FIRST-YEAR GROUP THEORY 7 LAGRANGE'S THEOREM EXAMPLE 7.1: Set G = D 3, where the elements of G are denoted as usual by e, a, a 2, b, ab, a 2 b. Let H be the cyclic subgroup of G generated by b; because b has order 2, this means that H consists of e and b. The point of this example is to illustrate the general method which will be used to prove Lagrange's theorem. Note that H has 2 elements. The idea is to divide the elements of G into disjoint (i.e. non-overlapping) subsets each with 2 elements, from which it will follow that the order of G is divisible by 2 (of course, in this example we know that G has order 6). For a fixed element x of G we consider the set consisting of the two elements xe = x and xb. Taking x = e gives the set fe; bg; taking x = a gives the set fa; abg; taking x = a 2 gives the set fa 2 ;a 2 bg; note that every element of G belongs to one, and only one, of these 2-element subsets. NOTATION 7.2: Let x be an element of a multiplicatively-written group G, and let H be a subgroup of G. Set xh = fxh : h 2 Hg; thus xh consists of all elements of G which can be formed by multiplying the fixed element x of G on the right by all possible elements of H. THEOREM 7.3: (Lagrange 1736-1813; Jordan and Jordan 10.3) Let G be a finite group with n elements, and let H be a subgroup of G with k elements. Then k divides n. PROOF: The idea is to divide the elements of G up into disjoint subsets each with k elements; suppose that there are r of these subsets; then each of the n elements of G belongs to exactly one of these r subsets each of which has k elements, so that n = kr. Let x 2 G and let xh be as in 7.2. Then x 2 xh because x = xe with e 2 H. Also xh has k elements, because we can pair each of the k elements h of H with the element xh of xh (note that if a, b 2 H with xa = xb then a = b). Now let y 2 G and suppose that xh and yh are not disjoint, i.e. that xh yh is non-empty; we shall show that xh = yh and that will complete the proof. Let g 2 xh yh. Then g = xa = yb for some a, b 2 H. Hence x = yba 1. Let h 2 H. Then xh = yba 1 h where b, a, h 2 H. Thus ba 1 h 2 H, so that xh 2 yh. But xh is a typical element ofxh. Therefore xh yh. By symmetry we alsohave yh xh, so that xh = yh. DEFINITION 7.4: (Index of a subgroup) Let G be a finite group of order n, and let H be a subgroup of G of order k. We know by Lagrange's theorem that k divides n, i.e. that n=k is an integer. The index of H in G is denoted by [G : H] and is defined by [G : H] =n=k = ord(g)=ord(h): COROLLARY 7.5: Let G be a finite group of order n, and let g 2 G. Then the order of g divides n, so that g n = e where e is the identity element ofg. PROOF: Let k denote the order of g, and let H be the subgroup of G generated by g. Then H has order k, by 4.2(2). Therefore k divides n, by Lagrange's theorem. Hence g n = e, by 3.13(1).
REMARKS 7.6: Let G be a finite group of order n. Then Lagrange's theorem shows that the order of every subgroup of G divides n; equivalently, if k is a positive integer which does not divide n then G has no subgroups of order k. Lagrange's theorem does not say that if k is a positive integer which divides n then G has a subgroup of order k; the group A 4 which will be defined in Section 8 has order 12 but it has no subgroups of order 6. Similarly Lagrange's theorem implies that the order of every element of G divides n; it does not imply that if k is a positive integer which divides n then G has an element of order k; for instance the Klein 4-group has order 4 but it has no elements of order 4. THEOREM 7.7: Let p be a prime number, and let G be a group of order p with identity element e. Then (1) G is cyclic and so is unique up to isomorphism; (2) Each non-identity element of G has order p and generates G as a cyclic group; (3) G has no subgroups other than Gitself and the subgroup which consists of e only. PROOF: (1) Let x 2 G with x 6= e, and set k = ord(x). Because x 6= e we have k 6= 1. But k divides p, by 7.5. Therefore k = p, and x generates G as a cyclic group by 4.14. (2) We have already done this in proving (1). (3) Let H be a subgroup of G of order r. Then r divides p, by Lagrange's theorem. Therefore either r = p in which case H = G, orr =1in which case H consists of e only. THEOREM 7.8: Let G be a group of order 4. Then G is Abelian, and either G is cyclic or G is the Klein 4-group. PROOF: Suppose that the operation in G is multiplication with identity element e. If G has an element oforder4theng is cyclic, by 4.14. So from now on we shall suppose that G has no elements of order 4. It follows from 7.5 that every element of G has order 1 or 2. Hence x 2 = e for all x 2 G. Let a, b, c be the non-identity elements of G. We have a 2 = e so that a = a 1. Similarly b = b 1 and c = c 1. The element ab of G must be one of the four elements e, a, b, c. We have b 6= a, i.e. b 6= a 1, i.e. ab 6= e. We have ab 6= a because b 6= e. We have ab 6= b because a 6= e. Therefore ab = c. Similarly ba = c, bc = a = cb, and ca = b = ac. Therefore G is the Klein 4-group. THEOREM 7.9: (Jordan and Jordan page 112 Theorem 9) Let G be a group with finite subgroups H and K such that the orders of H and K are relatively prime. Set I = H K. Then I consists of the identity element only. PROOF: We know by 2.11 that I is a subgroup of G, so that e 2 I. We must show that ord(i) = 1, i.e. that I contains nothing except e. Because I H G and both I and H are subgroups of G, we can regard I as being a subgroup of H. Hence ord(i) divides ord(h), by Lagrange's theorem. Similarly ord(i) divides ord(k). But ord(h) and ord(k) are relatively prime. Therefore ord(i) = 1. THEOREM 7.10: (Jordan and Jordan page 112 Theorem 9) Let p be a prime number, and let G be a group with distinct subgroups H and K such that both H and K have order p. Set I = H K. Then I consists of the identity element only.
