MATH 251, Handout on Sylow, Direct Products, and Finite Abelian Groups

MATH 251, Handout on Sylow, Direct Products, and Finite Abelian Groups 1. Let G be a group with 56 elements. Show that G always has a normal subgroup. Solution: Our candidates for normal subgroup are the p-sylow subgroups of G. Since G =2 3 7 then n 2 1 (mod 2) and n 2 7, and n 7 1 (mod 7) and n 7 8. So, the possible values for n 2 and n 7 are n 2 =1, 7 and n 7 =1, 8. Note that if any of these values is one then we obtain a normal subgroup. However, if n 2 = 7 and n 7 = 8 none of the p-sylow subgroups would be no normal subgroup. Let us see that this situation is impossible. Assume n 2 = 7 and n 7 = 8. Note that a 7 Sylow subgroup has order 7 and therefore it is cyclic so, two distinct 7-Sylow subgroups intersect trivially. We are assuming n 7 = 8 then the number of elements in the union of the 8 7-Sylow subgroups is 8 6 + 1 = 49. The group has 56 elements and 49 of them are in the 7 Sylows so we have only 7 elements to distribute in all the 2-Sylows (they are 7!!!!). Since each 2-Sylow has 8 elements we get a contradiction (we do not have enough number of elements in G). So, either n 2 = 1 or n 7 = 1 or both are equal to one. Anyway, there is a normal subgroup. 2. Classify all groups of order 21 up to isomorphism. Solution: G =3 7. First of all we compute the possible values for the number of p-sylow subgroups: n 3 =1, 7 and n 7 = 1. Then, the 7 Sylow is normal and (because of its order) it is isomorphic to Z 7. Also, we can easily check that the 3-Sylow subgroups are isomorphic to Z 3. Moreover, the Sylow subgroups intersect trivially. Since Z 7 is normal in G and G = Z 3 Z 7 then G is the semidirect product of Z 3 and Z 7. Recall that the direct product is a particular case of the semidirect product. In order to see all the possibilities for the group G we have to see all the possibilities for the function (the one that defines the semidirect product). Let s do that. Since : Z 3! Aut(Z 7 ) we need to figure out what Aut(Z 7 )is. Let g be the generator of Z 7 ( I am using multiplicative notation here, so the elements of Z 7 look like g i and g 7 = e). Then, an automorphism of Z 7 is defined by its action on g, which means that it is defined by g 7! g i some i. Since the function has to be bijective then g i has to generate Z 7. This happens whenever i and 7 are relatively prime so, i can be 1, 2,, 6. It might take some time to prove (DO IT!!!), but Aut(Z 7 ) = Z 6 is generated by : g 7! g 3. We need to find all the possible isomorphisms : Z 3! Z 6. As usual, we have to define in the generator of Z 3 (call it y). Since the order of y is three then the order of (y) can be 1 or 3. Then, (y) can be e (g 7! g) or 2 (g 7! g 9 = g 2 ). Finally, we see what the products are and we find the two non-isomorphic groups of order 21. One per each of the s. i G 1 = Z 3 n Z 7 where (y) =e. Another way to write this is G 1 = Z 3 Z 7 where ygy 1 = e(g) =g. That is, G 1 = Z3 Z 7 = Z21. ii G 2 = Z 3 n Z 7 where (y) = 2. Another way to write this is G 2 = Z 3 Z 7 where ygy 1 = 2 (g) = g 2. 3. Describe all Abelian groups of order 63. Prove or disprove that all groups of order 63 are Abelian. Solution: G = 63 = 3 2 7. We use (again) the classification theorem of finite Abelian groups to get that G is isomorphic to either Z 63 or Z 3 Z 21. Now consider any G of order 63. We use one of Sylow s theorems to get that n 7 1 mod 7 and that n 7 divides 9, thus n 7 = 1. It follows that there exists a unique 7-Sylow subgroup P 7, which is normal in G. 1

