Problem 8. If H and K are subgroups of finite index of a group G such that [G : H] and [G : K] are relatively prime, then G = HK.

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1 1 1.4 Problem 8. If H and K are subgroups of finite index of a group G such that [G : H] and [G : K] are relatively prime, then G = HK. Proof. We have that [G : H K] = [G : H][H : H K] [G : H][G : K] with equality iff [H : H K] = [G : K] iff G = HK by P4.8. Yet this is simple by our hypotheses; we must have that [G : H], [G : K] [G : H K] by T4.5, and since these indices are relatively prime we have that [G : H K] [G : H][G : K]. This yields equality and completes the proof. Problem 11. Let G be a group of order 2n; then G contains an element of order 2. If n is odd and G Abelian, there is only one element of order 2. Proof. Let S consist of the unordered sets of elements in G such that the elements in each pair are inverses of each other. We know that inverses are unique so that this collection S is well-defined, that there is one singleton set - namely, {e} - and that the maximum size of any set is two. Since G = 2n and every element has an inverse, it must be the case by sheer counting that there is another singleton set in S - say {a}. Then a is its own inverse and hence a has order 2. (This argument is due to Timbuc.) If h 1, h 2 are two distinct elements of G of order 2, then h 1 h 2 = {e, h 1, h 2, h 1 h 2 }, as is clear after a routine check using the fact that G is Abelian. This is clearly a subgroup, so h 1 h 2 = 4 G, which contradicts the fact that n is odd. Hence the element of order 2 is unique in G. Problem 12 will be taken for granted since it follows nearly directly from the proof of P4.8. Problem 13. If p > q are primes, a group of order pq has at most one subgroup of order p. Proof. If H, K are distinct subgroups of order p in a group G of order pq, then clearly H K = 1 by Lagrange, so that H K HK = H K H K = p2 > pq = G, a contradiction since H K < G. 1.5 Problem 16. If f : G H is a homomorphism, H is Abelian, and N is a subgroup of G containing Ker f, then N G. Proof. We have that f : G/ Ker f = Im f < H, so there exists an inverse isomorphism f 1. Note that f(n) is normal in Im f since f(g)f(n)f(g 1 ) = f(g)f(n)f(g) 1 = f(g)f(g) 1 f(n) = f(n) for g G (we used here that Im f < H is Abelian). Hence gng 1 = (g Ker f)n(g 1 Ker f) = f 1 (f(g))n f 1 (f(g 1 )) = f 1 (f(gng 1 )) = f 1 (N) = N Ker φ = N (using twice that N contains Ker f and that Ker f is a normal subgroup). 1.8 Problem 4. Give an example to show that the weak direct product is not a coproduct in the category of all groups.

2 2 Let us refresh on a few definitions: Definition 1. A coproduct for {A i } i I inside a category is an object of that category together with morphism {ι i : A i S} i I such that the following property holds. For any object B and family of morphisms {ψ i : A i B} i I, there is a unique morphism ψ : S B such that ψ ι i = ψ i for all i I. This differs from a product in that a coproduct supplies inclusion maps into the object of the coproduct, whereas the product supplies projection maps from the object of the coproduct. Definition 2. The weak direct product of a family of groups of {G i } i I, denoted { w } i I G i, is the set of all f i I G i such that f(i) = e i for all but a finite number of i I. That is, it is the set of all functions f : I i I A i such that (i) f(i) A i for all i, and (ii) f(i) = e i A i for all but finitely many i. Proof. It suffices to show that the obvious inclusion maps of some subgroups H, K into the weak direct product H K - i.e., ι H (h) = (h, e) and ι K (k) = (e, k) are insufficient to satisfy the definition. Using these two subgroups, we will also determine a group G with isomorphisms f H : H G and f K : K G for which there can be no homomorphism f : H K G such that f H = f H and f K = f K. Let G = H = K = S 3. Let f H, f G be the identity maps. We will attempt to establish a homomorphism f : H K G that fulfills the requirements above. Of course, we must mandate f((h, e)) = f H (h) for h H and f((e, k)) = Definition of product = f K (k) for k K. Using these requirements, we find that f((h, k)) f((h, e)(e, k)) Homomorphism = f((h, e))f((e, k)) = f H (h)f K (k). But we also have f((h, k)) = f((e, k)(h, e)) = f K (k)f H (h), so that f H (h)f K (k) = f K (k)f H (h). Recalling that f H, f K are identity maps, this gives us that hk = kh for any h H = S 3, k K = S 3. This implies that S 3 is Abelian, which is a contradiction. Problem 8. Corollary 8.7 is false if one of the N i is not normal. Proof. In general, the issue is that the product i I a i is no longer well-defined, since it depends on the order of the a i. Consider G = D 4, with subgroups H and K where H is the group of rotations and K is the group of flips. Clearly these subgroups generate G and have trivial intersection, but G is no longer a direct product of these subgroups. Problem 12. A normal subgroup H of a group G is said to a direct factor if there exists a normal subgroup K of G such that G = H K. (a) If H is a direct factor of K, and K is a direct factor of G, then H is normal in G. (b) If H is a direct factor of G, then every homomorphism H G may be extended to an endomorphism G G. However, a monomorphism H G need not be extendible to an automorphism G G. Proof. (a) We have that G = K N 2 = (H N 1 ) N 2 = H N 1 N 2 by the definition of meaningful product. Hence if g = h 0 n 1 n 2 G, we have ghg 1 = h 0 n 1 n 2 Hn 1 2 n 1 1 h 1 0 = h 0 Hh 1 0 = H.

