# understand all finite groups

MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 18 Cayley’s Theorem How can we understand all finite groups? One way, which we return to later, is to try and classify, i.e. list, all groups up to isomorphism. Here is another way to “understand all finite groups”: show that all such groups are isomorphic to subgroups of groups we understand. This can be done using the following theorem. § Cayley’s Theorem: Let G be a finite group, and |G| “ n. Then there is a 1-1 homomorphism φ ∶ G Ñ Sn . In particular, G is isomorphic to a subgroup of Sn . Before proving this theorem, let’s see why it does what we want. Given any homomorphism φ ∶ G Ñ G1 we can define a new homomorphism ψ ∶ G Ñ impφq by setting ψpaq “ φpaq for all a P G; we have only changed the definition of the target group. The homomorphism ψ ∶ G Ñ impφq has the advantage that it is onto. The 1st Isomorphism Theorem for ψ yields G{kerpφq – impφq In particular, when φ is 1-1, we have G – impφq. Now return to the statement of Cayley’s Theorem, which gives a 1-1 homomorphism φ ∶ G Ñ Sn . Then G – impφq, and impφq Ă Sn is a subgroup of Sn . Thus G is isomorphic to a subgroup of Sn . Proof. Label elements of G as a1 , a2 , . . . , an . Let a P G. (This a is among the ai , but we will just write a for it.) Define a function σa ∶ t1, . . . , nu Ñ t1, . . . , nu as follows: aai “ aσpiq In other words, given i, we take the product aai P G, and it is equal to some ak ; we let σa piq “ k. The map σa is 1-1: suppose σa piq “ σa pjq. Then aai “ aaj Ôñ ai “ aj Ôñ i“j The map σa is onto: suppose k P t1, . . . , nu. Let a “ ak a´1 i . Then aai “ ak a´1 i ai “ ak Ôñ σa piq “ k Thus σa is 1-1 and onto, and there a permutation, i.e. σa P Sn . Now we are in a position to define the sought after homomorphism: φ ∶ G ÐÑ Sn For a P G we declare φpaq “ σa . We check this is a homomorphism. Let a, b P G. We must compare φpaqφpbq “ σa ˝ σb with φpabq “ σab . We compute aσab piq “ pabqai “ apbai q “ aaσb piq “ aσa pσb piqq for any i P t1, . . . , nu, which shows that σab “ σa ˝ σb . To show φ is 1-1, we compute its kernel. Suppose φpaq “ e, i.e. σa piq “ i for all i P t1, . . . , nu. This means aai “ aσa piq “ ai which implies a “ e. Thus kerpφq “ teu and therefore φ is 1-1. 1 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 18 Let us see Cayley’s Theorem in action. To clarify the main point, take G to be an interesting group which is not defined using permutations. Let us introduce the quaternion group Q8 “ t 1, ´1, i, ´i, j, ´j, k, ´ku of order 8, with the relations that ´1 commutes with everything, p´1q2 “ 1, and also i2 “ j2 “ k2 “ ´1, and ij “ k. These relations generate the Cayley table of Q8 as follows: 1 ´1 i ´i j ´j k ´k 1 1 ´1 i ´i j ´j k ´k ´1 ´1 1 ´i i ´j j ´k k i i ´i ´1 1 k ´k ´j j ´i ´i i 1 ´1 ´k k j ´j j j ´j ´k k ´1 1 i ´i ´j ´j j k ´k 1 ´1 ´i i k k ´k j ´j ´i i 1 ´1 ´k ´k k ´j j i ´i ´1 1 Cayley’s Theorem gives us a way to realize Q8 as a subgroup of the symmetric group S8 . The construction is as follows. First we label the elements of Q8 as a1 , . . . , a8 . For example, we may use the following labelling (although any labelling will do): a1 “ 1, a2 “ ´1, a3 “ i, a4 “ ´i, a5 “ j, a6 “ ´j, a7 “ k, a8 “ ´k Next, given an element a P Q8 we define the associated permuation σa P S8 by the relation aai “ aj ðñ σa piq “ j Let us spell this out for the element a “ i P Q8 . We have aa1 “ piqp1q “ i “ a3 aa3 “ piqpiq “ ´1 “ a2 aa5 “ piqpjq “ k “ a7 aa7 “ piqpkq “ ´j “ a6 aa2 “ piqp´1q “ ´i “ a4 aa4 “ piqp´iq “ 1 “ a1 aa6 “ piqp´jq “ ´k “ a8 aa8 “ piqp´kq “ j “ a5 These relations then determine the permutation σa “ σi as follows: σa p1q “ 3, σa p2q “ 4, σa p3q “ 2, σa p4q “ 1, σa p5q “ 7, σa p6q “ 8, σa p7q “ 6, σa p8q “ 5 All together we may write our permutation σa “ σi in cycle notation as σi “ p1324qp5768q P S8 2 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 18 We may proceed to do this for each element a P Q8 . We get the following subgroup of S8 : te, p12qp34qp56qp78q, p1324qp5768q, p1423qp5867q, p1526qp3847q, p1625qp3748q, p1728qp3546q, p1827qp3645qu