On the other hand, n 3 1 mod 3 and that n 3 divides 7, so n 3 could be equal to 1 or 7. If n 3 =1then there is a unique 3-Sylow subgroup P 3, which is normal. It follows that G is the direct product of its Sylow subgroups, as groups of order 7 and 3 2 are Abelian, then G is abelian. If n 3 = 7, then there exists the possibility of a semi-direct product (as P 7 is normal). In order to see whether or not this actually happen, we have to see if we can construct a homomorphism : P 3! Aut(P 7 ), or equivalently : P 3! Z 6. Since we are looking for an example, consider P 3 = Z9 and the homomorphism by (1) = 2. It follows that there is at least one non-abelian group of order 63. : Z 9! Z 6 defined 4. Let G be a group of order 231. Show that G has subgroups of every possible order dividing 231. Solution: G = 231 = 3 7 11. By Cauchy s theorem, there are elements of order 3, 7, 11 in G, therefore there are subgroups of G having those orders. We use one of Sylow s theorems to find the number of p-sylow subgroups of G for p =3, 7, 11, we get: n 3 =1, 7, 77, n 7 = n 11 = 1, so the p-sylow subgroups P 7 and P 11 are normal in G. Let P 3 be a 3-Sylow subgroup of G (there might be one or more, we do not know). Since P 7 and P 11 are normal in G then P 7 P 3, P 11 P 3 and P 7 P 11 are subgroups of G, thus we obtain subgroups of G having orders 21, 77 and 33. We are done. 5. Any group of order 45 is Abelian. Solution: Let G have order 45. In G, a 3-Sylow subgroup has order 9 and a 5-Sylow subgroup has size 5. Using one of the Sylow theorems, n 3 5, n 3 1 (mod 3), thus n 3 = 1 and n 5 9, n 5 1 (mod 5) implies n 5 = 1. Therefore G has normal 3-Sylow and 5-Sylow subgroups. Denote them by P 3 and P 5 respectively, so P 3 = 9 and P 5 = 5. Then P 3 is Abelian (group of order p 2!!) and P 5 is cyclic (thus Abelian). Since P 3 and P 5 intersect trivially, then G = P 3 P 5, and thus G is Abelian. 6. Let G be a group of order p 2 q,wherep and q are distinct prime numbers. Show that G must contain a proper non-trivial normal subgroup. Solution: We know that n p 1 mod p and that n p q, soifq<pthen n p must be equal to one. Now consider the case when q>p. We know that n q p 2, which implies that n q =1,p, or p 2, but as n q 1 mod q and q>p,theneithern q = 1 or tq +1=p 2 for some t<q. If we assume that n q 6=1 then tq = p 2 1=(p 1)(p + 1) so, q divides (p 1)(p + 1). Hence, using that q is prime we have that q divides p 1 or p + 1. The only possibility (as q>p) is that q = p + 1, this forces q = 3 and p = 2, which means that G = 12. Let G be a group of order 12, then n 2 can be 1 or 3, and n 3 can be 1 or 4. If n 2 = 3, and n 3 = 4, then the number of elements in the Sylow 3-subgroups is 4 2 + 1 = 9, which leaves exactly 3 elements to be distributed among three groups of order 4, that is impossible. Contradiction. 7. Prove that if p is a prime number and Z p a = G1 G 2 then, either G 1 = Zp a or G 2 = Zp a. Solution: The easiest way to see this is to use that the representation of a finite abelian group into a product of cyclic p-groups is unique up to re-ordering of the factors. So, if Z p a = G1 G 2, then the groups G 1 and G 2 are cyclic p-groups themselves. Thus, this would yield two di erent representations for the same group (Z p a)intocyclicp-groups. 2

8. Using that a group of order 2p, for p a prime number, is either abelian or isomorphic to D p, and that there are exactly two non-abelian groups of order 8. List all non-isomorphic groups of orders less than 12. Solution: Using the hints, the non-abelian groups of order less than 12 are S 3,D 4,Q 8, and D 5. So, now we just have to list all the Abelian ones. Order 2 = Z2 Order 3 = Z3 Order 4 = Z4 = Z 2 Z 2 Order 5 = Z5 Order 6 = Z2 Z 3 = Z6 Order 7 = Z7 Order 8 = Z8 = Z 4 Z 2 = Z 2 Z 2 Z 2 Order 9 = Z9 = Z 3 Z 3 Order 10 = Z 2 Z 5 = Z10 Order 11 = Z 11 9. Find all Abelian groups of order 108. Solution: G = 108 = 2 2 3 3, then we can write 108 using powers of primes as 108 = 2 2 3 3 =2 2 3 2 3=2 2 3 3 3 108 = 2 2 3 3 =2 2 3 2 3=2 2 3 3 3 It follows that G is