3 3 (b) If φ : H G = H K is a homomorphism, let φ : G G be defined by φ((h, k)) = φ(h). Then φ((h 1, k 1 )(h 2, k 2 )) = φ((h 1 h 2, k 1 k 2 )) = φ(h 1 h 2 ) = φ(h 1 )φ(h 2 ) = φ((h 1, k 2 )) φ((h 2, k 2 )). This proves the first part of this statement. Now note that the monomorphism f : Z 2 S 3 Z 2 given by 1 (12) cannot be extended to an automorphism f : S 3 Z 2 S 3 Z 2. Such an automorphism would have to map ((12),0) to an element of order 2 such that (12) f((12), 0) also has order 2, since (1, (12)) has order 2. The only options that satisfy the first requirement are (13) and (23), and both generate 3-cycles, which do not satisfy the second requirement. 2.1 Problem 12. Let F be the free group on a set X and G the free group on a set Y. Let F < F be generated by {aba 1 b 1 a, b F } and similarly for G. (a) F F, and F/F is Abelian. (b) F/F is a free Abelian group of rank X. (c) F = G iff X = Y. Proof. (a) If g F, then gf g = {gaba 1 b 1 } = {ga(g 1 g)b(g 1 g)a 1 (g 1 g)b 1 (g 1 g)b 1 g 1 } = {(gag 1 )(gbg 1 )(gag 1 ) 1 (gbg 1 ) 1 } = F, so F is normal in F. Since (ab)(ba) 1 = aba 1 b 1 F, abf = baf, so F/F is Abelian. (b) We prove that {xf x X} is a basis for F/F. If y F, then there exist {a i } n 1 Z, {x i } n 1 X such that n i=1 a ix i = y, so ( n i=1 a ix i ) + F = n i=1 a ix i + F = y + F. Alternatively, if n i=1 a ix i + F = F, then n i=1 a ix i F. Hence there is an xyx 1 y 1 F such that this sum is 0. Hence ( n i=1 a ix i )xyx 1 y 1 = 0 (a i xyx 1 y 1 ) = 0 for all i, which implies that a i F = F for all i. This completes the proof. (c) ( ) is TI.7.8 by the definition of equivalence. ( ) Let φ : F G be an isomorphism and consider π : F G/G, the projection map. Clearly π is surjective. If f Ker π, then φ(f) G φ(f) = g 1 g 2 g1 1 2 for some g 1, g 2 G. Since φ is surjective, this means that φ(f) = φ(f 1 )φ(f 2 )φ(f1 1 1 )φ(f2 1f 2 f ), and shice φ is injective, f = f 1 f 2 f The reverse implications clearly hold; if f 1 f 2 f F, then φ(f 1 f 2 f )φ(f 2 )φ(f 1 ) 1 φ(f 2 ) 1 G = Ker π. So Ker π = F, and the First Isomorphism Theorem completes the proof. 2.2 Problem 5. If G is a finitely generated Abelian group such that G/G t has rank n and H < G is such that H/H t has rank m, then m n and (G/H)/(G/H) t has rank n m. Proof. It is worth proving that G = G/H H if H is a normal subgroup of G. Let G/H = {g 1 H = eh, g 2 H,..., g n H} where the sets are disjoint and {g i } n 1 G. Then every element of g can be written uniquely as g i h for some i [n] and h H, and the obvious map ensues. We have G = G/G t G t. We have that G/G t is finitely generated by T2.1 and Abelian, and since it contains no torsion elements it is free. The same holds for H = H/H t H t ; in particular, H = F, the maximal free group in H, so by identification H/H t < H < G, so H/H t is a subgroup of the maximal free group of G, G/G t. Hence by T1.6 m n.

