3.2.8 Prove that if H and K are finite subgroups of G whose orders are relatively prime then H K = 1.

Transcription

1 Prove that if G/Z(G) is cyclic then G is abelian. If G/Z(G) is cyclic with generator xz(g). Every element in G/Z(G) can be written as x k z for some k Z and z Z(G). Now, let g, h G, then g = x a z and h = x b w for z, w Z(G). We have gh = x a zx b w = x a+b zw = x b+a wz = x b wx a z = hg Show that if G = pq for some primes p, q (not necessarily distinct), then either G is abelian or Z(G) = 1. If G = pq and Z(G) 1, we have G/Z(G) = p, q or 1. If G/Z(G) = 1, then G = Z(G), we are done. If G/Z(G) = p or q, G/Z(G) is cyclic. This implies G is abelian by Exercise Prove that if H and K are finite subgroups of G whose orders are relatively prime then H K = 1. Since H K H and H K K and HK = H K H K. This implies that H K H, H K K. Since H and K are relatively prime, we have H K = 1, hence H K = Suppose H and K are subgroups of finite index in the (possible infinite) group G with G : H = m and G : K = n. Prove that l.c.m.(m, n) G : H K mn. Deduce that if m and n are relatively prime then G : H K = G : H G : K. Note that H K is a subgroup of H and K respectively. So there are a, b Z such that H K divides a H + b K. This implies that H K divides a G m mn + b G m. i.e. H K mn a G n + b G m an + bm g.c.d.(m, n) 1 divides = G = G = G, which implies that mn mn mn l.c.m.(m, n) G H K. Thus, we have l.c.m.(m, n) G : H K. l.c.m.(m, n) On the other hand, since H K H K = HK G by Proposition 13. We have H H K G K = n. Thus, G : H K G H H H K = m H = mn. Therefore, we have H K l.c.m(m, n) G : H K mn Now, if g.c.d.(m, n) = 1, l.c.m.(m, n) = mn. The above inequality becomes mn G : H K mn. Hence, G : H K = mn = G : H G : K Let G be a finite group, let H be a subgroup of G and let N G. Prove that if H and G : N are relatively prime then H N. Let f : G G/N with kernel ker f = N. Consider f H : H H/N, then Imf H G/N and Imf H H (image divides both domain and codomain). But G/N and H are relatively prime. So Im f H = 1 which implies that H N = ker f Prove that if N is a normal subgroup of the finite group G and ( N, G : N ) = 1 then N is the unique subgroup of G of order N. 1

2 Let H G with H = N. Then ( H, G : N ) = ( N, G : N ) = 1. By H N but H = N. So H = N If A is an abelian group with A G and B is any subgroup of G prove that A B AB. For g AB, A B. So x A and x B. Want to show that gxg 1 A B. For g AB, write g = ab for a A and b B. Then (ab)x(ab) 1 = abxb 1 a 1. Since x A, A G and B G, we have bxb 1 A which implies that abxb 1 a 1 A. Furthermore, since bxb 1 A and A abelian, x B, we have a(bxb 1 )a 1 = aa 1 (bxb 1 ) = bxb 1 B. Thus, gxg 1 A B for all g AB, i.e. A B AB Prove that if H is a normal subgroup of G of prime index p then for all K G either (i) K H or (ii) G = HK and K : K H = 1. Since H G, we have H HK G. So p = G : H = G : HK HK : H. Thus, either HK : H = 1 or p. In the first case, HK : H = 1 implies that HK = H, this gives K H. Otherwise, if HK : H = p, then HK : H = K : K H = p Let M and N be normal subgroups of G such that G = M N. Prove that G/M N = (G/M) (G/N). Consider MN M N M N 1 Define f : G G/M G/N by g (gm, gn). f is a homomorphism since f(g 1 g ) = (g 1 g M, g 1 g N) = (g 1 Mg M, g 1 Ng N) = (g 1 M, g 1 N)(g M, g N) = f(g 1 )f(g ). Also ker f = M N as follows: for x ker f, f(x) = (xm, xn) = (M, N). Hence x M and x N. On the other hand, for x M N, x M and x N, clearly x ker f since f(x) = (M, N). The result follows from the first isomorphism theorem Prove that G is an abelian simple group then G = Z p for some prime p (do not assume G is a finite group). Suppose that G is infinite. Let x 1 G be an element of infinite order. Then x is a subgroup of G and actually it must be G since G is simple. So x is a normal subgroup of G = x (with index ). This is a contradiction since G is normal. Now, suppose that G is finite. If the order of G is prime, then G = Z p. If G is not prime, say G = pm for some m. By Cauchy s Theorem, G has an element of order p, but this implies

3 x is a subgroup of G which is a contradiction. Hence, G = Z p by Corollary 10. In fact, every subgroup of an abelian group is normal. But G is simple, so G has no proper subgroup. So G must be prime, hence G = Z p Prove that subgroups and quotient groups of a solvable groups are solvable. Let G be a solvable group and H G a subgroup. Since G is solvable. There is a chain of subgroups 1 = G 0 G 1 G n = G such that G i+1 /G i is abelian. Then it is clear that H G i H G i+1. And in fact H G i H G i+1. Since for x H G i and g H G i+1 we have gxg 1 H. And gxg 1 G i because G i G i+1. Also it is clear that H G i+1 /H G i is abelian. Thus H is solvable. Similar way shows that G/H is solvable. Since G i /H G i+1 /H and in fact G i /H G i+1 /H. And (G i+1 /H)/(G i /H) is abelian If G is a finite group and H G prove that there is a composition series of G, one of whose terms is H. By Theorem (Jordan-Holder Theorem), every finite group has a composition series. So we have 1 = H 0 H 1 H n = H and 1 = K 0 /H K 1 /H K m /H = G/H By Lattice Isomorphism Theorem, this implies that Therefore, H = K 0 K 1 K m = G 1 = H 0 H 1 H K 1 K m = G Prove that if H is a nontrivial normal subgroup of the solvable group G then there is a nontrivial subgroup A of H with A G and A is abelian Prove (without using the Feit-Thompson Theorem) that the following are equivalent: (i) every group of odd order is solvable. (ii) the only simple groups of odd order are those of prime order. (i) (ii) Let G be a simple group of odd order. Suppose that G = n = pm for some m 1. If G is abelian, there exists an element x 1 G of order p, i.e. x G, a contradiction. If G is non-abelian, since G is solvable and every finite group has a composition series. So we have 1 = H 0 H 1 H n = G By Exercise 1, the length of the composition series must be at least. So H n 1 1. But this implies that H i 1 G, a contradiction since G is simple. Therefore, the only simple groups of 3