isomorphic to exactly ONE of the following groups G = Z 2 2 Z 3 3 = Z 2 2 Z 3 2 Z 3 = Z 2 2 Z 3 Z 3 Z 3 = Z 2 Z 2 Z 3 3 = Z 2 Z 2 Z 3 2 Z 3 = Z 2 Z 2 Z 3 Z 3 Z 3 We know get the other form by collecting largest p-groups available for all primes (in this case 2 and 3). We obtain that G is isomorphic to exactly ONE of the following groups G = Z 108 = Z 36 Z 3 = Z 12 Z 3 Z 3 = Z 54 Z 2 = Z 18 Z 6 = Z 6 Z 6 Z 3 10. Find all Abelian groups of order 900. Solution: G = 900 = 2 2 3 2 5 2. We repeat the process above to get 900 = 2 2 3 2 5 2 =2 2 3 2 5 5=2 2 3 3 5 5=2 2 3 3 5 2 3

900 = 2 2 3 2 5 2 =2 2 3 2 5 5=2 2 3 3 5 5=2 2 3 3 5 2 So, G is isomorphic to exactly ONE of the following groups G = Z 2 2 Z 3 2 Z 5 2 = Z 2 2 Z 3 2 Z 5 Z 5 = Z 2 2 Z 3 Z 3 Z 5 Z 5 = Z 2 2 Z 3 Z 3 Z 5 2 = Z 2 Z 2 Z 3 2 Z 5 2 = Z 2 Z 2 Z 3 2 Z 5 Z 5 = Z 2 Z 2 Z 3 Z 3 Z 5 Z 5 = Z 2 Z 2 Z 3 Z 3 Z 5 2 or G is isomorphic to exactly ONE of the following groups G = Z 900 = Z 180 Z 5 = Z 60 Z 15 = Z 300 Z 3 = Z 450 Z 2 = Z 90 Z 10 = Z 30 Z 30 = Z 150 Z 6 11. Find all Abelian groups of order 5 6. Solution: We get 5 6 = 5 5 5=5 4 5 2 =5 4 5 5=5 3 5 3 =5 3 5 2 5=5 3 5 5 5 = 5 2 5 2 5 2 =5 2 5 2 5 5=5 2 5 5 5 5=5 5 5 5 5 5 So, G is isomorphic to exactly ONE of the following groups G = Z 5 6 = Z 5 5 Z 5 = Z 5 4 Z 5 2 = Z 5 4 Z 5 Z 5 = Z 5 3 Z 5 3 = Z 5 3 Z 5 2 Z 5 = Z 5 3 Z 5 Z 5 Z 5 = Z 5 2 Z 5 2 Z 5 2 = Z 5 2 Z 5 2 Z 5 Z 5 = Z 5 2 Z 5 Z 5 Z 5 Z 5 = Z 5 Z 5 Z 5 Z 5 Z 5 Z 5 12. Let p > q be prime numbers such that q divides p 1. Let and be two injective homomorphisms of Z q into Aut(Z p ) = Z p 1. Prove that Z p o Z q = Zp o Z q Solution: In order for all to make better sense I will use only multiplicative notation... but let us not forget that both Z p and Z q are additive. I will not use [] p, nor [] q for elements, so let a, b 2 Z p and s, t 2 Z q. 4

We want to show that : Z p o Z q! Z p o Z q defines an isomorphism, where ((a, t) )=(a, t r ), and (t) = (t r ). Also note that this last equation means that t = t r. Since is a function between sets having the same number of elements, then the only thing we need to check is that it is a homomorphism. ((a, t) (b, s) ) = ((a t (b),ts) ) = (a t (b), (ts) r ) = (a t r(b), (ts) r ) = (a t r(b),t r s r ) = (a, t r ) (b, s r ) = ((a, t) ) ((b, s) ) Done. Note that from line 3 to line 4 we used that Z q is Abelian. 13. Show that the groups Z 30,D 30, Z 3 D 10, and Z 5 D 6 are mutually nonisomorphic. Solution: First of all, Z 30 is abelian and the other three groups are not. Let G be any of the groups Z 3 D 10, Z 5 D 6 or D 30. Note that an element of order 2 in any of the first two groups looks like (a, b) then, since 2 divides neither 5 nor 3, then the element has to have the form (0,b). We know that D 2n has n subgroups of order 2 (one per flip) then: Z 3 D 10 has 5 subgroups of order 2, Z 5 D 6 has 3 and D 30 has 15. Then, the groups are not isomorphic to each other. Other way to see this is to use that Z(A B) =Z(A) Z(B). Then it is easy to see that the centers of the four groups we are working with are all distinct. 14. Verify the following isomorphisms: (a) (Z 5 Z 3 ) o '1 Z 2 = D10 Z 3 (b) (Z 5 Z 3 ) o '1 Z 2 = Z5 D 6 (c) (Z 5 Z 3 ) o '1 Z 2 = D15 Solution: In the previous problem we found 4 non-isomorphic groups of order 30. In this problem we are given the only three non-abelian ones (this is something you should prove... there are no more possible ' s to consider). So, each of the groups we found in the previous problem has to be isomorphic to one of the ones given in this problem. I will give two di erent proofs for this problem: Proof 1. An element of order 2 in (Z 5 Z 3 ) o ' Z 2 looks like x =(a, b) wherea 2 Z 5 Z 3 and b 2 Z 2. Let s see what x 2 looks like. x 2 =(a, b) (a, b) =(a' b (a),b 2 ). Since x has order 2 then (e, e) =(a' b (a),b 2 ). We can see from here that b 2 = b, this holds always for b 2 Z 2. In order to analyze the first component we need to see what ' is... the first components tells us that a = a' b (a). Clearly the only interesting case to check is when b is not trivial. Let a =(x, y) wherex 2 Z 5 and y 2 Z 3. If ' = ' 1 : e = a' b (a) implies(0, 0) = (x, y)+( x, y) (Note that the operation in Z 5 Z 3 is addition). Then, x is free of condition and 2y = 0 forces y = 0. So, any element of the form [(x, 0),b] has order 2. These are 5. Then, the group corresponding to ' 1 is D 10 Z 3. If ' = ' 2 : Similarly, the number of groups of order 2 is 3. So, the group of ' 2 is D 6 Z 5. Since there is only one group left, ' 3 corresponds to D 30. Proof 2. Consider the first isomorphism (Z 5 Z 3 ) o ' Z 2 = D10 Z 3 where '(a, b) =( write D 10 = Z 5 o Z 2, then we need to define a homomorphism: a, b). If we :(Z 5 Z 3 ) o ' Z 2! (Z 5 o Z 2 ) Z 3 5

Since the same groups (Z 5, Z 2, Z 3 ) appear with di erent types of products in the domain and range, we try the obvious function: ((a, b),c) =((a, c),b). We need to show that is a homomorphism, then is when the way the semidirect products are defined will become important. [((a, b),c)((a 0,b 0 ),c 0 )] = [((a, b)+'(a 0,b 0 ),c+ c 0 )] read note below = [((a, b)+( a 0,b 0 ),c+ c 0 )] = [((a a 0,b+ b 0 ),c+ c 0 )] because of the sum in direct product = ((a(a 0 ) 1,cc 0 ),bb 0 ) definition of and switching to multiplicative = ((a, c)(a 0,c 0 ),bb 0 ) using ar = r 1 a (product in D 10 ) = ((a, c),b)((a 0,c 0 ),b 0 ) because of the product in direct product = [((a, b),c)] [((a 0,b 0 ),c 0 )] Note: At this point I am abusing that the group used for the semidirect product is Z 2 to simplify notation. The function is clearly onto, just look at the definition. Since the orders of the groups are the same, then the function has to be bijective. The other isomorphisms follow similarly. 15. Show that an abelian group is the direct product of its Sylow p-subgroups for primes p dividing G. Solution: We know this already, because we know that if G is abelian then G = Q Z p j where all the i p i s are, not necessarily distinct, prime numbers and the j s are positive integers... collect all the Z p j s i of order powers of the same prime number, let s say, p. That is a Sylow p-subgroup of G (just check the order). So, G is the direct product of these subgroups. 16. Let be the unique non-trivial homomorphism from Z 4 onto< j> Aut(Z 7 ). Show that Z 7 o Z 4 is generated by elements a and b satisfying a 7 = b 4 = 1 and bab 1 = a 1, and conversely, a group generated by elements a and b satisfying these relations is isomorphic to Z 7 o Z 4. Solution: Let : Z 4! Aut(Z 7 ), since Aut(Z 7 ) = Z 6,then (1) could be either 0 (which gives us the trivial homomorphism) or 3 (which is the only element of order dividing 4 in Z 6. It follows that, (x) =3x for all x 2 Z 7. Now switching to multiplicative notation, thus Z 4 =<b>and Z 7 =<a>we realize that the only element of order two in Aut(Z 7 )isdefinedbya 7! a 1 (here using that the group is Abelian). It follows that the group is generated by (a, e) and (e, b), which we can identify with a and b, and thus (e, b)(a, e)(e, b) 1 = (e, b)(a, e)(e, b 1 ) Since (e, b) 4 =(e, e) and (a, e) 7 =(e, e) we are done. = (e b (a),b)(e, b 1 ) = (a 1,b)(e, b 1 ) = (a 1 b (e),bb 1 ) = (a 1,e) = (a, e) 1 17. Show that D 28 and D 14 Z 2 are both groups of order 28 with Sylow 2-subgroups isomorphic to Z 2 Z 2. Give an explicit isomorphism D 28 = D14 Z 2. Solution: We know D 28 has 14 flips (order 2) and the rest are rotations, of the 13 non-trivial rotations there is one of order 2 but none may have order 4 because 4 does not divide 14. So D 14 does not have elements of order 4. Hence, the 2-Sylow subgroup of D 28 must be Z 2 Z 2. 6