4 4 (G/H) t is normal in G/H since G/H is Abelian. Assume for now that G/H is free so that (G/H) t = {e}. We claim that if G is generated by X and H by Y, then since G = H G/H we must have that the rank of G/H is n m. If G/H is not free, G/H/(G/H) t is the maximal free subgroup in G/H, and this completes the proof. Problem 7. Let G be an Abelian torsion group. (a) G(p) is the unique maximum p- subgroup of G. (b) G = G(p) where the sum is over all primes p such that G(p) 0. (c) If H is another Abelian torsion subgroup, then G = H iff G(p) = H(p) for all primes p. Proof. (a) If H < G is a p-subgroup, then by definition it is contained in G(p). (b) If g G is given and g = p n1 1 p nm m, let m i = u p n i. Then gcd({p i }) = 1, so by i the extended Euclidean Algorithm there exist {c i } m 1 Z such that m i=1 c im i = 1 m i=1 c im i u = u. Now m i u = p ni i, so c i m i u m i u and c i m i u G(p i ). Therefore G = G(2) + G(3) + + G(p) +, and since the G(p) are mutually disjoint and (trivially) normal in G we get the claim. (c) ( ) Given φ : G H, we can restrict φ to G(p) and land in H(p) since φ(g) must divide the order of g for any group element. Similarly for the inverse function from H(p) to G(p). If φ p : G(p) H(p) is given, then by (b) φ : G H can be given by (a 1, a 2, ) (φ 2 (a 1 ), φ 3 (a 2 ),... ). Problem 8. A finite abelian p-group G is generated by its elements of maximal order. Proof. Since G is finite, it is finitely generated, and G = Z m1 Z mn for m 1 m n. Clearly each m n is a power of p since otherwise (0,..., 0, 1) does not have order a power of p. Hence m i is a power of p for all i. We claim that X = {(z 1,..., z n ) z n 0} generates the group. But this is simple; given arbitrary {z i } 1 n 1, we have (z 1,..., z n 1, 0) = (z 1,..., z n 1, 1) + (0,..., 0, m n 1). Since X is exactly those elements of order m n, the maximal order of the group, this completes the proof. Problem 9. How many subgroup of order p 2 does the Abelian group Z p 3 Z p 2 have? Proof. We start with the cyclic subgroups. If we choose the component of the generator in Z p 2 to have order p 2, then the subgroups generated by (kp, 1) for 0 k < p 2 exhaust our options. This gives p 2 subgroups of order p 2. If we choose the component of the generator in Z p 2 to have order p, then (kp, p) for 0 k < p 2, gcd(k, p) = 1 exhaust our options. This yields p 1 subgroups; each one contains exactly p distinct elements of the form (kp, p) with requirements of k given above. Finally we may choose 0 to be the component of the generator in Z p 2, and (p, 0) gives another subgroup. This yields p 2 +p cyclic subgroups. Any non-cyclic subgroup of order p 2 would have to have all elements of order 1 or p. There are 48 elements of order p in this group, and they together with the identity form one remaining subgroup generated by (p, 0) and (0, p 2 ). Hence the total number comes to p 2 + p + 1 subgroups of order p 2. Problem 15. Every finite subgroup of Q/Z is cyclic. Proof. (Proof 1) Since S < Q/Z is finite, it has a smallest element s S (in the natural order of Q). We claim that s generates the group. If there is an element t not in s, then there exists a k such that ks < t < (k + 1)s (Archimedean Property). In particular, 0 < t ks < s, so t ks S and s is not the smallest element, a contradiction.

5 5 Proof. (Proof 2, the way they wanted us to do it) Let G be this group. Since G is finite, it is a torsion group, and by Exercise 7(b) G = p prime G(p) where all but finitely many of the summands are the identity. Note that G(p) consists of all the elements of G whose reduced fractions have p n in the denominator; hence G(p) < Z(p ). By Ex I.3.7(d), this implies that G(p) is cyclic group of order a power of p. Hence G = Z n p 1 Z 1 p nm m where each prime is distinct. Since the gcd of any two of these indices is 1, we may by induction show that this is congruent to Z n p 1, which completes our claim pnm m Problem 6. Let G be a group action on a set S containing at least two elements. Assume that G is transitive. Prove (a) for x S, the orbit x of x is S; (b) all the stabilizer G x are conjugate; (c) if G is faithful, N G, and N < G x for some x S, then N = e ; (d) for x S, S = [G : G x ]; hence S divides G. Proof. (a) This follows from the definition of transitive. (b) Given x, y S, let g G be such that g x = y. Then g G x g 1 y = g 1 G x x = g 1 x = y. So g G x g 1 G y. Similarly, g 1 G y g x = x, so g 1 G y g G x G y gg x g 1. (c) Since G is faithful, x S G x = {e}, so in particular if there is e n N there is some y S such that n / G y. Yet since G x and G y are conjugate by (b), n cannot be in N. Hence N = {e}. (d) Let g 1 G x = eg x,..., g n G x be the distinct cosets of G x in G. Then g i x g j x for i j (otherwise g 1 i g 2 G x ) Clearly g i G x x = g i x, and this exhausts all of the elements of G in the acting set. But G is transitive, so {g i G x x} n 1 = S. Finally, note that S [G : G x ] G x = G. Problem 10. Show that the center of S 4 is {e}; conclude that S 4 is isomorphic to the group of all inner automorphisms of S 4. Proof. Any non-transitive non-trivial cycle is clearly not in C(S 4 ) since it does not commute with any cycle with elements both inside and outside the set of numbers that cycle operates on. Otherwise, (s 1 s 2 )(s 3 s 4 ) does not commute with (s 2 s 3 ), and (s 1 s 2 s 3 s 4 ) does not commute with (s 1 s 2 ) since (s 1 s 2 )(s 1 s 2 s 3 s 4 )(s 1 s 2 )[s 1 ] = s 3. The second statement follows from the proof of C4.7(ii) since the homomorphism is given by conjugation and hence has Inn(S 4 ) as its image. The kernel is C(G) = {e}, so this is an isomorphism. Problem 13. If a group G contains a proper subgroup of finite index, it contains a proper normal subgroup of finite index. Proof. Let [G : H] = n. Let G act on S, the set of cosets of H, by translation. This action induces a homomorphism G A(S), where A(S) = n! The kernel of this homomorphism then has index at most n!. Problem 14. If G = pn, p > n, p prime, and H is a subgroup of order p, then H G. Proof. Let G act on the set S of the n cosets of H by translation. This induces a homomorphism φ : G A(S) of order n! Note that p n! since p > n is prime. Hence Ker φ Im φ = G Ker φ l = pn where l n! l = n Ker φ = p. Since Ker φ < H by P4.8, Ker φ = H and H is normal. Problem 15. If a normal subgroup N of prime order p is contained in a group G of order p n, then N is in the center of G.