PROOF: As in the proof of 7.9 we know that ord(i) divides ord(h) and ord(k), and we must show thatord(i) = 1. Suppose that ord(i) 6= 1. Because ord(i) divides ord(h) and ord(h) = p, it follows that ord(i) =p. Thus I H and ord(i) =ord(h), so that I = H. Similarly I = K. Therefore H = K, which is a contradiction. THEOREM 7.11: (Cauchy 1789-1857; Jordan and Jordan 14.2) Let p be a prime number, and let G be a finite group of order n. Suppose that p divides n. Then G has at least one element of order p. REMARK 7.12: We shall not prove Cauchy's theorem in these notes, but from now on you may use it. THEOREM 7.13: Let p be a prime number and let G be a group of order 2p. Then either G is cyclic or G ο = D p. PROOF: The case in which p =2was done in 7.8. From now on we suppose that p is an odd prime number. By Cauchy's theorem we know thatg has an element x of order 2 and an element y of order p. Suppose that x is the only element of G of order 2. We know from an exercise that y 1 xy has the same order as x, sothaty 1 xy = x. Thus xy = yx, and it follows from 3.14 that xy has order 2p. Therefore G is cyclic, by 4.14. Let H be the subgroup of G generated by y. Then H has order p because y has order p. Suppose that K is a subgroup of G such that K has order p and K 6= H. Then all the nonidentity elements of H or K have order p (by 7.7), and H K consists of the identity element only (by 7.10). Therefore H [ K consists of 2p 1 elements each of which has order p or 1, so that x is the only other element ofg and hence is the only element of order 2; we dealt with this case in the previous paragraph. Therefore we shall suppose from now on that H is the only subgroup of G which has order p. In particular H contains all elements of G which have order p. But y has order p, and hence so also does x 1 yx. Therefore x 1 yx 2 H. Thus x 1 yx is a non-identity element of the cyclic subgroup of G generated by y, so that x 1 yx = y s for some integer s which is not divisible by p (for if p divides s then y s = e). If s 1 mod(p) then y s = y, i.e. x 1 yx = y, i.e. xy = yx and, as before, it follows that G is cyclic. Now suppose that s is not congruent to 1 mod(p). We have x 2 = e, so that y = eye = x 2 yx 2 = x 1 (x 1 yx)x = x 1 y s x = (x 1 yx) s = (y s ) s = y t where t = s 2. Hence y t 1 = e, where y has order p. Therefore p divides t 1, by 3.13(1). Thus p divides s 2 1, and we are supposing that p does not divide s 1. Hence p divides s +1,i.e. s 1 mod(p), i.e. y s = y 1. Therefore x has order 2, y has order p, x 1 yx = y 1, and G has order 2p. It follows that G ο = D p. COROLLARY 7.14: The only groups of order 6 are Z 6 and D 3. REMARKS 7.15: Results such as 7.13 which classify all groups of a given order tend to be very complicated to prove. For more information see Jordan and Jordan Chapter 17. We have stated 7.14 in the usual abbreviated form; what it really means is that any group of order 6 is either cyclic or is isomorphic to D 3. Later on we shall be able to replace D 3 by S 3 in this statement. Some classification results tell us that certain groups are made up from smaller ones; for instance D 6 ο = D 3 Z 2. We shall next list all groups of order at most 15, but not all of these assertions will be justified in these notes.