4 odd order are those of prime order. (ii) (i) Prove that the (unique) subgroup of order 4 in A 4 is normal and is isomorphic to V 4. As shown in the Figure 8 on page 111, the unique subgroup of order 4 in A 4 is H = (1)(34), (13)(4). Let σ = (1)(34) and τ = (13)(4). Since στ = (1)(34)(13)(4) = (14)(3) and all three elements σ, τ, στ have order. So H = V 4. Now, show that H is normal. Only need to check on the generators σ, τ and only need to check that gσg 1 V 4 and gτg 1 V 4 for g all three cycles containing 3. i.e. (13)(1)(34)(13) = (14)(3) V 4 and Thus, H G. (13)(13)(4)(13) = (1)(34) V 4 (134)(1)(34)(143) = (14)(3) V 4 (134)(13)(4)(143) = (1)(34) V Find a composition series for A 4. Deduce that A 4 is solvable. Consider the chain of subgroups 1 (1)(34) V 4 A 4 Note that (1)(34) V 4 because of index. So each subgroup is a normal subgroup of the group on the right-hand side. And we just proved in previous exercise that V 4 A 4. Furthermore, the composition factors are S 4 /A 4 = Z A 4 /V 4 = Z3 V 4 / (1)(34) = Z (1)(34) /1 = Z are all simple. So this is a composition series, hence A 4 is solvable Let G act on the set A. Prove that if a, b A and b = g a for some g G, then G b = gg a g 1 (G a is the stabilizer of a). Deduce that if G acts transitively on A then the kernel of the action is g G gg a g 1. For x G b, x b = b. So g 1 xg a = g 1 xb = g 1 b = a. Hence g 1 xg G a which implies that x gg a g 1. On the other hand, for x G a we have gxg 1 b = gxg 1 ga = gx a = g a = b. Thus, G b = gg a g 1. Let K be the kernel of the action, i.e. K = {g G g a = a for all a A}. Show that K = g G gg a g 1. For x K, x a = a. Since g 1 xg a = g 1 g a = a for all g G. We have g 1 xg G a for all g G. Hence, x gg a g 1 for all g G. i.e. x g G gg a g 1. On the other hand, let x g G gg a g 1. Then x gg a g 1 for all g G. So for any b A we have x b = ghg 1 b for some h G a (i.e. h a = a). But G acts transitively on A, 4

5 i.e. g a = b. Thus, x b = ghg 1 b = gh a = g a = b. So x K. Therefore, K = g G gg a g Let G be a permutation group on the set A (i.e., G S A ), let σ G and let a A. Prove that σg a σ 1 = G σ(a). Deduce that if G acts transitively on A then σ G σg a σ 1 = 1. Let x σg a σ 1. And let x = σgσ 1 for g G a. Then x σ(a) = σgσ 1 σ(a) = σg a = σ(a). Hence x G σ(a). On the other hand, let x G σ(a). So x σ(a) = σ(a). We have σ 1 xσ a = σ 1 (σ(a)) = a. Hence, σ 1 xσ G a, i.e., x σg a σ 1 Therefore σg a σ 1 = G σ(a). Now, if G acts transitively on A, by previous exercise the kernel is g G ggg 1. Let ϕ : G S A be the homomorphism associated to the action. Since G S A, ϕ is injective, the kernel is Assume that G is an abelian, transitive subgroup of S A. Show that σ(a) a for all σ G {1} and all a A. Deduce that G = A. [use the preceding exercise.] Suppose that σ(a) = a for some a A and some σ G. This implies that G a 1 hence σ G σg aσ 1 1, a contradiction to the result in preceding exercise. For a fixed a A, and any b A, there is σ G such that σ(a) = b. To show this σ is unique, suppose there is τ G such that σ(a) = τ(a) = b. Then τ 1 σ(a) = a, so τ 1 σ G a = 1. Thus, σ = τ. Hence we have an injective map ϕ a : A G sending a to σ. Hence A G. Similarly, if we define ψ : G A by ψ(σ) = σ a. Then ψ is injective since σ(a) = τ(a) implies that τ 1 σ G a = 1 which again implies thatσ = τ. Therefore, G = A Prove that if H has finite set index n then there is a normal subgroup K of G with K H and G : K n!. Let G act by left multiplication on the set A of left cosets (i.e. G/H). And let π H be the associated permutation representation afforded by this action π H : G S A By Theorem 3, K = ker π H = x G xhx 1 and K is the normal subgroup of G contained in H. By first isomorphism theorem, G/K = Imπ H S A. Since H has index n, S A = n!. Hence, we have [G : K] = G K S A = n! Prove that if p is a prime and G is a group of order p α for some α Z +, then every subgroup of index p is normal in G. Deduce that every group of order p has a normal subgroup of order p. Since G = p α. Since p is the smallest prime dividing G. By Corollary 5, any subgroup of index p is normal. In particular, let α =, then G = p. By Cauchy s Theorem, G has an element x of order p. So H = x and [G : H] = p. H has index p, hence H is normal Let G be a finite group and let π : G S G be the left regular representation. Prove that if x is an element of G of order n and G = mn, then π(x) is a product of m n-cycles. Deduce that π(x) is an odd permutation if and only if x is even 5

6 and G x is odd. The permutation representation π : G S G is defined by π(x) = σ x and for each g G, the permutation of G is σ x (g) = x g. Let H = x. Since H is cyclic. For each g G, the stabilizer of g in H is 1 (G H (x) = 1), so the action is faithful (i.e. the kernel of the action is 1). So each orbit of H in G has order n and is the cycle containing g in the decomposition of π(x). i.e. for each g, Hg = (g xg x g x n 1 g) Since G = mn, G = Hg 1 Hg m. Hence the image of x of left regular representation is a product of m of n-cycles. Now, if x is even and G x is odd, by Proposition 5 π(x) is an odd permutation. Conversely, if π(x) is odd and x is odd, the number of cycles of even length in the decomposition is zero, which is even. By Proposition 5 again, π(x) is even, a contradiction Let G and π be as in the preceding exercise. Prove that if π(g) contains an odd permutation then G has a subgroup of index. [use Exercise 3 in Section 3.3] Prove that if G = k where k is odd then G has a subgroup of index. [Use Cauchy s Theorem to produce an element of order and then use the preceding two exercises.] By Cauchy s Theorem, there is an element x G of order, so x =. By Exercise 11, π(x) is an odd permutation, then by Exercise 1 G has a subgroup of index Let G be a finite group of composite order n with the property that G has a subgroup of order k for each positive integer k dividing n. Prove that G is not simple. Let p be the smallest prime dividing n. So n = pm for some m Z. By assumption, G has a subgroup H of order m. Then H has index p, hence is normal by Corollary 5. Therefore G is not simple Prove that if S G and g G then gn G (S)g 1 = N G (gsg 1 ) and gc G (S)g 1 = C G (gsg 1 ). (1) gn G (S)g 1 = N G (gsg 1 ) ( ) Let gag 1 N G (gsg 1 ), for a N G (S) = {x G xgsg 1 x 1 = gsg 1 }. We have (gag 1 )gsg 1 (gag 1 ) 1 = gag 1 gsg 1 ga 1 g 1 = gasa 1 g 1 = gsg 1 since asa 1 = S. So gag 1 N G (gsg 1 ). ( ) Let x N G (gsg 1 ). Then xgsg 1 x 1 = gsg 1. This implies that g 1 xgsg 1 x 1 g = S, i.e. (g 1 xg)s(g 1 xg) 1 = S. Thus g 1 xg N G (S), so x gn G (S)g 1. () gc G (S)g 1 = C G (gsg 1 ) 6