6 6 Proof. Let G act on the elements of N by conjugation. This is a well-defined action since N is normal in G. We claim this leads to a homomorphism φ from G to Aut(N) where g τ g and τ g (n) = gng 1 for n N. Clearly these are automorphisms on N since they are injective homomorphisms (τ g (n 1 n 2 ) = gn 1 n 2 g 1 = gn 1 g 1 gn 2 g 1 = τ g (n 1 )τ g (n 2 )). Now any automorphism on N must send the identity to itself, and since N is cyclic there are p 1 possible automorphisms on N. But since Ker φ Im φ = p n and p is prime, Im φ = 1, so every conjugation acts trivially on the elements of N and N < C(G). 2.5 Problem 1. If N G and N, G/N are both p-groups, then G is a p-group. Proof. Let G G/N be the canonical projection. Then any element of G maps to some element of order a power of p - say g pk - so this element is in the kernel of the projection - i.e., g pk N. Hence (g pk ) pl = g pk+l = e since N is a p-group, and since g G is arbitrary, G is a p-group. Problem 2. If G is a p-group, H G, and H {e}, then H C(G) {e}. Proof. First, if H is normal in G, then H = {ghg 1 h H} is the union of conjugacy classes of G. Since each conjugacy class has order equal to the index of the centralizer of some element in G, its order divides G = p n. Hence if H 1, then p H, and hence at least p of the conjugacy classes in G making up H must have order 1 (since {e} is one of those conjugacy classes) by the class equation. Yet any conjugacy class of order 1 has entries in C(G), so H C(G) {e}. (Thanks to stackexchange user Arturo Magidin for the hint.) Problem 3. Let G = p n. For each k, 0 k n, G has a normal subgroup of order p k. Proof. This is clear for k = 0; we shall show an induction step and apply mathematical induction. Let H k be a normal subgroup of order p k in G. Then G/H k is also a p-group and hence has a non-trivial center, also a p-group. Let g k+1 H k be an element of order p in the center; then {g k+1 h : h H k } =: H k+1 is a subgroup of G of order p k+1 since H commutes with g k+1. Now for g G, g gh k, so it suffices to show that gh k is normal in G/H k. But this is clear since g k+1 H k is in the center of G/H k. Problem 4. If G is an infinite p-group, then either G has a subgroup of order p n for each n 1 or there exists m N {0} such that every finite subgroup of G has order p m. Proof. This is equivalent to showing that every finite subgroup of order p n in G contains a subgroup of order p n 1. This is shown in the First Sylow Theorem. Problem 9. If G = p n q, with p > q primes, then G contains a unique normal subgroup of index q. Proof. The existence of a (not necessarily normal) subgroup of index q is given by First Sylow Theorem. Call this subgroup H. The number of Sylow p-subgroups must be (kp + 1) for some k N {0} and must divide G. Since kp + 1 does not divide p for k > 0, it would have to divide q; but kp + 1 > p > q, so this is impossible, k = 0, and by C5.8 H is the unique Sylow p-subgroup.