The elements are listed in the order corresponding the above ordering of the elements of Q8 . In summary, Q8 is isomorphic to the subgroup of S8 given by the subgroup displayed above. Cayley’s Theorem is conceptually very important: it says that every finite group, however abstractly defined, is isomorphic to a subgroup of some symmetric group, a very concrete type of group that we understand how to work with. On the other hand, Cayley’s Theorem is often not practical: it realizes our group as a subgroup of a generally very large group. For example, in the example above, Q8 has order 8 and we realized it as a subgroup of S8 which is of order 8! “ 40320. The following is a generalization of Cayley’s Theorem. § Let G be a group, H Ă G a subgroup, and suppose rG ∶ Hs “ n is finite. Then there is a homomorphism φ ∶ G Ñ Sn . Furthermore, kerpφq is equal to the largest normal subgroup of G which is contained in H. To obtain Cayley’s Theorem as a special case, let H “ teu. Then n “ rG ∶ teus “ |G|. The largest normal subgroup contained in teu is of course teu, and so kerpφq “ teu, i.e. φ is 1-1. To proof of this generalization is similar to that of Cayley’s Theorem. We list the distinct cosets in G{H as a1 H, . . . , an H. Then define φ ∶ G Ñ Sn by φpaq “ σa where σa P Sn is determined by the relation: σa piq “ j if and only if aai H “ aj H, where i, j P t1, . . . , nu. The details for the rest of the proof are left as an exercise. § Let G be a finite group, H Ă G a proper subgroup, and suppose |G| does not divide rG ∶ Hs!. Then H contains a non-trivial normal subgroup. This is a corollary of the above theorem. Suppose H does not contain a non-trivial normal subgroup. Then the homomorphism φ ∶ G Ñ Sn from the above theorem, where n “ rG ∶ Hs, must have kerpφq “ teu. Thus G – impφq is isomorphic to a subgroup of Sn . By Lagrange’s Theorem, we must have that |G| divides |Sn | “ n! “ rG ∶ Hs!. 3 MTH 461: Survey of Modern Algebra, Spring 2021 Homework 7 Homework 7 1. Consider the following group, which you may think of as GL2 pZ2 q: “ˆ ˙ * a b G“ ∶ a, b, c, d P Z2 , ad ´ bc ” 1 pmod 2q c d The operation is matrix multiplication, although we now use addition and multiplication in Z2 instead of R. You may verify that this is a group, although you do not need to prove it on your homework. Show that G is isomorphic to the group S3 . 2. In class we applied the construction of Cayley’s Theorem to the quaterion group Q8 to produce a 1-1 homomomorphism φ ∶ Q8 Ñ S8 . Apply the same construction to the following groups, to get homomorphisms into symmetric groups: (a) pZ3 , `q (b) pZˆ 8 , ˆq (c) S3 3. Consider the quaternion group Q8 “ t1, ´1, i, ´i, j, ´j, k, ´ku. (a) List all subgroups of Q8 , and explain why your list is complete. (b) Show that every subgroup of Q8 is normal. (c) Construct an onto homomorphism Q8 Ñ Z2 ˆ Z2 . From your homomorphism, what does the 1st Isomorphism Theorem imply? 4. Use the last result stated in Lecture 18 to show the following: a group of order p2 , where p is a prime, must have a normal subgroup of order p. 1 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 16 First isomorphism theorem, and symmetries of a cube In this lecture we prove the “1st Isomorphism Theorem” and discuss some consequences. We also explore the rotational symmetries of a cube in 3-dimensional Euclidean space. § 1st Isomorphism Theorem: Let φ ∶ G Ñ G1 be an onto homomorphism. Then there is a naturally induced map which we write as ψ ∶ G{kerpφq ÐÑ G1 and this map ψ is an isomorphism of groups. Proof. Write N “ kerpφq. Define the map ψ as follows: for any coset aN P G{N we let ψpaN q “ φpaq. Let us check this is well-defined. Suppose aN “ bN . This means ab´1 P N , i.e. e1 “ φpab´1 q “ φpaqφpbq´1 . Thus φpaq “ φpbq. In particular, φpaq “ ψpaN q “ ψpbN q “ φpbq This tells us ψ is well-defined map, independent of how the coset aN is written. Next, we check ψ is a homomorphism. For aN, bN P G{N we simply compute ψpaN qψpbN q “ φpaqφpbq “ φpabq “ ψpabN q “ ψpaN bN