7 ( ) Let gag 1 gc G (S)g 1, for some a C G (S) = {x G xsx 1 = s for all x S}. Then for any s S we have (gag 1 )gsg 1 (gag 1 ) 1 = gag 1 gsg 1 ga 1 g 1 = gasa 1 g 1 = gsg 1 since asa 1 = s. Thus, gag 1 N G (gsg 1 ). ( ) Let x C G (gsg 1 ). Then xgsg 1 x 1 = gsg 1 for all s S. This implies that g 1 xgsg 1 x 1 g = s, i.e. (g 1 xg)s(g 1 xg) 1 = s for all s S. Thus g 1 xg C G (S), so x gc G (S)g If the center of G is of index n, prove that every conjugacy class has at most n elements. By Orbit-Stabilizer Theorem [G : G g ][G g : Z(G)] = [G : Z(G)] = n, we see that [G : G g ] divides n. When the action is conjugation, the number of conjugates of g is precisely G : C G (g). Thus, G : C G (g) divides n, which implies that every conjugacy class has at most n elements Assume G is a non-abelian group of order 15. Prove that Z(G) = 1. Use the fact that g C G (g) for all g G to show that there is at most one possible class equation for G. [Use Exercise 36, Section 3.1] Center Z(G) is a subgroup of G. Since G is non-abelian, Z(G) can only be 1, 3, 5. Suppose Z(G) = 5. By Orbit- Stabilizer Theorem [G : G g ][G g : Z(G)] = [G : Z(G)] = 15/5 = 3. So the number of conjugates of any non-center element g G, i.e. G : C G (g), must divides 3. And it must be 3, otherwise g would be in the center. However, by Class Equation, r G = Z(G) + G : C G (g i ) we have r i=1 G : C G(g i ) = 15 5 = 10 which is not divisible by 3, a contradiction. A quicker way: suppose Z(G) is prime, by Exercise 36 in Section 3.1, G/Z(G) is cyclic, so G is abelian, a contradiction. Hence Z(G) = 1. Now, since g C G (g) for all g G (Exercise 6 in Section.). Let g G be any non-identity element in G. Since Z(G) = 1 g C G (g) G. C G (g) can only be 3 or 5. Since G = 15, by Lagrange Theorem, we must have g = C G (g). So the number of conjugates in every conjugacy class is 3 or 5. By Class Equation, r i=1 G : C G(g i ) = 15 1 = 14. Thus, we have the relation 3a + 5b = 14 where a denotes the number of conjugacy classes of order 3 and b denotes the number of conjugacy classes of order 5. The only one solution is a = 3 and b = 1. Therefore, there is at most one possible class equation for G. i.e. G = 15 = Find all finite groups which have exactly two conjugacy classes. Every element in an abelian group is a conjugacy of its own. So abelian groups having exactly two conjugacy classes must have two element hence is isomorphic to Z. Now, let G be nonabelian group. Note that the identity in G forms a conjugacy class of its own. So Z(G) = 1. Since we assume that G has exactly two conjugacy classes, there is an element g Z(G) such that G = 1 + G : C G (g) 7 i=1

8 by class equation. And G : C G (g) is the number of elements in the other conjugacy class. So we have G 1 = G : C G (g) this implies that the number of conjugates in the other conjugacy class divides G 1. But this happens only when G =. Hence G is isomorphic to Z Recall that a proper subgroup M of G is called maximal if whenever M H G, either H = M or H = G. Prove that if M is a maximal subgroup of G then either N G (M) = M or N G (M) = G. Deduce that if M is a maximal subgroup of G that is not normal in G then the number of non-identity elements of G that are contained in conjugates of M is at most ( M 1) G : M. Since M N G (M) G by Exercise 6 in Section. and the fact that N G (M) is a subgroup of G. But M is a maximal subgroup of G, so it is clear that either N G (M) = M or N G (M) = G. If M is not normal, i.e. N G (M) G, then M = N G (M). So we have G N G (M) = G M By Proposition 6, the number of conjugates of M, gmg 1, is [G : N G (M)]. And each conjugate of M has cardinality gmg 1 = M. The largest number of non-identity elements in the conjugates of M occurs when these conjugates intersect trivially. Hence, the number of nonidentity elements of G that are contained in the conjugates of M is at most ( M 1) G : M Assume H is a proper subgroup of the finite group G. Prove that G g G ghg 1, i.e. G is not the union of the conjugates of any proper subgroup. [Put H in some maximal subgroup and use the preceding exercise.] Let M be a maximal subgroup of G containing H. If M is normal in G, we have g G ghg 1 g G gmg 1 = M G. If M is not normal in G, we still have g G ghg 1 g G gmg 1. By previous exercise, we know that g G gmg 1 contains at most ( M 1) G : M non-identity elements of G. Hence, since G M > 1. g G ghg 1 ( M 1) G : M = G G M + 1 < G < Let g 1, g,, g r be representatives of conjugacy classes of the finite group G and assume these elements pairwise commute. Prove that G is abelian. If g 1, g,, g r are pairwise commute, then g i g j = g j g i which implies that g j g i gj 1 = g i. Thus, for each i, we have g j C G (g i ) for all j = 1,, r, so C(g i ) r. By Class Equation, G is the sum of all distinct conjugacy classes. Each conjugacy class has G : C G (g i ) elements, and there are r of them. Hence G = r G : C G (g i ) = i=1 r i=1 8 G r C G (g i ) r=1 G r = G

9 This implies that C G (g i ) = r. i.e. C G (g i ) = {g 1, g,, g r }. We claim that G = C G (g i ). G C G (g i ) is clear. And every element x G is in the centralizer of itself, hence G C G (g i ). Therefore, G = C G (g i ), hence G is abelian Let p be a prime and let G be a group of order p α. Prove that G has a subgroup of order p β, for every β with 0 β α. [Use Theorem 8 and induction on α.] Suppose G = p, then G has a subgroup order 1 and p, so this is clear. Suppose any group of order p a has a subgroup of order p b for every b with 0 b a. Let G be a group of order G = p a+1, want to prove that G has a subgroup of order p b for every 0 b a + 1. By Theorem 8, Z(G) 1. Then H = Z(G) is a normal subgroup of G. For any subgroup A of G containing H, we have the following correspondence: G/H G A/H A 1 H Since G/H = p c for some c with 0 c b. By induction hypothesis, G/H has subgroup of order p b for every b with 0 b c. Hence by Lattice Isomorphism Theorem, G has a subgroup of order p b for every 0 b n If G is a group of odd order, prove for any nonidentity element x G that x and x 1 are not conjugate in G. First, note that x x 1. since if x = x 1, then x = 1 and x =, a contradiction since G is even and cannot have an element of even order. Suppose x and x 1 are conjugate, want to show that C G (x) has even order, a contradiction. By definition, C G (x) = {g G gxg 1 = x} Let G be a group of order 03. Prove that if H is a normal subgroup of order 7 in G then H Z(G). Deduce that G is abelian in this case. G = 03 = 7 9. Since H is normal, by Corollary 13 G/C G (H) is isomorphic to a subgroup of Aut(H). And Aut(H) = 6, so G/C G (H) must divide 6. Since G = 7 9, the only possibility is G/C G (H) = 1. Hence G = C G (H). i.e. H Z(G). So H divides Z(G). Since P = 7, we have Z(G) = 7 or 7 9. This implies that G/Z(G) = 1 or 9, hence G/Z(G) is cyclic. Therefore, G is abelian Prove that a group of order 56 has a normal Sylow p-subgroup for some prime p dividing its order. 9

