Converse to Lagrange s Theorem Groups

Converse to Lagrange s Theorem Groups Blain A Patterson Youngstown State University May 10, 2013

History In 1771 an Italian mathematician named Joseph Lagrange proved a theorem that put constraints on what types of subgroups a group is allowed to have. Today we call it Lagrange s theorem and it states: Let G be a group and H G, then H G. About 100 years later, group theorists began to wonder about a converse to Lagrange s theorem. That is, if G is a group and n is an integer such that n G, does G have a subgroup H of order n? If this holds true for the group G, then we call G a Converse to Lagrange s theorem group, or CLT group for short.

Introduction We will cover the following questions concerning CLT groups:

Introduction We will cover the following questions concerning CLT groups: 1 What are some examples of CLT groups and what are some examples of groups that are not CLT groups?

Introduction We will cover the following questions concerning CLT groups: 1 What are some examples of CLT groups and what are some examples of groups that are not CLT groups? 2 If a group is CLT group, are all of it s subgroups and quotient groups CLT groups? 3 Can we find wide classes of groups that are CLT groups?

Lagrange s Theorem Below is an example of Lagrange s Theorem:

Lagrange s Theorem Below is an example of Lagrange s Theorem: Z 6 = {0, 1, 2, 3, 4, 5}, so Z 6 = 6. Now all subgroups will divide the order of the group. The subgroups of Z 6 include: Z 6, < 2 >= {0, 2, 4}, < 3 >= {0, 3}, and the trivial subgroup, < 0 >, which have orders of 6, 3, 2, and 1 respectively. Thus the order of each subgroup divides the order of Z 6.

Converse to Lagrange s Theorem Group Let G be a group, then G is a Converse to Lagrange s Theorem Group (CLT group), if for all n G there exists H G such that H = n.

Converse to Lagrange s Theorem Group Let G be a group, then G is a Converse to Lagrange s Theorem Group (CLT group), if for all n G there exists H G such that H = n. Example S 3 = {1, (123), (132), (12), (13), (23)} is a CLT group since S 3 = 6 and there exists at least one subgroup of each divisor of 6, namely 6, 3, 2, and 1. S 3, < (123) >= {1, (123), (132)}, < (12) >= {1, (12)}, and {1} are all subgroups of S 3 with orders of each of the divisors of 6, namely orders of 6, 3, 2, and 1 respectively.

Example Below is another example of a CLT group:

Example Below is another example of a CLT group: Example S 4 is a CLT group since S 4 = 24 and there exists at least one subgroup of each divisor of 24. Observe {1} = 1, < (12) > = 2, < (123) > = 3, < (1234) > = 4, a copy of S 3 which has order 6, a copy of D 4 which has order 8, A 4 = 12, and S 4 = 24. Notice that all the groups listed are subgroups of S 4. Thus there exists a subgroup of every order of the divisors of S 4, namely 24, 12, 8, 6, 4, 3, 2, and 1.

Theorems Theorem 1.1 Let G be a group and a in G. Then a G = 1.

Theorems Theorem 1.1 Let G be a group and a in G. Then a G = 1. Theorem 1.2 Let G be a group and H G such that then H G. G H = 2

Proposition Claim: A 4 is not a CLT group since there exists no subgroup of A 4 with order 6.

Proposition Claim: A 4 is not a CLT group since there exists no subgroup of A 4 with order 6. Suppose there exists a subgroup of A 4 with order 6, call this subgroup H. Then A 4 H = 12 6 = 2

Proposition Claim: A 4 is not a CLT group since there exists no subgroup of A 4 with order 6. Suppose there exists a subgroup of A 4 with order 6, call this subgroup H. Then A 4 H = 12 6 = 2 So, by Theorem 1.2, H A 4. Thus A 4 /H is a group with order 2. Let φ A 4. Then φh A 4 /H. Now (φh) A 4 H = 1H (φh) 2 = 1H

Proposition Continued Hence φ 2 H = 1H. Now, 1 1 φ 2 H so φ 2 H for all φ A 4.