7 7 Problem 10. Every group of order 12, 28, 56, 200 must contain a normal Sylow subgroup and hence is not simple. Proof. The Third Sylow Theorem shows that the number of 7-Sylow subgroups or 5-Sylow subgroups in any group of order 28 or 200 respectively must be unique; applying C5.8 yields the statement. If there are 4 3-Sylow subgroups in a group of order 12, each have empty intersection pairwise, so there are 8 elements of order 3 in this group, leaving three elements which together with the identity must build the unique Sylow 2-subgroup. If there are 8 Sylow 7-subgroups in a group of order 56, each have empty intersection pairwise, so there are 48 elements of order 7 in this group, leaving seven elements which together with the identity must buidl the unique Sylow 2-subgroup. Third Sylow Theorem and C5.8 complete the proof. Problem 12. Show that every automorphism of S 4 is an inner automorphism, and hence S 4 = Aut S4. Proof. Our finding that the center of S 4 is the identity yielded that S 4 = Inn S4 in Exercise II.4.10, so the first part of the statement is all that is necessary to prove. Any automorphism of S 4 must permute the four Sylow 3-subgroups of S 4 : (123), (124), (134), (234). (There are no more by Third Sylow Theorem.) Note that any automorphism that fixes (123) must fix 4 - similarly for the other three Sylow 3-subgroups of S 4. Hence if f, g are automorphisms of S 4 such that f(p i ) = g(p i ) for {P i } 4 1 the Sylow 3-subgroups of S 4, we have g 1 f(p i ) = P i for all i [4], implying that g 1 f fixes each element of [4] and hence that g 1 f is the identity f = g. Therefore there are at most 24=4! automorphisms of S 4 - those that permute the Sylow 3-subgroups - and since there are 24 inner automorphisms of S 4 and Inn S 4 < Aut S 4 we have Aut S 4 = Inn S 4. This leads to a problem interesting in its own right: Problem. Any group of order 24 is not simple. Proof. G has either 1 or 3 Sylow 2-subgroups. If it has 3, then conjugation sends these Sylow 2-subgroups to themselves, inducing an action of G on these subgroups P i and further inducing a homomorphism φ : G S(P i ) = S 3 given by how G interacts with these subgroups. The kernel of this homomorphism must have order 4 by order arguments. But Ker φ G. The next few problems are from Artin s Algebra, with solutions offered here as a way to continue practicing group actions. Problem. A nonabelian simple group cannot operate nontrivially on a set with fewer than five elements. Proof. Once it has been determined that the smallest nonabelian simple group is of order 60, the proof of this follows the same process as above. Here is another solution summarized from a McGill s professor s website: it must be the case that the kernel of the homomorphism G S 4 induced by this action on a set of four (or fewer) elements is trivial, so the homomorphism is injective and we may consider G < S 4 by inclusion. Now consider the determinant map φ : G {±}. If ker φ = G, then G is contained in A 4, and we have

8 8 already shown above that A 4 is not simple. Orders 6 and 10 are products of two primes, and orders 8 and 9 are powers of primes, exhausting all options for G; hence ker φ G. If ker φ = {e}, all nontrivial elements of G consist of odd permutations of order 2 and hence is Abelian. This completes the proof. Problem. If H is a proper subgroup of a finite group G, then G is not a union of conjugates of H. Proof. Let G act on N G (H) by conjugation. Now N G (H) H, so G = O H G H O(H) = G G H G H. Yet since every conjugate of H has at least a trivial intersection, the union of the conjugates of H can have at most O(H) ( H 1) + 1 elements, which is strictly less than G since H is proper. Problem. Prove that no group of order 224 is simple. Proof. 224 = It is easily check that there are either 1 or 7 2-Sylow subgroups. If there are seven, let φ : G S 7 be the induced homomorphism from the action of G on S 7 by conjugation. Note that S 7 = 5040 = ; in particular, G S 7, so φ is not injective and Ker φ {e}. Problem. Let G be a group of order 30. (a) Prove that either the Sylow 5-subgroup K of the Sylow 3-subgroup H is normal. (b) Prove that HK is a cyclic subgroup of G. (c) Classify groups of order 30. Proof. (a) Omitted; it is a Third Sylow Theorem argument. (b) Let H G. Note that (HK) 5 = H(KHK)H(KHK)HK = H 5 K = K. Let φ : HK K send g HK to g 5. Clear φ is not trivial - try letting h = e - so Ker φ = 5 and hk HK such that h e and (hk) 5 K \{e}. We claim that this element has order 15. If hk = 3, then (hk) 2 K \{e}. That is, hkhk K hkh K. This shows that hkh = hh 1 k by the fact that H G, so hkh = k. Hence h is in the normalizer of K, and since h e, N G (K) = 15 and K G. Hence H K G, and any element (h, k) where h, k e is of order 15. (This is one possible argument; can you find another?) Now let K G. Note (HK) 3 = HK(HKH)K = HK 3 = H. Let φ : HK H map g g 3. By similar reasoning to the above, h H, k e such that (hk) 3 H \ {e}. If hk has order 5, then (hk) 6 = hk H \ {e}, so k = e, a contradiction. Hence hk = 15, and hk generates G. (c) By (b), we have that G = Z 2 φ Z 15 where φ : Z 2 Aut(Z 15 ) = Aut(Z 3 ) Aut(Z 5 ) = Z 2 Z 4. Hence G is uniquely determined by where φ maps 1 Z 2. Since φ(1) {(0, 0), (1, 0), (0, 2), (1, 2)}, these each classify the groups of order 30. Hence there are at most four groups of order 30, and since Z 2 Z 3 Z 5, Z 5 S 3, Z 3 D 5, and D 15 are pairwise non-isomorphic, this completes the proof. We return now to Hungerford. 2.6 Problem 1. Let G, H be groups, θ : H Aut G a homomorphism. Let G θ H be the set G H with the binary operation (g, h)(g, h ) = (g[θ(h)(g )], hh ). Show that G θ H is a group.