10 G = 56 = 3 7. Since n 1 (mod ), n 7 so n = 1, 7 n 7 1 (mod 7), n 8 so n 7 = 1, 8 If n 7 = 1, the Sylow 7-subgroup is normal in G. Then we are done. If n 7 = 8, G has 8 6 = 48 non-identity elements of order 8 in these Sylow 7-subgroups. The rest = 8 are precisely elements in Sylow -subgroup. This implies that G has a unique Sylow -subgroup Prove that a group of order 31 has a normal Sylow p-subgroup for some prime p dividing its order. G = 56 = Since So Syl 13 is normal in G. n 3 1 (mod 3), n so n 3 = 1, 4, 13, 5 n 13 1 (mod 13), n so n 13 = Prove that a group of order 351 has a normal Sylow p-subgroup for some prime p dividing its order. G = 351 = Since n 3 1 (mod 3), n 3 13 so n 3 = 1, 13 n 13 1 (mod 13), n so n 13 = 1, 7 If n 13 = 1, the Sylow 13-subgroup is normal in G. Then we are done. If n 13 = 7, G has 7 1 = 34 non-identity elements of order 13 in these Sylow 13-subgroups. The rest = 7 are precisely elements in Sylow 3-subgroup. This implies that G has a unique Sylow 3-subgroup Prove that if G = 105 then G has a normal Sylow 5-subgroup and a normal Sylow 7-subgroup. (The solution basically follow the proof of the example on page 143 where G = 30). G = 105 = Since n 5 1 (mod 5), n so n 5 = 1, 1 n 7 1 (mod 7), n so n 7 = 1, 15 Let P = Syl 5 and Q = Syl 7. If n 5 = 1 and n 7 = 1, then we are done. If either n 5 = 1 or n 7 = 1, i.e., either P or Q is normal. Let H = P Q. Then H G by Corollary 15, and both P and Q are characteristic subgroups of H by statement () on page 135 that every subgroup of a cyclic group is characteristic (H is cyclic by Example on page 143), or it is easy to see that n 5 = 1 and n 7 = 1, hence P H and Q H. So P and Q are characteristic in H. Since H G by Corollary 5. By exercise or the statement (3) on page 135, both P and Q are normal subgroup of G. Now, assume that neither Sylow subgroup is normal, i.e., n 5 = 1 and n 7 = 15. Then G has 1 4 = 84 elements of order 5 and 15 6 = 90 elements of order 7. So total elements would be = 174 > 105, a contradiction. Therefore, one of P or Q, hence both, must be normal in G. 10

11 4.5.4 Prove that if G is a group of order 31 then Z(G) contains a Sylow 11- subgroup of G and a Sylow 7-subgroup is normal in G. G = 31 = Since n 3 1 (mod 3), n so n 3 = 1 n 7 1 (mod 7), n so n 7 = 1 n 11 1 (mod 11), n so n 11 = 1 It is clear that Syl 7 G. Let P = Syl 11. Since G/C G (P ) = a subgroup of Aut(P ) by Corollary 13. And Aut(P ) = 11 1 = 10, so G/C G (P ) must divides 10. But G = , the only possibility is that G/C G (P ) = 1. Hence, G = C G (P ). i.e. P Z(G) Prove that if G is a group of order 385 then Z(G) contains a Sylow 7- subgroup of G and a Sylow 11-subgroup is normal in G. G = 385 = Since n 5 1 (mod 5), n so n 5 = 1, 11 n 7 1 (mod 7), n so n 7 = 1 n 11 1 (mod 11), n so n 11 = 1 It is clear that Syl 11 G. Let P = Syl 7. Since G/C G (P ) = a subgroup of Aut(P ) by Corollary 13. And Aut(P ) = 6. So G/C G (P ) must divides 6. But G = , the only possibility is that G/C G (P ) = 1. Hence, G = C G (P ). i.e. P Z(G) Let G be a group of order 105. Prove that if a Sylow 3-subgroup of G is normal then G is abelian. G = 385 = Since n 3 1 (mod 3), n so n 3 = 1, 7 n 5 1 (mod 5), n so n 5 = 1, 1 n 7 1 (mod 7), n so n 7 = 1, 15 If P = Syl 3 is normal. By Corollary 13, G/C G (P ) is isomorphic to a subgroup of Aut(P ). Since Aut(P ) =, so G/C G (P ) must divides. But G = 3 5 7, the only possibility is that G/C G (P ) = 1. Hence, G = C G (P ). i.e. P Z(G). Similarly, Q = Syl 5 is normal by Exercise 17. Thus we can also prove that Q Z(G). By Lagrange, Z(G) = 3 5 or So G/Z(G) = 1, 7 hence G/Z(G) is cyclic. Therefore, G is abelian Let G be a group of order 315 which has normal Sylow 3-subgroup. Prove that Z(G) contains a Sylow 3-subgroup of G and deduce that G is abelian. G = 315 = Since n 3 1 (mod 3), n so n 3 = 1, 7 n 5 1 (mod 5), n so n 5 = 1, 1 11

12 n 7 1 (mod 7), n so n 7 = 1, 15 If Sylow 3-subgroup, P = Syl 3, is normal, G/C G (P ) is isomorphic to a subgroup of Aut(P ) by Corollary 13. Since Aut(P ) = 6 or 48. And P is a square of a prime, P is an abelian group, hence P C G (P ) by Exercise 6(b) in Section.. It follows that C G (P ) is divisible by 9, which implies that G/C G (P ) = 1, 5, 7, 35. Together these imply G/C G (P ) = 1. Hence, G = C G (P ). i.e. P Z(G). So G/Z(G) =1, 5, 7 or 35. If G/Z(G) = 1, then G is abelian. If G/Z(G) = 5 or 7, G/Z(G) is cyclic hence G is abelian. If G/Z(G) = 35, it is easy to see that n 5 = 1 and n 7 = 1 then G/Z(G) is cyclic by Example on page 143. Therefore, G is abelian Let P be a Sylow p-subgroup of H and let H be a subgroup of K. Let P H and H K, prove that P is normal in K. Deduce that if P Syl p (G) and H = N G (P ), then N G (H) = H. If P is a Sylow p subgroup of H and assume that P H, then P is characteristic in H by Corollary 0. Since H is normal in K, by the statement (3) P is normal in K. Hence, if P Syl p (G) and H = N G (P ) (i.e., P H), then by above argument P K. This implies that K = N K (P ) N G (P ) = H Thus, H = K. Since H K (i.e. K = N G (H)), we have H = N G (P ) K which implies that N G (H) = N G (N G (P )) = K = N G (P ) = H, as desired. (the solutions was not correct, will come back later!) Let P be a normal Sylow p-subgroup of G and let H be any subgroup of G. Prove that P H is the unique Sylow p-subgroup of H. It suffices to show that P H is Sylow p subgroup of H. Then by Exercise 4 in Section 3.1, P H H. Let G = p α m with p m. Then H = p β m where 0 β α and P = p α. Since P G, HP G. By Proposition 13 in Section 3., we have H H P = HP P divides G P = pα m p α = m Hence P H = p β. Thus, P H is the unique Sylow p-subgroup of H. Alternatively, since P H H and P H is a p-subgroup of H. Suppose that P H is not a Sylow p-subgroup of H. There is an element x H\P of order p. But this is not possible since P is the unique Sylow p-subgroup of G, that contains all elements of order p, a contradiction! Let P Syl p (G) and assume N G. Use the conjugacy part of Sylow s Theorem to prove that P N is a Sylow p-subgroup of N. Deduce that P N/N is a Sylow p-subgroup of G/N Suppose Q is a Sylow p-subgroup of N. Q is also a p-subgroup of G, by Theorem 18(), there exists g G such that Q gp g 1. Since Q N, we have Q gp g 1 N. But N is normal. so Q gp g 1 gng 1 = g(p N)g 1. This implies that g 1 Qg P N. Now, g 1 Qg is a Sylow p-subgroup of N since g 1 Qg = Q. The group P N is a p-subgroup of N, so g 1 Qg = P N. This shows that P N is a Sylow p-subgroup of N. 1