Similarly, we can see that D 14 has elements of order 2 but no elements of order 4. Since all elements in Z 2 have order dividing two, then there are no elements of order 4 in D 14 Z 2. In order to define the isomorphism we use that D 28 = Z14 o Z 2 = (Z7 Z 2 ) o Z 2 and that D 14 Z 2 = (Z7 o Z 2 ) Z 2. Then, the function :(Z 7 Z 2 ) o Z 2! (Z 7 o Z 2 ) Z 2 is defined by: ((a, b),c)=((a, c),b) We need to show that is a homomorphism, then is when the way the semidirect products are defined will become important. Assume b, c non-trivial [((a, b),c)((a 0,b 0 ),c 0 )] = [((a, b) (a 0,b 0 ),c+ c 0 )] = [((a a 0,b b 0 ),c+ c 0 )] = ((a a 0,c+ c 0 ),b b 0 ) = ((a, c)(a 0,c 0 ),b+ b 0 ) = ((a, c),b)((a 0,c 0 ),b 0 ) = [((a, b),c)] [((a 0,b 0 ),c 0 )] If either b or c is trivial, the proof is even easier. So, is a homomorphism. It is clearly onto and since both groups have the same order, then must be bijective. 18. Let N and A be groups, : A! Aut(N) a homomorphism and ' 2 Aut(A). Show that N o A = A o ' A. Solution: Define the function : N o A! N o ' A by: (n, a) =(n, ' 1 (a)) We need to show that is a homomorphism, as usual, this is when the way the semidirect products are defined will become important. Assume b, c non-trivial on the other hand [(n, a)(n 0,a 0 )] = [(n a (n 0 ),aa 0 )] = (n a (n 0 ), ' 1 (aa 0 )) = (n a (n 0 ), ' 1 (a)' 1 (a 0 )) [(n, a)] [(n 0,a 0 )] = (n, ' 1 (a))(n 0, ' 1 (a 0 )) = (n( ') ' 1 (a)(n 0 ), ' 1 (a)' 1 (a 0 )) = (n a (n 0 ), ' 1 (a)' 1 (a 0 )) So, is a homomorphism. It is clearly onto and since both groups have the same order, then must be bijective. 19. Classify the non-abelian group(s) of order 20. Solution: We easily see that n 5 = 1 and that n 2 =1or 5. So, P 5 is normal, and isomorphic to Z 5 because it has 5 elements. The Sylow 2-subgroups could be isomorphic to either Z 2 Z 2 or Z 4. Let s study the case when there are 5 Sylow 2-subgroups. This subdivides in two cases. In any case Aut(Z 5 ) = Z 4. Consider f 2 Aut(Z 5 )definedbyf(1) = 2. We note that (f f)(1) = 4, so f does not have order 2, then it must have order 4. So, Aut(Z 5 )=<f>={id,f,f 2,f 3 }. Now we analyze the cases. 7

I If P 2 = Z2 Z 2. Then G = (Z 2 Z 2 ) n Z 5. We need to analyze all homomorphisms : Z 2 Z 2! Aut(Z 5 ) = Z 4. Since the elements of Z 2 Z 2 have all order 2, then their images have order 1 or 2. That is, the image of is (isomorphic to) Z 2... we have done this case before in part d of the previous problem. So, all (non-trivial) semidirect products are isomorphic to each other. We know D 20 has a Sylow 2-subgroup that is isomorphic to Z 2 Z 2. So, that is the group we were looking for. II We need to analyze homomorphisms : Z 4!<f>= Aut(Z 5 ). Since goes from Z 4 to itself, we don t have any restrictions besides that 1 has to be sent to a non-trivial element in <f>. Hence, we have the functions: (i) : Z 4!<f> defined by (i) (1) = f i for i =1, 2, 3 Let '(1) = 3, note that (3) (1) = f 3 = f f f = (1) (3) = ( (1) ')(1). Since these functions are all homomorphisms, it follows that (3) = (1) '. Hence, the semidirect product induced by (1) and (3) yield isomorphic groups. Remark 1 Note that ' has to be bijective. Be careful with that. For example, ' defined by '(1) = 2 does not work because it is not bijective. So, trying to mimic what we did above with (1) and (2) or (3) and (2) is not going to work. So, it seems there are two di erent non-abelian group of order 20 with 2-Sylow subgroups isomorphic to Z 4, one corresponding to (1) and the other corresponding to (2). Note that, using (1),(a, b) 2 =(a(1 + 2 b ), 2b) and, using (2),(a, b) 2 =(a(1 + 2 2b ), 2b). It follows that there is only one element of order 2 using (2) and 5 using (1). Thus, the groups are non-isomorphic to each other. 