Proposition Continued Hence φ 2 H = 1H. Now, 1 1 φ 2 H so φ 2 H for all φ A 4. Thus we have 1 2 = 1 (123) 2 = (132) (132) 2 = (123) (124) 2 = (142) (142) 2 = (124) (134) 2 = (143) (143) 2 = (134)

Proposition Continued Hence φ 2 H = 1H. Now, 1 1 φ 2 H so φ 2 H for all φ A 4. Thus we have 1 2 = 1 (123) 2 = (132) (132) 2 = (123) (124) 2 = (142) (142) 2 = (124) (134) 2 = (143) (143) 2 = (134) Therefore, we get H 7, which is a contradiction. Thus, A 4 does not have a subgroup of order 6 and is not a CLT group.

Behavior of CLT Groups Recall that S 4 is a CLT group and A 4 is not a CLT group.

Behavior of CLT Groups Recall that S 4 is a CLT group and A 4 is not a CLT group. Note that A 4 S 4.

Behavior of CLT Groups Recall that S 4 is a CLT group and A 4 is not a CLT group. Note that A 4 S 4. So a subgroup of a CLT group might not necessarily be a CLT group itself.

Theorems Theorem 1.3 Let G 1 and G 2 be groups with H 1 G 1 and H 2 G 2. Then H 1 H 2 G 1 G 2.

Theorems Theorem 1.3 Let G 1 and G 2 be groups with H 1 G 1 and H 2 G 2. Then H 1 H 2 G 1 G 2. Theorem 1.4 Let G 1 and G 2 be groups with H 1 G 1. Then H 1 {1} G 1 G 2.

Theorems Theorem 1.3 Let G 1 and G 2 be groups with H 1 G 1 and H 2 G 2. Then H 1 H 2 G 1 G 2. Theorem 1.4 Let G 1 and G 2 be groups with H 1 G 1. Then H 1 {1} G 1 G 2. Theorem 1.5 Let G 1 and G 2 be groups. Then G 1 G 2 G 1 {1} = G 2

Behavior of CLT Groups Look at the groups Z 2 A 4 and Z 2 {1}, and note that Z 2 A 4 is a CLT group. Let us define H = {1, (12)(34), (13)(24), (14)(23)}. Observe that H A 4. Below is a list of subgroups of Z 2 A 4 with their respective orders. {(0, 1)} = 1 < (1, 1) > = 2 < (0, (123)) > = 3 {0} H = 4 < (1, (123)) > = 6 Z 2 H = 8 {0} A 4 = 12 Z 2 A 4 = 24

Behavior of CLT Groups We have Z 2 {1} Z 2 A 4 from Theorem 1.4.

Behavior of CLT Groups We have Z 2 {1} Z 2 A 4 from Theorem 1.4. Thus we can mod out Z 2 {1} and get by Theorem 1.5 Z 2 A 4 Z 2 {1} = A 4

Behavior of CLT Groups We have Z 2 {1} Z 2 A 4 from Theorem 1.4. Thus we can mod out Z 2 {1} and get by Theorem 1.5 Z 2 A 4 Z 2 {1} = A 4 So a quotient group of a CLT group may not be a CLT group.

Theorems Theorem 1.6 Let G be a group and a G an element of finite a = n. Then for any k Z, a k = 1 if and only if n k.

Theorems Theorem 1.6 Let G be a group and a G an element of finite a = n. Then for any k Z, a k = 1 if and only if n k. Theorem 1.7 Let G be a group and a G an element of finite order a = n. Then < a >= {1, a, a 2,..., a n 1 }, also < a > = a.

Cyclic Groups are CLT Theorem Let G be a cyclic group such that G = n. Then G is a CLT group.