13 Let P = p a and Q = p b, so a Sylow p-subgroup of G/N has order p a b. By nd Isomorphism Theorem, P N N = P P N So P N/N = P / P N = p a /p b = p a b. Thus, P N/N is a Sylow p-subgroup of G/N. G/N G P N/N P N 1 N For any left ideal I of R define IM = { a i m i a i I, m i M} finite to be the collection of all finite sums of elements of the form am where a I and m M. Prove that IM is a submodule of M. First, prove that IM is a subgroup. Note that the identity exists since 0 M and take m = 0, then 0m IM. For x, y IM, x = a i m i, y = b i m i, we have x+y = (a i +b i )m i IM. Finally, let y = a i m i, then x + y = (a i a i )m i = 0, so y is the inverse of x. Thus, IM is a subgroup. IM is closed under the action of ring elements: let r R, rx = r a i m i = ra i m i IM since ra i I. Hence, IM is a submodule Let N 1 N be an ascending chain of submodules of M. Prove that i=1 N i is a submodule of M. We prove by Proposition 1. First, i=1 N i φ since N i φ for each i. Furthermore, for r R and x, y i=1 N i, we have x N m and y N k for some m, k. Without loss of generality, assume m k. Then x N k and x + ry N k hence x i=1 N i If N is a submodule of M, the annihilator of N in R is defined to be {r R rn = 0 for all n N}. Prove that the annihilator of N in R is a -sided ideal of R. (1) Prove that Ann(N) is a subring. Note that 0 N, so Ann(N) is nonempty. For x, y Ann(N), xn = 0 and yn = 0 for all n N. So (x y)n = xn yn = 0 for all n N. Hence, x y Ann(N). Now, (xy)n = x(yn) = x 0 = 0. Hence Ann(N) is a subring. () Prove that Ann(N) is closed under right and left multiplication by element of r R. Let x Ann(N), xn = 0 for all n N. (rx)n = r(xn) = r 0 = 0, hence rx Ann(N). Similarly, (xr)n = x(rn) = x 0 = 0. Therefore, Ann(N) is a -sided ideal of R. 13

14 Let F = R, let V = R and let T be the linear transformation from V to V which is projection onto the y-axis. Show that V, 0, the x-axis and the y-axis are the only F [x]-submodules for this T. Since the F [x]-submodules of V are precisely the T -invariant subspace of V. We see that T (0) = 0 V, T (V ) V, T (x-axis) = 0 x-axis and T (y-axis) = y-axis y-axis. Hence, these are F [x]-modules. In general, for L be any 1-dimensional subspace on R, neither x-axis nor y-axis. It is easy to see that T (L) L. Hence the only F [x]-submodules for T are 0, V, x-axis and y-axis Let A be any Z-module, let a be any element of A and let n be a positive integer. Prove that the map ϕ a : Z/nZ A given by ϕ a ( k) = ka is a well-define Z-module homomorphism if and only if na = 0. Prove that Hom Z (Z/nZ, A) = A n, where A n = {a A na = 0} (so A n is the annihilator in A of the ideal (n) of Z). Since ka = ma if and only if (k m)a = 0 if and only if a nz if and only if na = 0. This proves that ϕ a is well-defined if and only if na = 0. For x, ȳ Z/nZ, r Z, we have ϕ a ( x + ȳ) = (x + y)a = xa + ya = ϕ a ( x) + ϕ a (ȳ). And ϕ a (r x) = rxa = rϕ a ( x). Hence, ϕ a is Z-module homomorphism. Define f : Hom(Z/nZ, A) A n by f(ϕ a ) = ϕ a ( k) = ka for k Z/nZ. First, f is well-defined since ϕ a is. For ϕ a, ϕ b Hom (Z/nZ, A), we have f(ϕ a + ϕ b )( k) = ϕ a ( k) + ϕ b ( k) = ka + kb = ϕ a ( k) + ϕ b ( k) = f(ϕ a ) + f(ϕ b ). Furthermore, let r R, then we have f(rϕ a ) = rϕ a ( k) = rka = r(ka) = rϕ a ( k) = rf(ϕ a ). So f is a group homomorphism. To show that f is surjective, for a A n with na = 0, let k = 1, then we have ϕ a ( 1) = 1 a = a. Suppose ϕ a ( k) = ϕ b ( k), then ka = kb which implies that a = b (by cancellation property since A is an abelian group). i.e. ϕ a = ϕ b. Therefore, f is an isomorphism Prove that Hom Z (Z/nZ, Z/mZ) = Z/(n, m)z. The homomorphism between them will be determined by where the generator in Z/nZ goes. Let ϕ : Z/nZ Z/mZ be a Z-module homomorphism, then ϕ(0) = 0. Hence, 0 = ϕ(0) = ϕ( ) = ϕ(1) + + ϕ(1) = nϕ(1) To have a homomorphism, we must have nϕ(1) mz and this is true if and only ϕ(1) = m m, or its multiples. These elements form a group generated by and is cyclic gcd(m, n) gcd(m, n) of order gcd(m, n). Therefore, Hom Z (Z/nZ, Z/mZ) = Z/(m, n)z Let z be a fixed element of the center of R. Prove that the map m zm is an R-module homomorphism from M to itself. Show that for a commutative ring R the map from R to End R (M) given by r ri is a ring homomorphism (where I is the identity endomorphism). Let ϕ : R M M be defined by ϕ(m) = zm. For x, y M and r R, we have ϕ(rx + y) = z(rx + y) = z(rx + y) = zrx + zy = rzx + zy = rϕ(x) + ϕ(y). Hence, ϕ is an R-module homomorphism. 14

15 Recall that End R (M) =Hom R (M, M). Let ψ : R End R (M) be defined by ψ(r) = ri. For r, s R, we see that ψ(r +s) = (r +s)i = ri +si = ψ(r)+ψ(s). Also, ψ(rs) = rsi = ri si = ψ(r)ψ(s). Therefore, ψ is a R homomorphism Let A 1, A,, A n be R-modules and let B i be submodule of A i for each i = 1,,, n. Prove that (A 1 A n )/B 1 B n ) = (A 1 /B 1 ) (A n /B n ). Define a map ϕ : A 1 A n (A 1 /B 1 ) (A n /B n ) by (a 1,, a n ) (a 1 + B 1,, a n + B n ) for a i A i, i = 1,, n. Since on each component A i A i /B i is a natural projection hence is an R-module homomorphism by Proposition 3. Therefore, ϕ is an R-module homomorphism. Prove that ϕ is a bijection. First, note that ϕ is surjective since each component ϕ i : A i A i /B i is a natural projection hence surjective (well, I thought it is clear that the product of a surjective map is surjective). It suffices to prove that ker ϕ = B 1 B n. (1) ker ϕ B 1 B n If a = (a 1,, a n ) ker ϕ, then ϕ(a) = (a 1 + B 1,, a n + B n ) = ( 0,, 0) (A 1 /B 1 ) (A n /B n ). So we have a i + B i = 0 hence a i B i for each i. Thus, a = (a 1,, a n ) B 1 B n. () ker ϕ B 1 B n For b = (b 1,, b n ) B 1 B n, we have ϕ(b) = ϕ(b 1,, b n ) = (b 1 + B 1,, b n + B n ) = (B 1,, B n ) = ( 0,, 0) in (A 1 /B 1 ) (A n /B n ). Hence, b ker ϕ. Therefore, ker ϕ = B 1 B n. By first isomorphism theorem for modules (Theorem 4 in Section ), (A 1 A n )/B 1 B n ) = (A 1 /B 1 ) (A n /B n ) Let I be a nilpotent ideal in a commutative ring R, let M and N be R- modules and let ϕ : M N be an R-module homomorphism. Show that if the induced map ϕ : M/IM N/IN is surjective, then ϕ is surjective. M ϕ N p M/IM ϕ N/IN Given n N, want to show that there is m M such that ϕ(m) = n by showing that q(ϕ(m) n) = 0 N/IN. Indeed, by commutativity of the diagram we have q(ϕ(m) n) = q ϕ(m) q(n) = ϕ p(m) q(n) = ϕ(m + IM) q(n) = n + IN (n + IN) = 0 15 q