9 9 Proof. Closure is clear from definition. Associativity takes a bit of bookkeeping: [(g, h)(g, h )](g, h ) = (gθ(h)(g ), hh )(g, h ) = (gθ(h)(g )θ(hh )(g ), hh h ) θ hom = (gθ(h)(g )θ(h)(θ(h )(g )), hh h ) θ(h) Aut G = (gθ(h)(g θ(h )(g )), hh h ) = (g, h)(g θ(h )(g ), h h ) = (g, h)[(g, h )(g, h )]. We note that (g, h)(e, e) = (gθ(h)(e), h) θ(h) Aut G = (g, h) and (e, e)(g, h) = (θ(h)(g), h) = (g, h), yielding that (e, e) is the identity element of G θ H. Finally, (θ(h 1 )(g 1 )θ(h 1 )(g), e) θ(h 1 ) Aut g = (e, e) and (g, h)(θ(h 1 )(g 1, h 1 )) = (gθ(h)(θ(h 1 (g 1 )), e) θ hom = (e, e), yielding that (θ(h 1 )(g 1 ), h 1 ) and an inverse element of (g, h). Problem 9. Classify up to isomorphism all group of order 18. Do the same for order 20. Proof. We see that the Sylow 3-subgroup Sy 3 must be normal in the group, so G = Sy 3 φ Z 2. Since any group of order p 2 for p prime is Abelian, Sy 3 is one of two groups: Case 1: Sy 3 = Z9. Then φ : Z 2 Aut(Z 9 ) = (Z 9 ) = Z 2 Z 3. Hence 1 (0, 0) or 1 (1, 0). In the first, case G = Z 2 Z 9 ; in the second, the order 2 elements conjugates with the generator of Z 9 by sending it to its inverse, and this group is easily seen to be D 9. Case 2: Sy 9 = Z3 Z 3. First we note that Aut(Z 3 Z 3 ) = GL 2 (F 3 ). Z 3 Z 3 is generated by (1, 0) and (0, 1). In any automorphism, (1, 0) may map to any of its eight non-identity elements, and (0, 1) may map to anything outside of the cyclic group generated by that element, yielding six choices for (0, 1). Hence there are 48 elements in Aut(Z 3 Z 3 ). By a similar line of reasoning, there are eight choices for the first column of M GL 2 (F 3 ) and six choices for the second column (columns must be linearly independent; that is, the second column cannot be contained within the cyclic group generated by the first column), so there are 48 elements in GL 2 (F 3 ). Finally, GL 2 (F 3 ) acts naturally on Z 2 3 as a vector space, inducing an isomorphism. Any element of order 2 must have minimal polynomial (in the linear algebra sense) of x 2 1. Hence its eigenvalues can be either {1, 1} or { 1, 1}. (Thanks to stackexchange user Derek Holt for this tip.) This leads naturally to the matrices which send generators g, h of each copy of Z 3 to the following: g g, h h (trivial), g g 1, h h, or g g 1, h h 1. The first yields G = Z 2 Z 3 Z 3, the second G = Z 3 S 3 (note the inverse flip under conjugation between the generators of Z 2 and one copy of Z 3 ), and a group G = x, y, z x = 2, y = z = 3, xyx 1 = y 1, xzx 1 z. This exhausts all cases, so these are the five groups of order 18. Now for groups of order 20. The Sylow 5-subgroup is normal, so G = Z 5 φ Sy 2. Again, there are two cases for Sy 2 : Case 1: Sy 2 = Z4. Then φ : Z 4 Aut(Z 5 ) = Z 4 can map 1 to the automorphism 1 1, 1 2, 1 3, and 1 4. We claim the middle two are isomorphic. This stems from the

10 10 fact that Z 5 φ1 2 Z 4 = x, y x = 4, y = 5, xyx 1 = y 2 = x 1, y x 1 = 4, y = 5, x 1 yx = y 3 = Z 5 φ1 3 Sy 2. The other two are neither isomorphic to each other - the group generated from 1 1 is Abelian while 1 4 is not - nor to the group in question (x in the group in question can only map to x and x 1, y to y or y 1, and none of these give 1 4), so this generates the groups Z 5 Z 4, G 2 := x, y x = 4, y = 5, xyx 1 = y 2, and G 4 := x, y x = 4, y = 5, xyx 1 = y 1. It is easy to check that these are indeed groups. Case 2: Sy 2 = Z2 Z 2. Then φ : Z 2 Z 2 Aut(Z 5 ) = = (Z 5 ) = Z 4 can either map trivially - in which case G = Z 2 Z 2 Z 5 - or it can map one of the generators of Z 2 Z 2 to the automorphism taking 1 4 in (Z 5 ) and the other to the identity. This yields Z 5 Z 2 G and the element of order 2 in the copy of Z 2 that is not normal in G to act on Z 5 Z 2 by conjugation by sending elements to their inverse. This is D 10. This exhausts all options, and hence the groups are Z 4 Z 5, Z 2 Z 2 Z 5, D 10, G 2, and G 4. It is worth a note that there is one more map to consider in discussing possible maps φ : Z 2 Z 2 (Z 5 ), which is the one that has both generators of Z 2 Z 2 act on Z 5 by conjugation. We show that this case reduces to one of our other cases. Note that this would generate the group x, y, z x = y = 2, z = 5, xzx 1 = z 1 = yzy 1. Since x = x 1, this implies that z = xyz(xy) 1 and hence that z C(xy) Z 5 < N( xy ) xy G (since all of Z 2 Z 2 is in the normalizer of this as well). Since xy = 2, we have that Z 5 Z 2 G, so this reduces to the case of D 10. Here are some interesting problems from previous A& M exams: Problem 1. (J14.4, modified) Let R be a commutative ring. Let L, M, N, A be R-modules. Suppose that 0 L f M g N 0 is a short exact sequence. Prove 0 Hom(A, L) f Hom(A, M) g Hom(A, N) and