and m. Now, since the operation in G H is componentwise, (a, b) r = (a r, b r ). Thus, as (a r, b r ) = (e G, e H ), we get a r = e G and b r = e H. Thus, by what we know about the order of an element, the order of a divides r and the order of b divides r. This means both n and m divide r. Thus, r is a common multiple of n, m, and so l r. On the other hand, since a n = e G and b m = e H, because l is a multiple of both n and m, we see that a l = e G and b l = e H. Thus, (a, b) l = (e G, e H ). Because r is the smallest positive integer whose power of (a, b) gives the identity, r l. The two inequalities then give r = l, which means the order of (a, b) is the lcm of n and m. 10. Let G be a group of order p 2 for some prime p. Prove that each proper subgroup of G is cyclic. Solution. Let H be a proper subgroup of G. By Lagrange s theorem, H divides G = p 2. Since p is a prime, the only positive divisors of p 2 are 1, p, p 2. Now, H = p 2 forces H = G, which is false since H is proper. Thus, H = 1 or H = p. If H = 1, then H = {e} = e is cyclic. If H = p, then the order of H is prime, so H is cyclic by a corollary to Lagrange s theorem. 3
11. Let r, n be positive integers with r n. Let H = {σ S n : σ(r) = r}. Prove that H is a subgroup of S n. Solution. First of all, H is not empty because it contains the identity. Next, let σ, τ H. Then σ(r) = r and τ(r) = r. Thus, (σ τ)(r) = σ(τ(r)) = σ(r) = r. Thus, σ τ H. Therefore, H is closed under composition. Finally, let σ H. Then σ(r) = r. Since the inverse function σ 1 can be described by σ 1 (x) = y exactly when σ(y) = x, from σ(r) = r we see that σ 1 (r) = r. Thus, σ 1 H. Therefore, H is a subgroup of S n. 12. Let A be an n n matrix over R. If ϕ : R n R n is the matrix transformation given by ϕ(x) = AX, prove that ϕ is a group homomorphism. Also prove that ker(ϕ) = {0} if and only if det(a) 0. Solution. If X, Y R n, then ϕ(x + Y ) = A(X + Y ) = AX + AY = ϕ(x) + ϕ(y ), so ϕ is a group homomorphism. Recall that det(a) 0 iff A 1 exists. If this happens, then the only solution to AX = 0 is X = 0, since we can multiply both sides by A 1 to get this conclusion. Conversely, if the kernel is { 0}, then AX = 0 has only the trivial solution X = 0. One way to see that det(a) 0 is to recall from linear algebra that, viewing AX = 0 as a system of equations, the system has only the trivial solution iff A can be reduced to the identity matrix. And, when this happens, if we perform exactly the same row operations starting with the identity matrix, we obtain A 1. Thus, A 1 exists, and so det(a) 0. 13. Let v R 3, and let ϕ : R 3 R be given by ϕ(w) = v w, the dot product of v and w. Prove that ϕ is a group homomorphism and determine the kernel of ϕ. Solution. From Calculus 3 we have the identity v (w + u) = v w + v u. Thus, ϕ(w + u) = ϕ(w) + ϕ(u). Thus, ϕ is a group homomorphism. To determine the kernel, we can either use the formula for the dot product or call the fact that v w = v w cos(θ), where θ is the angle between v and w. To have v w = 0 we must have cos(θ) = 0 (or w = 0). Thus, w must be perpendicular to v. Therefore, ker(ϕ) is the plane perpendicular to v, at least when v is nonzero. If v = 0, then the kernel is all of R 3. 4