20. Classify the non-abelian group(s) of order 18. Solution: Let G = 18 = 2 3 2. n 2 1 (mod 2) and n 2 9, so, n 2 =1, 3, 9. Also, we know that P 2 = Z2. n 3 1 (mod 3) and n 3 2, so, n 3 = 1. It follows that P 3 / G. So, we have a semi-direct product here. We need to analyze all the homomorphisms : Z 2! Aut(P 3 ) We will take two cases, as P 3 = Z9 or P 3 = Z3 Z 3. Case P 3 = Z9. We know that Aut(Z 9 ) = Z 9, which has 6 elements. Thus Aut(Z 9 )iseitherz 6 or S 3. But, since the elements in Aut(Z 9 ) are of the form : g 7! g i then Aut(Z 9 ) is Abelian, forcing it to be isomorphic to Z 6. So, we are looking for : Z 2! Z 6 It follows that (1) = 0, 3, since the former case yields an Abelian group then we will stick to (1) = 3. Since we are in the case P 2 = Z2 and P 3 = Z9,thenthisunique non-abelian group must be D 18. Case P 3 = Z3 Z 3. We know that Aut(Z 3 Z 3 ) = GL(2, 3), which has (3 2 1)(3 2 3) = 48 elements. So, we are looking for : Z 2! GL(2, 3) It follows that (1) could be zero (yielding an Abelian group) or any of the many elements of order two in GL(2, 3). Moreover, since Aut(Z 2 ) is trivial then we will not have any isomorphism of the associated 8

semi-direct products. Thus there are as many groups as elements of order 2 in GL(3, 2). Just using apple 2 apple a b a = 2 + bc b(a + d) c d c(a + d) d 2 + bc and assuming such a matrix has order two we get the equations in Z 3 b(a + d) =c(a + d) =0 a 2 + bc = d 2 + bd =1 which, taking cases, yields the fourteen elements of order 2 in GL(2, 3), they are all the matrices in the following four sets apple a 0 ; a, d 2 Z 0 d apple a 0 ; a, c 2 Z c 3 a apple 0 b ; b 2 Z b 0 apple a b ; a, b 2 Z 0 3 a So, there should be fourteen non-abelian groups of order 18... well, in fact there are only three, two of them coming out of these 14 (recall that D 18 was already found). 21. Exercise 4.5.17. Solution: We know that G = 105 = 3 5 7. Then n 3 1 (mod 3) and n 3 35 =) n 3 =1, 7 n 5 1 (mod 5) and n 5 21 =) n 5 =1, 21 n 7 1 (mod 7) and n 7 15 =) n 7 =1, 15 If we assume that n 5 = 21 and n 7 = 15 then the Sylow 5-subgroups contain 4 21 = 84 non-identity elements. Similarly, the Sylow 7-subgroups contain 6 15 = 90 non-identity elements. Since all these elements must be distinct, then the union of all these Sylow subgroups contain 84 + 90 + 1 > 105 elements. Contradiction. It follows that n 5 or n 7 must be equal to one. Assume assume one of them is equal to 1, and let P 5 and P 7 be a Sylow 5-subgroup and a Sylow 7-subgroup respectively. Since one of these groups is normal then P 5 P 7 apple G. Moreover, since 5 does not divide 7 1 then P 5 P 7 = Z35. We notice that [G : P 5 P 7 ] = 3, which is the least prime dividing G, and thus P 5 P 7 E G. It follows that gp i g 1 apple P 5 P 7, for i =5, 7. But, since P 5 P 7 is cyclic there is exactly one subgroup of order 5 in P 5 P 7, and exactly one subgroup of order 7 in P 5 P 7. Hence, there is a unique Sylow p-subgroup, for both p =5, 7. 22. Exercise 4.5.24. Solution: The phrasing of this problem is just awkward (to me). Anyway, G = 231 = 3 7 11. We look at the n p s: n 3 1 (mod 3) and n 3 77 n 7 1 (mod 7) and n 7 44 n 11 1 (mod 11) and n 11 21 Hence the options for these values are n 3 =1, 7, n 7 = n 11 = 1. So, there is a unique Sylow p-subgroup for p =7, 11. Let us say that P 7 2 Syl 7 (G) and P 11 2 Syl 11 (G). Note that both P 7 and P 11 are normal 9