LIST 7.16: We shall give (upto isomorphism) all groups of order» 15. Order 1: The only group of order 1 is Z 1 (which consists of the identity element only). Order 2: The only group of order 2 is Z 2. Order 3: The only group of order 3 is Z 3. Order 4: There are two groups of order 4, namely Z 4 and the Klein 4-group, and they are both Abelian. Order 5: The only group of order 5 is Z 5. Order 6: There are two groups of order 6, namely Z 6 and D 3 (or S 3 ; note that D 3 is the smallest non-abelian group. Order 7: The only group of order 7 is Z 7. Order 8: There are five groups of order 8. Three of them are Abelian, namely Z 8, Z 4 Z 2, and Z 2 Z 2 Z 2. The two non-abelian ones are D 4 and the quaternion group. Order 9: The only groups of order 9 are Z 9 and Z 3 Z 3. If p is a prime number then any group of order p 2 is Abelian and is either cyclic or isomorphic to Z p Z p. Order 10: The only groups of order 10 are Z 10 and D 5. Order 11: The only group of order 11 is Z 11. Order 12: There are five groups of order 12. Two of them are Abelian, namely Z 12 and Z 2 Z 6 ; note that Z 12 ο = Z 3 Z 4 and Z 2 Z 6 ο = Z 2 Z 2 Z 3. The three non-abelian groups of order 12 are D 6, A 4 (see Section 8), and a third one which we shall not meet in this course. Order 13: The only group of order 13 is Z 13. Order 14: The only groups of order 14 are Z 14 and D 7. Order 15: The only group of order 15 is Z 15. EXAMPLE 7.17: Set G = D 4. We shall find all the subgroups of G. We have ord(g) = 8. Hence any subgroup of G, other than G and the trivial subgroup, has order 2 or 4. We consider these cases separately. Case (1): Any subgroup of G of order 2 consists of e (the identity element) together wih an element of order 2. But G has five elements of order 2: geometrically these are rotation through 180 degrees and the four reflections; in the usual algebraic notation they are a 2, b, ab, a 2 b, a 3 b. Therefore G has five subgroups of order 2. Case (2): Let H be a subgroup of G of order 4. Then H is either cyclic or the Klein 4-group. Sub-Case (a): Suppose that H is cyclic. Then H is generated by an element x of G of order 4. The only possibilities for x are x = a and x = a 3 = a 1, and these generate the same subgroup. Therefore G has a unique cyclic subgroup of order 4, and it consists of e, a, a 2, a 3. Sub-Case (b): Suppose that H is the Klein 4-group. Then H consists of e together with three elements of order 2. There are five elements of order 2 in G, but we can not pick three of them at random because we need H to be closed under multiplication. It can be checked that there are two possibilities for H; each contains e and a 2 ; one also contains b and a 2 b; the other contains ab and a 3 b. THEOREM 7.18: (Fermat's Little Theorem) Let p be a prime number and let a be an integer which is not divisible by p. Then a p 1 1mod(p). PROOF: The condition that p does not divide a means that a is non-zero as an element of Z=pZ. Let G be the group of non-zero elements of Z=pZ under multiplication. Then a 2 G and G has order p 1. Therefore by 7.5 we have a p 1 =1in G, i.e. a p 1 =1in Z=pZ, i.e. a p 1 1 mod(p).
EXAMPLE 7.19: Suppose that we want to calculate 7 100 mod(31). All congruences will be mod(31). By Fermat's Little Theorem we have 7 30 1. Hence 7 90 (7 30 ) 3 1, so that 7 100 7 10. But 7 3 =343=341+2=11:31 + 2, so that 7 3 2. Therefore 7 100 7 10 (7 3 ) 3 :7 2 3 :7 56 31+25 25. EXAMPLE 7.20: The 17th Mersenne number M 17 is defined by M 17 = 2 17 1. It is a non-trivial problem to determine whether such numbers are prime, but the following argument reduces considerably the amount of calculation involved. Suppose that p is a prime factor of M 17. Then p divides 2 17 1, i.e. b 17 1 mod(p). Let G be the group of non-zero elements of Z=pZ under multiplication. Then 2 2 G, and in G we have 2 6= 1 and 2 17 = 1. Hence 2 has order 17 as an element ofg. But 7.5 or Fermat's Little Theorem gives 2 p 1 =1inG. Therefore 17 divides p 1, by 3.13(1). Because M 17 is odd, so also is p. Thus p 1 is divisible by 2 and 17 and hence also by 34. This means that every prime factor of M 17 is of the form 34n +1for some positive integer n. Therefore, in order to determine whether or not M 17 is prime, we need only test whether it is divisible by any prime numbers in the list 35, 69, 103, 137,... up to the square root of M 17.