Hom(L, A) g Hom(M, A) f Hom(N, A) 0 are short exact sequences where f, g are the contextually-derived induced homomorphisms of f, g, respectively (for instance, if f : L M, then f : Hom(A, L) Hom(A, M) is defined by g fg and f : Hom(M, A) Hom(L, A) is defined by g gf). Proof. We start with the first case. We want to show two things: f is injective, and im f = ker g. Let f (φ) = fφ = fψ = f (ψ). Then since f is injective, g = h. If h im f, there exists φ Hom(A, L) such that f (φ) = fφ = h. Since im f = ker g, we have gh = gfφ = 0, so h ker g and im f ker g. If h ker g, then gh(x) = 0 for any x. Hence h(x) is in the kernel of g, so for any element x M, there exists an element l in L such that f(l) = h(x) since im f = ker g. This element l is unique since f is injective. Define φ Hom(A, L) by g(x) = l. This is a homomorphism, and fg(x) = f(l) = h(x), so h im f, and this completes the first part. Now we consider the second case. We want to show two things: g is injective, and im g = ker f. (This case is the dual of the first case, but there are interesting differences, so we include the proof.) If g (φ) = φg = ψg = g (ψ), then φ(x) = φ(g(y x )) for some y x M since g is surjective, which in turn equals ψ(g(y x )) = ψ(x), so φ = ψ. If h im g, there exists an element φ Hom(N, A) such that φg = h hf = φgf = 0 since im g = ker f, so h ker f. If h ker f, then hf = 0, so 0 = h(ker f) = h(im g). We can then form h : M/ ker g A mapping m + ker f h(m), and we can also write the canonical

11 11 isomorphism ḡ : M/ ker g N; this modding out by the kernel yields an isomorphism since g is surjective. Hence for any n there is a unique element m+ker g such that f(m+ker g) = n. Define ψ(n) = h(m). (Note that this is well-defined; this function is defined by composing the inverse map of ḡ with f.) Then ψ is a homomorphism, and g (ψ) = ψg = h by definition of ψ. So h im g and we are done. Problem 2. Let R be an integral domain with field of fractions K, and let K be an algebraic closure of K. Fix α K. Suppose that M K is a finitely generated R-submodule such that αm M. Prove that there is a monic polynomial f R[x] such that f(α) = 0. Proof. We have that det(a Iα) = f(α) for some matrix A M n n (R) and f R[x]. We will find a specific characteristic polynomial and then apply Cayley-Hamilton to complete the proof. Let {m i } n 1 be a set of generators for M. Then since αm M, αm i M for all i [n], so αm i = n j=1 c ijm j for any i [n]. Define A := [a ij ]. Then A[m i ] = [αm i ] (A αi)[m i ] = 0. Hence det(a αi) = f(α) = 0. Problem 3. Let R be a commutative ring, and let M be a noetherian R-module. I := {r R m M, rm = 0}. Prove that R/I is a noetherian ring. Set Proof. Since the terms noetherian ring and noetherian module are not always defined in some algebra courses, we define them quickly. A noetherian module is a module such that every submodule is finitely generated iff it satisfies the ascending chain condition: if A 1 A 2 are modules, then there exists an n such that A m = A n for all m n. A noetherian ring is a ring such that every prime ideal is finitely generated iff it satisfies the ascending chain condition (for ideals). Principal ideal domains (as rings and as modules over themselves) are Noetherian. Let {m i } n 1 be a finite set of generators for M, and define a map R M n by r (rm 1,..., rm n ). Then the kernel of this map is exactly the annihilator of M, so R/I M n as a submodule. Since M n is finitely generated (naturally), R/I is finitely generated. Problem 4. Let K be a field. Prove that the polynomial ring K[x] has infinitely many maximal ideals. Proof. This problem reduces to showing that K[x] has infinitely many irreducible elements, even after multiplication by a unit. If K is infinite, (x + a) for a K is sufficient. If K is finite, chark = p for p prime. If there are only finitely many irreducible polynomials in K[x], then the algebraic closure of K is finite dimensional over K and hence is finite of degree n. But then this algebraic closure contained the splitting field of the separable polynomial x pn x over K, a contradiction. Since K[x] is a principal ideal domain, every ideal of K[x] not contained in K contains one irreducible element, so there are infinitely many maximal ideals in K[x]. (You can also use the theorem that for any n there is an irreducible polynomial of degree n over a finite field - namely, x pn x.) Problem 5. Let V be a finite dimensional vector space over C, and suppose we have C- linear maps A 1,..., A k : V V such that for all i, j we have A i A j = A j A i. Show that there exists a non-zero vector in V that is simultaneously an eigenvector for each of these maps (with possible different eigenvalues).