in G. Now we look at P 11. Since P 11 E G then G/C G (P 11 ) =apple Aut(P 11 ). Moreover, we can use that P 11 = Z11 to get G/C G (P 11 ) =apple Z 10, which implies that G/C G (P 11 ) divides 10. Now, we know that P 11 is Abelian, and so P 11 apple C G (P 11 ). It follows that G/C G (P 11 ) =1, 3, 7, 3 7. The only one of these numbers dividing 10 is 1, and thus G = C G (P 11, which means that P 11 Z(G). 23. Exercise 4.5.35. Solution: Let P 2 Syl p (G), H apple G, and Q 2 Syl p (H). Since H is a p-subgroup of H then it is a p-subgroup of G. It follows (from Sylow II) that there is a g 2 G such that Q apple gpg 1. Hence, (gpg 1 ) \ H = Q 2 Syl p (H). For the counterexample consider G = S 3, p = 2, P =< (12) > and H =< (13) >. 24. Exercise 4.5.44. Solution: We know that N G (P )/C G (P ) =apple Aut(P ). Since P is cyclic then Aut(P ) = Aut(Z p a). It follows that N G (P )/C G (P ) divides p a 1 (p 1). Moreover, since P is Abelian, then P apple C G (P ), then p does not divide N G (P )/C G (P ). Hence, N G (P )/C G (P ) divides p 1. However, p is the smallest prime dividing G, and thus it is the smallest prime that could possibly divide N G (P )/C G (P )... which divides p 1. Impossible unless N G (P )/C G (P ) is equal to 1, and thus N G (P )=C G (P ). 25. Exercise 5.1.5. Solution: The group we are looking for cannot be of the form P Q, as this will definitely yield a normal subgroup. So, we look at H =< (i, 1) >= {(i, 1), ( 1, 2), ( i, 3), (1, 0)}. We notice that which is not in H. (k, 0)(i, 1)(k, 0) 1 =( i, 1) 26. Exercise 5.4.10. Solution: Let G be a finite Abelian group with Sylow subgroups P 1,P 2,...,P k. We know that P 1 E G, for all i, and that P 1 P 2 E G, implying that (P 1 P 2 )P 3 apple G, and thus normal. Continuing this way we get that P 1 P i E G, for all i. Similarly, the orders of the Sylow subgroups of G being pairwise relatively prime, and some inductive argument prove that P 1 P i = P 1 P i for all i. Moreover, then again, similar arguments prove that (P 1 P i 1 ) \ P i = 1. It follows that G = (((((P 1 P 2 ) P 3 ) P k ) = P 1 P 2 P k 27. Exercise 5.4.11. Solution: Let 2 Aut(H K). H and K being characteristic means that H 2 Aut(H) and K 2 Aut(K). We define the function : Aut(H K)! Aut(H) Aut(K) by ( )=( H, K). This function is well-define because of the comment before the definition. Note that ( ) =(( ) H, ( ) K )=( H H, K K )=( H, K)( H, K )= ( ) ( ) 10

Also, if were the identity for both H and K then must be the identity automorphism, which means that is injective. Also, for any 2 Aut(H) and 2 Aut(K) welet (x, y) =( (x), (y)), for all (x, y) 2 H K. It is easy to show that is in Aut(H K, and that ( )=(, ), implying that is onto. The result for products is obtained by induction on the number of factors and Exercise 5.4.10. 11

5.3 Applications of Sylow s Theorems One of the standard applications of Sylow s Theorems is to prove that there are no simple groups of a certain order. We illustrate a couple of techniques. If G is a finite group and p is a prime dividing G, weputn p = Syl p (G). By Sylow s Theorems, we get that n p divides G and n p 1 mod p. So we can easily calculate the possible candidates for n p. We assume that G is not a p-group (if G is a p-group then G is simple if and only if G = p). (1) n p = 1 immediately (without counting) If n p =1thenP G by Proposition 5.11 where P 2 Syl p (G) andsog is not simple. (2) small index argument Let P 2 Syl p (G). Then n p =[G : N G (P )]. So if G is simple and G does not divide n p!, then by Proposition 5.12, N G (P )=G; so P G, a contradiction. +1X apple n Note that if p is a prime, then ord p (n!) =. p i (3) counting elements whose order is a prime power i=1 If n p 6=1,wemightbeabletocountthenumberofelementsin[{S 2 Syl p (G)}. This is particularly easy if ord p ( G ) =1: ifp, Q 2 Syl p (G) withp 6= Q, then P \ Q =1and so [ {S 2 Syl p (G)} =1+n p (p 1). Sometimes it is possible this way to establish that n p =1forsomeprimep and so G is not simple. (4) playing p-subgroups o against each other Let P 2 Syl p (G). Then n p = G : N G (P ) and so N G (P ) = G n p. Suppose that q 6= p is aprimewithq dividing N G (P ). Sometimes we can find a q-subgroup of G with a big normalizer. Then by the small index