16 Thus, ϕ(m) n IN. This is true for all n N, hence ϕ(m) N = IN. i.e. N = ϕ(m)+in. This implies that N = ϕ(m) + I(ϕ(M) + IN) = ϕ(m) + Iϕ(M) + I N = ϕ(m) + I N Inductively, we obtain N = ϕ(m) + I r N. Since I is nilpotent, I k = 0 for some k 1. Therefore, N = ϕ(m) and ϕ is surjective Assume R is commutative. Prove that R n = R m if and only if n = m, i.e. two free R-modules of finite rank are isomorphic if and only if they have the same rank. [Apply Exercise 1 of Section with I a maximal ideal of R. You may assume that F is a field, then F n = F m if and only if n = m.] By Exercise 1 in Section 10., we have R n /IR n = I/IR I/IR. If I R is a maximal ideal, then R/I is a field. Also, IR = I. Hence we have Let F = R/I we obtain Similarly, we have Therefore, R n /IR n = R/I R/I R n /IR n = F F = F n R m /IR m = R/I R/I = F m m = n F m = F n R m /IR m = R n /IR n R m = R n Let N be a submodule of M. Prove that if both M/N and N are finitely generated then so is M. Let f : M M/N be a projection homomorphism defined by m m + N. Let {y 1,, y n } be a basis for N and {x 1,, x m } be a basis for M/N. For m M, its image m in M/N can be written as m = i a i x i. This implies that m + N = i a i(x i + N) = i a ix i + N. Thus, m i a ix i N, which implies that m i a ix i = j b jy j. Thus, m = i a ix i + j b jy j. Note that M = N M/N, therefore M is finitely generated (a) Show that M is irreducible if and only if M 0 and M is cyclic module with any nonzero element as generator. ( ) Suppose M is not cyclic, let M = a 1,, a n for n > 1. i.e. M is finitely generated by a 1,, a n with a i 0 for all i. Then M has a submodule N 0, and N M. This is because we can let A = {a 1,, a k } for k < n, then N = RA = {r 1 a 1,, r k a k r i R, a i A} is a submodule. This proved that M is cyclic. Now, wuppose M is irreducible, pick a nonzero element m M, then 0 Rm M. Clearly Rm is a submodule. Hence Rm = M since M is irreducible. Therefore, M is a cyclic module. ( ) Suppose M is a cyclic module with any nonzero element as generator. i.e. M = Ra for any a M. Want to prove that M is irreducible. Assume that M has a proper submodule N, then 16

17 0 N M. Pick a nonzero element n in N, so we have 0 Rn N M. But Rn = M, so N = M hence M is irreducible. Now, determine all the irreducible Z-modules. Let M be an irreducible Z-module. Note that Z-module can be seen as an abelian group, so by the structure of an abelian group, M has torsion part and free part. Since M is irreducible hence does not have any proper submodule. But Tor(M) is a submodule of M. So Tor(M) is either 0 or Tor(M) = M. Suppose Tor(M)=0, take m M and m 0. We have am bm for a, b Z, a b. So we have 0 Zm Zm M. Contradiction to the fact that M is irreducible. On the other hand, suppose Tor(M) = M, take m M and m 0. By what we just proved in ( ), M = Zm. Now, since Z is a PID, so Ann(m) = (k) for some k Z. Then M = Z/kZ. We claim that k is a prime. If not, k = qr for some q, r Z, then 0 Z/qZ Z/kZ M. Contradiction to the fact that M is irreducible. So k is a prime. Hence we proved that M = Z/pZ some some prime p. Finally, if p is a prime, by what we proved in ( ), Z/pZ is irreducible Assume R is commutative. Show that an R-module M is irreducible if and only if M is isomorphic (as a R-module) to R/I where I is a maximal ideal of R. ( ) Suppose M is irreducible. Define ϕ : R M by r rm for m 0, by previous exercise, M is cylic, i.e. M = Rm for some m 0. So ϕ is actually surjective. Prove that ker ϕ =Ann(m). For r ker ϕ, ϕ(r) = rm = 0 implies that r Ann(m). The converse is also clear. Thus we have R/I = M where I = ker ϕ =Ann(M). Now, prove that I is a maximal ideal. First, I is an ideal since for any s R, s(rm) = s 0 = 0. So SI I. To show that I is maximal, suppose I is not a maximal ideal, there is an ideal J such that I J R. R/I = M R J/I J 1 I Then M = R/I has a submodule J/I which contradicts to the irreducibility of M. Hence, I must be a maximal ideal. ( ) Suppose M = R/I for a maximal ideal I of R. If M is not irreducible, there is a submodule N such that N M. By Lattice Theorem again, there is an ideal J with I J R. This implies that J is not maximal. 17

18 If M 1 and M irreducible R-modules, prove that any nonzero R-module homomorphism from M 1 to M is an isomorphism. Let ϕ : M 1 M be a nonzero R-module homomorphism. Want to prove that ϕ is surective and injective hence is a bijection. First, note that ker ϕ is a submodule of M 1 and im ϕ is a submodule of M by in last homework assignment. But M 1 and M are irreducible, so ker ϕ = 0 (note that ker ϕ can not be M, or it would imply that ϕ is a zero map which contradicts to the assumption), so ϕ is injective. On the other hand, we obtained that im ϕ = M (note that im ϕ 0, if it were, ϕ is a zero map which again contradicts to the assumption). This implies that ϕ is surjective. Therefore, ϕ is a bijection, hence is an isomorphism. Now, if M is irreducible, prove that End R (M) is a division ring. Let ϕ : M M be a nonzero R-module homomorphism. So ϕ End R (M). But we have just proved that ϕ is a bijection in this case, which implies that ϕ has an inverse. In other words, every nonzero element in End R (M) has a multiplicative inverse. Note that the multiplication in End R (M) is the composition. i.e. there exists ψ End R (M) such that ϕ ψ = ψ ϕ = 1 M. Therefore, End R (M) is a division ring Show that Q Z Q and Q Q Q are isomorphic left Q-modules. [Show they are both 1-dimensional vector spaces over Q.] By definition on page 367, since Q is a left Z-module and a (Q, Z) bimodule, hence Q Z Q is a left Q-module. Also, since Q is a left Q-module and a (Q, Q) bimodule, thus Q Q Q. Now, show that they are isomorphic to 1-dimensional vector spaces over Q. For a b Q Q Q for a, b Q, we have a b = a 1 b = a(1 b) = a(1 b 1) = a(1 b 1) = a(b 1 1) = ab(1 1). Thus, Q Q Q is Span Q {1 1} = Q. Hence, Q Q Q is a 1-dimensional vector space over Q. On the other hand, since Q Z Q is a left Q-module defined by So s( a b) = sa b for s Q and a, b Q a b = a p q = ap 1 q = ap q q 1 q = ap q 1 = ab 1 Thus, Q Z Q is a group generated by α 1 for α Q. And α Z 1 = α(1 1) in Q-module Q Z Q. Hence Q Z Q is a 1-dimensional vector space over Q. (Example: Q Z Q: 3 1 Z = Z = 1 3 Z 1 = 1 3 Z 1. However, Q Q Q: 3 1 Q = Q = 1 Q 3 1 = 1 1 Q 3.) Let A be a finite abelian group of order n and let p k be the largest power of the prime p dividing n. Prove that Z/p k Z Z A is isomorphic to the Sylow p- subgroup of A. 18