Cyclic Groups are CLT Theorem Let G be a cyclic group such that G = n. Then G is a CLT group. Proof Let d G. Since G is cyclic, there exists a G such that G =< a >. Then d G, where G = < a > = a

Cyclic Groups are CLT Theorem Let G be a cyclic group such that G = n. Then G is a CLT group. Proof Let d G. Since G is cyclic, there exists a G such that G =< a >. Then d G, where G = < a > = a Then there exists k Z such that dk = a. Let H =< a k >. Then H G. By Theorem 1.7 H = < a k > = a k

Proof Continued Now (a k ) d = a kd = a a = 1

Proof Continued Now (a k ) d = a kd = a a = 1 Therefore, a k d, by the minimality of a k. Also 1 = (a k ) ak = a k ak

Proof Continued Now (a k ) d = a kd = a a = 1 Therefore, a k d, by the minimality of a k. Also 1 = (a k ) ak = a k ak By Theorem 1.6 a k a k or dk k a k, thus d a k, so d a k. Hence, d = a k. Therefore, we have H = < a k > = a k = d

Proof Continued Now (a k ) d = a kd = a a = 1 Therefore, a k d, by the minimality of a k. Also 1 = (a k ) ak = a k ak By Theorem 1.6 a k a k or dk k a k, thus d a k, so d a k. Hence, d = a k. Therefore, we have Thus G is a CLT group. H = < a k > = a k = d

Abelian Groups are CLT Theorem Let G be a group such that G is Abelian. Then G is a CLT group.

p-groups Below we define a p-group:

p-groups Below we define a p-group: Let G be a group and p be a prime. Then G is a p-group if there exists n Z + {0} such that G = p n.

p-groups Below we define a p-group: Let G be a group and p be a prime. Then G is a p-group if there exists n Z + {0} such that G = p n. Example D 4 is a 2-group since D 4 = 8 = 2 3.

Theorems Theorem 1.8 Let G be a p-group such that G {1}. Then Z(G) {1}.

Theorems Theorem 1.8 Let G be a p-group such that G {1}. Then Z(G) {1}. Theorem 1.9 (Cauchy s theorem for Abelian groups) Let G be a finite Abelian group and p be a prime dividing the order of G. Then G has an element of order p. Theorem 1.10 Let φ : G G be a homomorphism. Then if K is a subgroup of G, then φ 1 (K) = {x G φ(x) K} is a subgroup of G.

p-groups are CLT Theorem Let G be a p-group. Then G is a CLT group.

p-groups are CLT Theorem Let G be a p-group. Then G is a CLT group. Proof (Induction on G ) Suppose G = p. Then G is a CLT group since G G and {1} G.

p-groups are CLT Theorem Let G be a p-group. Then G is a CLT group. Proof (Induction on G ) Suppose G = p. Then G is a CLT group since G G and {1} G. Suppose this holds true for all p-groups of order less than G. Let G = p n and d G. Then d = p α where 0 α n. If α = 0, then d = 1 and {1} = 1. If α = n then d = p n and G = p n. So without loss of generality assume 0 < α < n.

Proof Continued Now since G is a p-group, Z(G) 1 by Theorem 1.7. Let 1 z where z is in Z(G) such that z p = 1 by Theorem 1.9. Then < z > G, so G/ < z > is a p-group since G is a p-group. Now G < z > = G < z > < G Thus G/ < z > is a CLT group, by induction.

Proof Continued Also note that 0 α 1 < n 1, so p α 1 p n 1, where p n 1 G G =. Since is a CLT group, there exists <z> <z> H G such that H = <z> pα 1.

Proof Continued Also note that 0 α 1 < n 1, so p α 1 p n 1, where p n 1 G G =. Since is a CLT group, there exists <z> <z> H G such that H = <z> pα 1. Now let φ : G G be defined by φ(g) = g < z >, which is <z> a homomorphism by the Natural Map Theorem. Now < z > φ 1 (H) G by Theorem 1.10. Thus < z > φ 1 (H) so φ 1 (H) is a group <z>

Proof Continued Then φ 1 (H) = φ 1 (H) < z > < z > = H < z > = p α 1 p = p α = d

Proof Continued Then φ 1 (H) = φ 1 (H) < z > < z > = H < z > = p α 1 p = p α = d Thus there exists a subgroup of order d, so G is a CLT group.

How common are CLT groups? We can define what is known as a CLT number:

How common are CLT groups? We can define what is known as a CLT number: A number, n, is a CLT number if every group G such that G = n is a CLT group.

References 1 Papantonopoulou, A. (2002). Algebra Pure and Applied. Upper Saddle River, NJ: Pearson Prentice Hall. Special thanks to Dr. Neil Flowers for excellent Advisement!