argument, G is not simple. A candidate for this q-subgroup is a Sylow-q-subgroup of N G (P ). (5) studying normalizers of intersections of Sylow-p-subgroups Again, we try to find a subgroup of G with a big normalizer but this time, a candidate is the intersection of two Sylow-p-subgroups (the method in 4. doesn t work if N G (P ) = P ). Iftheintersection ofany two Sylow-p-subgroups is trivial, we typically try a counting argument to deduce that G is not simple. So suppose that P and Q are two di erent Sylow-p-subgroups with P \ Q 6= 1. Sometimes,P \ Q P, Q (because [P : P \ Q] =p). P Q Then P, Q apple N G (P \ Q). So N G (P \ Q) PQ = P \ Q. Since N G(P \ Q) divides G and is divisible by P, this often leads to N G (P \ Q) =G (using the small index argument) and so G is not simple. 28

We illustrate these methods with the following exercises (the numbering of the exercises corresponds to the numbering of the methods). Examples : Prove that there are no simple groups of the following orders : (1) 200 Let G be a group of order 200 = 2 3 5 2 n2 2 {1, 5, 25}. We get n 5 =1 So G is not simple since n 5 =1. (2) 5103 Let G be a simple group of order 5103 = 3 6 7. We get n3 2 {1, 7} n 7 2 {1, 729} Suppose that G is simple. Then n 3 =7andn 7 = 729. By the Small Index Argument, 5103 divides 7!, a contradiction since 7! = 5040 (or since ord 3 (7!) = 2). (3) 380 8 < n 2 2 {1, 5, 19, 95} Let G be a group of order 380 = 2 2 5 19. We get n 5 2 {1, 76} : n 19 2 {1, 20} Suppose that n 5 6=1andn 19 6=1. ThenG contains 76 (5 1) = 304 elements of order 5and20 (19 1) = 360 elements of order 19, a contradiction since 304 + 360 > 380. Hence n 5 =1orn 19 =1. SoG is not simple. (4) a) 4389 8 n 3 2 {1, 7, 19, 133} >< n Let G be a group of order 4389 = 3 7 11 19. We get 7 2 {1, 57} n 11 2 {1, 133} >: n 19 2 {1, 77} Suppose that G is simple. Then n p 6=1forp =3, 7, 11, 19. So n 7 =57andn 11 =133. Hence N G (P 7 ) =77foranyP 7 2 Syl 7 (G) and N G (P 11 ) =33foranyP 11 2 Syl 11 (G). Let P 2 Syl 7 (G). Since 11 divides N G (P ), we can pick Q 2 Syl 11 (N G (P )). Then Q =11. SoQ 2 Syl 11 (G). However, a group of order 77 has a unique Sylow-11-subgroup (n 11 =1forsuchagroup). SoQ N G (P ). Hence N G (P ) apple N G (Q), a contradiction since N G (P ) =77doesnotdivide N G (Q) =33. SoG is not simple. b) 3675 8 < n 3 2 {1, 7, 25, 49, 175, 1225} Let G be a group of order 3675 = 3 5 2 7 2. We get n 5 2 {1, 21} : n 7 2 {1, 15} Suppose that G is simple. Then n p 6=1forp =3, 5, 7. So n 5 =21andn 7 =15. Let P 2 Syl 5 (G). Then N G (P ) =175=5 2 7. Let Q 2 Syl 7 (N G (P )). Then Q N G (P ) since a group of order 175 has only one Sylow-7-subgroup. So P apple N G (P ) apple N G (Q). 29

Pick Q 2 Syl 7 (G) withq apple Q. Since Q =5and Q =25,wehavethat[Q : Q] =5 and so Q Q. Hence Q apple N G (Q). So PQ N G (Q). But PQ = P Q P \ Q =52 7 2. Hence [G : N G (Q)] apple 3. By the Small Index Argument, G = N G (Q). So Q G, a contradiction. (5) 144 Let G be a group of order 144 = 2 4 3 2. We get n2 2 {1, 3, 9} n 3 2 {1, 4, 16} Suppose that G is simple. Then n 2 6=1andn 3 6= 1. By the Small Index Argument, n 2 6=3andn 3 6=4. Son 2 =9andn 3 =16. Then N G (P 2 ) =16foranyP 2 2 Syl 2 (G) and N G (P 3 ) =9foranyP 3 2 Syl 3 (G). Suppose that P \ Q =1foranyP, Q 2 Syl 3 (G) with P 6= Q. Then G has 16 (9 1) = 128 elements of order a power of three. Since G 128 = 16, we see that G has only one Sylow-2-subgroup (so n 2 =1),acontradiction. Hence there exist P, Q 2 Syl 3 (G) withp 6= Q and P \ Q 6= 1. Since P = Q =3 2, we get that P \ Q =3. So[P : P \ Q] =3=[Q : P \ Q]. Hence P \ Q P and P Q P \ Q Q. So PQ N G (P \ Q). But PQ = P \ Q =27. SoN G(P \ Q) apple G, 9 divides N G (P \ Q) and N G (P \ Q) 27. So [G : N G (P \ Q)] 2 {1, 2, 4}. By the Small Index Argument, G = N G (P \ Q). So P \ Q G, a contradiction. 30