19 Let P be a Sylow p-subgroup of A with P = p k. Define ϕ : Z/p k Z A P by (xmodp k Z, a) xa. ϕ is bilinear: for r 1, r Z, x 1, x Z/p k Z and a 1, a A ϕ(r 1 x 1 +r x, a) = ϕ(r 1 x 1 +r x, a) = (r 1 x 1 +r x )a = r 1 x 1 a+r x a = r 1 ϕ(x 1, a)+r ϕ(x, a). ϕ(x 1, r 1 a 1 + r a ) = x 1 (r 1 a 1 + r a ) = x 1 r 1 a 1 + x 1 r a = r 1 x 1 a 1 + r x 1 a = r 1 ϕ(x 1, a 1 ) + r ϕ(x 1, a ). Thus, ϕ is bilinear. Note that A is subgroup of finite abelian group, hence is abelian. So P is a Z-module. Since Z/p k Z and A are Z-modules and let Z/p k Z Z A be the tensor product of Z/p k Z and A over Z. By Corollary 1 there is a Z-module homomorphism Φ : Z/p k Z Z A P. Z/p k Z A ι Z/p k Z Z A ϕ Φ P Since Φ maps x a to xa P which is an element of order p k. Finally, Z/p k Z Z A has order at least p k because p k ( x a) = p k x a = 0 a = 0. Hence, Φ is an isomorphism If R is any integral domain with quotient field Q, prove that (Q/R) R (Q/R) = 0. R is an integral domain and Q is the quotient field. So Q = { m n m 1, m, n 1, n 1 R with n i 0 we have m, n R}. For since (m mod R) = 0. ( m 1 n 1 mod R) ( m n mod R) = n ( m 1 n 1 n mod R) ( m n mod R) = ( m 1 n 1 n mod R) n ( m n mod R) = ( m 1 n 1 n mod R) (m mod R) = Let {e 1, e } be a basis of V = R. Show that the element e 1 e + e e 1 in V R V cannot be written as a simple tensor v w for any v, w R. Suppose v = ae 1 + be, w = ce 1 + de for a, b, c, d R. Then v w = (ae 1 + be ) (ce 1 + de ) = ac(e 1 e 1 ) ad(e 1 e ) + bc(e e 1 ) + bd(e e ) If v w = e 1 e + e e 1, then ac(e 1 e 1 ) (ad 1)(e 1 e ) + (bc 1)(e e 1 ) + bd(e e ) = 0 Since {e 1 e 1, e 1, e e 1, e e } is a basis, we have ac = 0, ad 1 = 0, bc 1 = 0 and bd = 0. Thus, a = 0 and c = 0, but this implies that 1 = 0, a contradiction! 19

20 Let V be a vector space over the field F and let v, v be nonzero elements of V. Prove that v v = v v in V F V if and only if v = av for some a F. Let {v 1, v,, v n } be a basis for V. (Here we assume that V is finite dimensional so that for any element in V, its expression of a FINITE linear combination of the basis makes sense. If V is infinite, then let s choose a finite dimensional subspace W V containing v and v to proceed the proof). So V F V = v i v j i, j = 1,,, n. Let v, v V, then v = Σa i v i and v = Σa i v i. So we have v v = (Σa i v i ) (Σa i v i) = Σ i Σ j (a i a i )v i v j. Similarly, we have v v = Σ i Σ j (a i a i)v i v j. Hence we have v v = v v if and only if a i a j = a i a j for all i, j. Now, for fixed j, we have a i a j = a i a j for all i if and only if a i = a ia j (a j) 1 if and only if a i = (a j (a j) 1 )a i since F is commutative and every element in F has a multiplicative inverse. Since this is is true for some fixed j, let s denote the term a = (a j (a j) 1 ), note that a F. So for this fixed j, we have a i = a a i, i = 1, n. Hence we have v = Σa i v i = Σ(aa i)v i = aσa iv i = av. Therefore, we proved that v v = v v if and only if v = av for some a F Show that tensor products do not commute with direct products in general. Consider Q Z ( Z/ i Z), this implies i=1 Q Z ( Z/ i Z) = (Q Z Z/Z) (Q Z Z/ Z) (Q Z Z/ 3 Z) Since for each i, Z/ i Z is a torsion abelian group (i.e. every element has a finite order), by exercise (d) (prove later) each term on the right hand side is 0, hence the direct product is 0. However, the left hand side is not zero since Z/ i Z is not a torsion abelian. This is because there is no integer m such that m( Z/ i Z) = 0 when i goes arbitrarily large. Hence, we conclude that tensor products do not commute with direct products in general. Now we prove (d): If A is an abelian group group, show that Q Z A = 0 if and only if A is a torsion abelian group (i.e. every element of A has finite order). ( ) Suppose Q Z A = 0. For a A and q Q, q = m n for m, n in Z and n 0, we have 0 = q a = m n a = m 1 n a = 1 n m a. Since 1 n 0, so m a must be zero. So a has finite order m. Since a A is arbitrary, so every element in A has finite order hence A is a torsion group. ( ) Suppose a A has finite order, say m. i.e. ma = 0 with some m Z +. Let q Q. Then we have q a = m q m a = q m m a = q m 0 = 0. Hence Q Z A = Let P 1 and P be R-modules. Prove that P 1 P is a projective R-module if and only if both P 1 and P are projective. 0