12 12 Proof. We use induction on k. The base case is trivial. If there exists an element v which is an eigenvector for A i v for i [k 1], note that A i (A k v) = A k (A i v) = A k (λ i v) = λ i (A k v), so that multiplication by A k preserves the joint eigenspace of {A 1,..., A k 1 }. Now restrict A k to this joint eigenspace (well-defined by this preservation). Since C is algebraically closed, we may consider this eigenspace as the set of linear combinations of some basis of eigenvectors. This intersects non-trivially with the eigenspace of A k restricted to this joint eigenspace, and this subspace is as desired. Mathematical induction completes the proof. Problem 6. (Part (b) only) Let R be a commutative ring, and let P, F be left R-modules. Assume P, F are finitely generated, that P is projective, and that F is free. Prove that Hom R (P, F ) is a projective R-module. Proof. Since P is projective, F = P K for F free and some R-module K and where F can be chosen to be finitely generated (by the proof of this equivalence). We first claim Hom( F, F ) is free. If {f i } n 1 forms a basis for F, {g i } m 1 forms a basis for F, it is not hard to show that {δ ij } n,m i,j=1, the Kronecker deltas, form a basis for this Hom group. Clearly each of these exist in the Hom group. The rest follows since any element in Hom( F, F ) is completely and uniquely determined by how each f i maps to some f j for i [n] and j [m]. Then it is easy to see that Hom( F, F ) = Hom(P, F ) Hom(K, F ) by the restriction maps. Problem 7. Let R be an integral domain. R-module is projective. Show that R is a field if and only if every Proof. If R is a field, every R-module is (by definition) a vector space and is therefore free and hence projective. If R is not a field, let I be a maximal ideal of R and consider R/I as an R-module. We claim that following diagram cannot commute: h π R/I id R R/I 0 for any h an R-module homomorphism. Since (0) I, then for 0 a R, 0 i I, π(a + i) = π(a). But then h must map a + I to both a + i and to a in R, contradiction. Problem 8. Let k be a field, a k, and let p be a prime number. Prove that the polynomial x p + a is either irreducible or has a root in k. Proof. Let E be the splitting field of x p + a over k. Then x p + a = (x z 1 ) (x z p ) for {z 1,..., z p } roots of the polynomial in E. If x p + a = fg for f, g of degree 1, f factors into (x z 1 ) (x z n ) for some n < p. We have z := n 1 z i is equal to the additive inverse of the constant term of f and hence is an element in k. Furthermore z p = ( a) n, since (z 1 z n ) p = z p 1 zp n = ( a) n. Since n, p are relatively prime, there exist c, d such that cn + dp = 1. Then (z c ( a) d ) p = z cp ( a) dp = ( a) cn+dp = a. This is an element in k, so our polynomial has a root in k. Problem 9. Suppose z is any generator for the unit group of F 4 k. Prove that x 2k +x+z 2k +z has exactly 2 k roots in F 4 k.

13 13 Proof. First note that F 4 k has order a power of 2, so it is of characteristic 2. Hence x 2k + x + z 2k + z = (x + z) 2k (x + z) by Freshman s Dream. This polynomial divides (x+z) 22k +(x+z); indeed, this polynomial is its square. Now the roots of (x+z) 22k +(x+z) and x 22k + x are clearly in bijective correspondence and are all contained in F 4 k, so this former polynomial has 4 k roots, so each factor of (x + z) 2k + (x + z) must have 2 k roots exactly. Problem 10. (b only) A group is said to be locally finite if every finitely generated subgroup of the group is finite. Suppose that G is a group containing a normal subgroup K such that K and G/K are both locally finite. Show that G is locally finite. Proof. Map G G/K in the canonical way. Any finitely generated subgroup H in G has finitely many generators in its image through this map, so its index over K is finite. We use the fact that every subgroup of finite index in a finitely generated subgroup is itself finitely generated to get that, since [H : H K] < by this fact, H K is itself finitely generated and is hence finite. Since H = [H : H K] H K, we are done.