14. Let v R 3, and let ϕ : R 3 R 3 be given by ϕ(w) = v w, the cross product of v and w. Prove that ϕ is a group homomorphism and determine the kernel of ϕ. Solution. From Calculus 3 we have the identity v (w + u) = v w + v u for all u, v, w R 3. Thus, ϕ(w + u) = ϕ(w) + ϕ(u). Thus, ϕ is a group homomorphism. To determine the kernel, we can either use the formula for the cross product or recall the fact that v w = v w sin(θ), where θ is the angle between v and w. If w ker(ϕ), then v w = 0, so its norm is 0. This forces the angle between v and w to be 0 (or w = 0), which means w is a multiple of v. Therefore, ker(ϕ) = {αv : α R}, when v is nonzero. If v = 0, then the kernel is all of R 3. Note: There was a typo in the problem: the map ϕ maps to R 3, not to.r. 15. Define ϕ : G G by ϕ(g) = g 1. If G is Abelian, prove that ϕ is a group homomorphism. Also, find an example of a non-abelian group G such that ϕ is not a group homomorphism. Solution. Suppose that G is Abelian. If g, h G, then ϕ(gh) = (gh) 1 = h 1 g 1 = g 1 h 1 = ϕ(g)ϕ(h), the second to last equality because G is Abelian (e.g., commutative). Therefore, ϕ is a group homomorphism. Consider G = S 3. We claim that ϕ isn t a group homomorphism in this case. For, let a = ( 1 3 2 ) (, b = 2 1 3 ). Then so ab = ( 3 1 2 ϕ(ab) = (ab) 1 = ), ( 2 3 1 ). Note that ab (ab) 1. On the other hand, as a, b are both transpositions, a 1 = a and b 1 = b. Thus, ϕ(a)ϕ(b) = a 1 b 1 ab, so ϕ(ab) ϕ(a)ϕ(b). 16. Let G and H be groups. Define π : G H G by π(g, h) = g. Prove that π is a group homomorphism and determine its kernel. 5
Solution. Let x, x G H. Then we may write x = (g, h) and x = (g, h ) for some g, g G and h, h H. Note that π(x) = g and π(x ) = g. Thus, π(xy) = π((g, h)(g, h )) = π(gg, hh ) = gg = π(g, h)π(g, h ). Therefore, π is a group homomorphism. We have ker(π) = {(g, h) G H : π(g, h) = e G } = {(g, h) G H : g = e G } = {e G } H. 17. Let G and H be groups. Define ϕ : G G H by ϕ(g) = (g, e H ). Prove that ϕ is a group homomorphism. Solution. Let g, g G. Then ϕ(gg ) = (gg, e H ) = (g, e H )(g, e H ) = ϕ(g)ϕ(g ) since the operation on G H is componentwise. Thus, ϕ is a group homomorphism. 18. Let n be a positive integer. Define ϕ : S n S n+1 by defining, for σ S n, ϕ(σ) to be the permutation satisfying { σ(r) if r n ϕ(σ)(r) = n + 1 if r = n + 1. Prove that ϕ(σ) is indeed an element of S n+1 and that ϕ is a group homomorphism. Solution. Let σ S n. Then ϕ(σ) maps {1, 2,..., n + 1} to itself. Since the image of σ is {1, 2,..., n}, the formula for ϕ(σ) shows that each number in {1, 2,..., n + 1} is in the image. Thus, ϕ(σ) is onto. To see it is 1-1, if ϕ(σ)(r) = ϕ(σ)(s), then if this common value is n+1, then r = s = n+1. Otherwise, we must have r, s {1, 2,..., n}. Thus, ϕ(σ)(r) = σ(r) and ϕ(σ)(s) = σ(s). Thus, σ(r) = σ(s). Since σ is 1-1, we get r = s. Thus, ϕ(σ) is both 1-1 and onto, and so is a permutation. Thus, ϕ(σ) S n+1, and so ϕ is a function from S n to S n+1. To see that ϕ is a homomorphism, we need to show, for all σ, τ S n, that ϕ(σ τ) = ϕ(σ) ϕ(τ). To do this, since both sides of this equation are functions from {1, 2,..., n + 1} to itself, we need to check that they have the same function value for each domain value. First, suppose 1 r n. Then ϕ(σ τ)(r) = (σ τ)(r) = σ(τ(r)) since ϕ(ρ)(r) = ρ(r) for each r n. Similarly, ϕ(σ)(r) = σ(r) and ϕ(τ)(r) = τ(r). Thus, (ϕ(σ) ϕ(τ))(r) = σ(τ(r)). 6