21 ( ) Suppose P 1 and P are projective modules, by Proposition 30(4), P 1 and P are direct summand of free R-modules. i.e. P 1 Q 1 and P Q are free for some modules Q 1 and Q. But free module is isomorphic to a direct sum of regular modules, so P 1 Q 1 P Q is also a direct sum of modules, hence is free. Since P 1 Q 1 P Q = (P1 P ) (Q 1 Q ), therefore, P 1 P is a direct summand of a free module hence is projective. ( ) Suppose P 1 P is projective. For R-modules M, N, let M ϕ N 0 be an exact sequence, and f : P 1 N an R-module homomorphism. Want to prove that f lifts to an R-module homomorphism into M. i.e. f lifts to F : P 1 M. Let π : P 1 P P 1 be the natural projection (x, y) x for x P 1 and y P. Then the composition f π is an R-module homomorphism. Let n x be the image of x in N. i.e. f(x) = n x. Since ϕ is surjective (by exactness), so there exists an element m x M such that ϕ(m x ) = n x. Now, since P 1 P is a projective R-module, f π lifts to F : P 1 P M such that ϕ F = f π. i.e. the big diagram commutes and F : (x, y) m x. Now, want to prove that f lifts to an R-module homomorphism F : P 1 M and the small diagram commutes. i.e. f = ϕ F. Let i 1 : P 1 P 1 P be an inclusion and define F = F i 1 : P 1 M by sending x to m x. Certainly F is an R-module homomorphism since F is hence the composition with the inclusion is. Also we claim that f = ϕ F. Indeed, for x P 1, f(x) = n x N. On the other hand, ϕ F (x) = ϕ(f (x)) = ϕ(m x ) = n x. Hence the small diagram commutes and therefore P 1 is a projective R-module. Similarly we can prove that P is a projective R-module by replacing P 1 by P in the proof Let A be a nonzero finite abelian group. (a) Prove that A is not a projective Z-module. (b) Prove that A is not a injective Z-module. (a) By Example (3) on page 391, free Z-modules have no nonzero elements of finite order so no nonzero finite abelian group can be isomorphic to a submodule of free module. Since A is abelian, A is not isomorphic to a submodule of a free module. So A is not a direct summand of a free module. Hence A is not projective. (b) Since A is a finite abelian group, A = Z p α 1 1 a Z α p i i is not divisible. Therefore A is not injective. Z α p k. Since each Z α k p i i and for n = p α i 1, there is no x such that xn = a since x pα i i = 0. Hence Z p α 1 1 is not divisible: for Z p α k k Let Q be a nonzero divisible Z-module. Prove that Q is not a projective Z-module. Deduce that the rational numbers Q is not a projective Z-module. Lemma If F is a any free module then n=1nf = 0. 1

22 Let {x 1,, x k } be a basis for F. Suppose x n=1 nf. Note that {nx 1,, nx k } be a basis for nf. So if x n=1, x F, x F, x 3F,, hence x = a 1 x 1 + a x + + a k x k = a 1(x 1 ) + + a (x ) + + a k (x k). = a n 1 (nx 1 ) + + a n (nx ) + + a n k (nx k) (Note: the notation a j i, here j is an index, not an exponent!) Thus, 0 = (a 1 na n 1 )x (a k na n k )x k for all k. Hence a 1 = na n 1, i.e. a 1 = a 1 = 3a 3 1 = 4a4 1 =. So n a 1 for all n. Similarly, a k = na n k which implies that n a k for all n. Therefore, a 1,, a k must be zero, hence x = 0. Now, since Q is a divisible Z-module, nq = Q for all n. Suppose Q is a projective Z-module. Then Q is a direct summand of a free module F, i.e. Q K = F. So Q is isomorphic to a submodule of a free module. But n=1 nf = 0 while n=1 nq = n=1 Q = Q a contradiction! Therefore, Q is not projective Assume R is commutative with 1. (a) Prove that the tensor product of two free R-modules is free. [Use the fact that tensor products commute with direct sums.] (b) Use (a) to prove that the tensor product of two projective R-modules is projective. (a) By Corollary 19 in Sec (b) Suppose M, N are projective. Then M P = F 1 and N Q = F for some P, Q and free modules F 1 and F. Thus, F 1 R F = (M P ) (N Q) = (M P ) N (M P ) Q = (M R N) (P N) (M Q) (P Q) By (a) F 1 R F is free. Thus, M R N is the direct summand of a free module, hence is projective Let M be a left Z-module and let R be a ring with 1. (a) Show that Hom Z (R, M) is a left R-module under the action (rϕ)(r ) = ϕ(r r). (b) Suppose that 0 A ψ B is an exact sequence of R-modules. Prove that if every Z-module homomorphism f from A to M lifts to a Z-module homomorphism F from B to M with f = F ψ, then every R-module homomorphism f from A to Hom Z (R, M) lifts to an R-module homomorphism F from B to Hom Z (R, M) with f = F ψ. (c) Prove that if Q is an injective Z-module then Hom Z (R, Q) is an injective R- module. (a) For ϕ Hom Z (R, M), r, r R, define rϕ : R M by rϕ(r ) = ϕ(r r). First prove that rϕ Hom Z (R, M).

23 For r 1, r R, and a R, we have - (rϕ)(r 1 + r ) = ϕ((r 1 + r )r) = ϕ(r 1 r + r r) = ϕ(r 1 r) + ϕ(r r) = rϕ(r 1 ) + rϕ(r ) - (rϕ)(ar 1 ) = ϕ((ar 1 )r) = ϕ(a(r 1 r)) = aϕ(r 1 r) = arϕ(r 1 ) by the definition of rϕ and that ϕ is a homomorphism. Hence rϕ Hom Z (R, M). Now, prove that this action of R on Hom Z (R, M) makes it into a left R-module. Let ϕ, ψ Hom Z (R, M). First, by Proposition in 10., define ϕ + ψ by (ϕ + ψ)(r) = ϕ(r) + ψ(r), then ϕ + ψ Hom Z (R, M) and with this operation Hom Z (R, M) is an abelian group. Prove that Hom Z (R, M) satisfies the R-module criteria: let r, r, s R, (1) (r + s)ϕ = rϕ + sϕ (r + s)ϕ(r ) = ϕ(r (r + s)) = ϕ(r r + r s) = ϕ(r r) + ϕ(r s) = rϕ(r ) + sϕ(r ) = (rϕ + sϕ)(r ). () (rs)ϕ = r(sϕ) (rs)ϕ(r ) = ϕ(r (rs)) = ϕ((r r)s) = sϕ(r r) = r(sϕ(r )) since sϕ Hom Z (R, M). (3) r(ϕ + ψ) = rϕ + rψ r(ϕ + ψ)(r ) = (ϕ + ψ)(r r) = ϕ(r r) + ψ(r r) = rϕ(r ) + rψ(r ) = (rϕ + rψ)(r ) since ϕ + ψ Hom Z (R, M). (4) 1 ϕ = ϕ 1 ϕ(r ) = ϕ(r 1) = ϕ(r ). Hence, Hom Z (R, M) is a left R-module. (b) Given an R-module homomorphism f : A Hom Z (R, M) and a A, define f : A M by a f (a)(1 R ). f is well-defined since f is. Also, f is a Z-module homomorphism since for a, b A and r Z, we have - f(a + b) = f (a + b)(1 R ) = ((f (a) + f (b))(1 R ) = f (a)(1 R ) + f (b)(1 R ) = f(a) + f(b). - f(ra) = f (ra)(1 R ) = rf (a)(1 R ) = rf(a). Now, since ψ is injective, let b B be the image of a. Since by assumption f lifts to F : B M and f = F ψ. So F : b f(a). Now, define F : B Hom Z (R, M) by F (b)(r) = F (rb). Show that F is an R-module homomorphism. For b 1, b B, r, s, R, we have - F (b 1 + b )(r) = F (r(b 1 + b )) = F (rb 1 + rb ) = F (rb 1 ) + F (rb ) = F (b 1 )(r) + F (b )(r) - F (sb)(r) = F (r(sb)) = F ((rs)b) = F (b)(rs) = sf (b)(r) by the definition of F and the fact that B is an R-module, F is an R-module. The last equality holds because F (b) Hom Z (R, M) hence by (a) we have F (b)(rs) = sf (b)(r). Hence F is an R-module homomorphism. Finally, prove that the diagram commutes, i.e. f = F ψ. Indeed, f (a)(r) = f (a)(r 1 R ) = rf (a)(1 R ) = rf(a) since f (a) is a homomorphism. On the other hand, (F ψ)(a)(r) = F (ψ(a))(r) = F (b)(r) = F (rb) = rf (b) = rf(a). Hence, the 3