On the other hand, if r = n + 1, then both ϕ(σ τ) and ϕ(σ) ϕ(τ) map n + 1 to n + 1. Therefore, these two functions agree at each domain value, so ϕ(σ τ) = ϕ(σ) ϕ(τ). Thus, ϕ is a group homomorphism. 19. Let G = a be a cyclic group of order n. Define ϕ : G S 1 by ϕ(a r ) = e 2πir n. Prove that ϕ is well defined; that is, if a r = a s, then ϕ(a r ) = ϕ(a s ). Then show that ϕ is a group homomorphism. Solution. The issue of well defined is something we ll talk about more in Section 14. Suppose that a r = a s. Then, by what we proved about finite cyclic groups in Section 6, we have r s (mod n), which means r = s + qn for some integer q. Then e 2πis n = e 2πi(r+qn) n = e 2πir n e 2πi = e 2πir n. Thereore, ϕ is a (well-defined) function. To see it is a homomorphism, if x, y G, we may write x = a k and y = a l for some integers k, l. Then ϕ(xy) = ϕ(a k a l ) = ϕ(a k+l ) = e 2πi(k+l) n = e 2πik+2πil n = ϕ(x)ϕ(y). Thus, ϕ is a group homomorphism. = e 2πik 2πil n e n = ϕ(a k )ϕ(a l ) 20. Let ϕ : G G be an onto group homomorphism. If G is cyclic, prove that G is cyclic. Solution. Since G is cyclic, there is a G with G = a. By definition, this means G = {a n : n Z}. Since ϕ is a homomorphism, we have ϕ(a n ) = ϕ(a) n for each n Z. Therefore, ϕ[g] = {ϕ(g) : g G} = {ϕ(a n ) : n Z} = {ϕ(a) n : n Z}. This means the image ϕ[g] is the cyclic group generated by ϕ(a). Since ϕ is onto, G = ϕ[g]. Thus, G is cyclic (and is generated by ϕ(a). 21. Let ϕ : G G be an onto group homomorphism. If G is Abelian, prove that G is Abelian. Solution. Let x, y G. Since ϕ is onto, we may write x = ϕ(g) and y = ϕ(h) for some g, h G. Then, as G is Abelian and ϕ is a homomorphism, we have Thus, G is Abelian. xy = ϕ(g)ϕ(h) = ϕ(gh) = ϕ(hg) = ϕ(h)ϕ(g) = yx. 7
22. Let G be a group with more than 1 element in which G has no subgroups other than G and {e}. Prove that G is a finite group of prime order. Solution. Let a G with a e; such an element exists since G has more than 1 element. Because a is a subgroup of G and contains both a, e, it must be equal to G. If G is infinite, then there is no power of a equal to e except for a 0. Consequently, the subgroup a 2 = {a 2n : n Z} is a proper subgroup of G else a lies in it, which would mean a = a 2n for some n, and so a 2n 1 = e, meaning some nonzero power of a is e. This proves that if G is infinite, it has at least one nontrivial subgroup a 2 (as this subgroup isn t G, and isn t {e} else a 2 = e). Thus, G must be finite. Finally, suppose that the order of G is n, and suppose that n isn t prime. Then we can factor n = ml for some integers m, l with 1 < m, l < n. Because the order of a is n, the order of a l is m (a fact from Section 6). Thus, a l has m elements, fewer than that of G. This is then a subgroup other than G or {e}. Since no such subgroup can exist, the assumption that n isn t prime is false. Thus, n = G is prime. 8