q and thus ψ is a homomorphism. Finally, we check that ψ is 1-1 and onto. Let a1 P G1 . Because φ is onto, there is some a P G such that φpaq “ a1 . Then also ψpaN q “ φpaq “ a1 . Therefore ψ is onto. Finally, suppose aN, bN P G{N are such that ψpaN q “ ψpbN q. This implies φpaq “ φpbq, or φpab´1 q “ e1 , implying ab´1 P N . In particular, aN “ bN . Thus ψ is 1-1, and ψ is an isomorphism. Examples 1. Let φ ∶ Z20 Ñ Z5 be the homomorphism given by φpk pmod 20qq “ k pmod 5q. Then kerpφq “ x5y Ă Z20 . The 1st Isomorphism Theorem gives Z20 {x5y – Z5 2. Consider the exponential map φ ∶ pR, `q Ñ pU p1q, ˆq, where U p1q Ă Cˆ is the circle group, defined by φpθq “ e2πiθ . This is an onto homomorphism. The kernel is kerpφq “ tθ P R ∶ e2πiθ “ 1u “ Z Ă R By the 1st isomorphism theorem we conclude we have an isomorphism R{Z – U p1q 1 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 16 3. Consider the group of upper triangular matrices in GL2 pRq given by ” ˆ ˙ * a b ∶ a, b, c P R, ac ‰ 0 G“ A“ 0 c Define a map φ ∶ G Ñ Rˆ ˆ Rˆ by φpAq “ pa, cq. This is easily checked to be an onto homomorphism. The kernel of φ is given by the subgroup H Ă G of upper triangular matrices with a “ c “ 1. Thus H is normal. The 1st Isomorphism Theorem gives G{H – Rˆ ˆ Rˆ In particular, G{H is abelian. § Let G be cyclic. If |G| “ 8 then G – Z. If |G| is finite then G – Zn where n “ |G|. Proof. Since G is cyclic, G “ xay “ tak ∶ k P Zu for some a P G. Define a map φ ∶ Z ÐÑ G by setting φpkq “ ak . Then this is onto, since a generates G. The kernel of φ consists of m P Z such that am “ e in G. There are two cases. If G is finite, then |G| “ n is the order of a and kerpφq “ nZ Ă Z. The 1st Isomorphism Theorem gives Zn “ Z{nZ “ Z{kerpφq – G In the other case, G is infinite, and there are no m aside from m “ 0 such that am “ e, and thus the kernel of φ is trivial. Then G – Z. Symmetries of a cube Now let us turn to the group G which consists of the rotational symmetries of a cube situated in 3-dimensional Euclidean space. This group has 24 elements, as shown on the next page. We first follow a familiar strategy of describing this group: label the vertices 1 to 8, and associate to each rotation a P G a corresponding permutation φpaq P S8 based on how the vertices are moved around. This gives a homomorphism φ ∶ G ÐÑ S8 This homomorphism φ is 1-1, but it is not onto: indeed, |G| “ 24 but the target group S8 has 8! “ 40320 elements. 2 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 16 Mapping G into S8 is not the most ideal scenario. After all, |S8 | “ 40320 whereas G has order only 24. A better way of representing G as a group of permutations is as follows. There are 4 diagonal axes that pass through the cube; each one goes through two vertices that are as far apart as possible. Label these 4 diagonal axes 1, 2, 3, 4. Then for a rotation a P G we define ψpaq P S4 to be the permutation determined by how the diagonal axes are moved around by the rotation a. This gives a homomomorphism ψ ∶ G ÐÑ S4 It is straightforward to verify that ψ is 1-1, i.e. that a rotation of the cube is entirely determined by how these 4 diagonal axes are permuted. Then, since |G| “ 24 “ |S4 |, we know ψ must also be onto. Therefore ψ is an isomorphism! The construction of the map ψ is illustrated on the next page. Finally, let us return to the 1st Isomorphism Theorem. To this end, we define a map µ ∶ S4 ÐÑ S3 as follows. Let σ P S4 . Then a “ ψ ´1 pσq P G is a rotational symmetry of the cube. Consider the six pictures of the cube below, each with a distinguished pair of diagonal axes chosen. These six pictures of the cube are divided into 3 columns labelled 1, 2, 3. You may verify that the rotation a takes the two pictures in column 1 to either the two pictures in column 2, or the two pictures in column 3, and so on. Thus the columns 1, 2, 3 above are permuted. We obtain a permutation of t1, 2, 3u which we call µpσq P S3 . Further, kerpµq “ H “ te, p12qp34q, p13qp24q, p14qp23qu, is a normal subgroup, and µ is onto. The 1st Isomorphism Theorem then tells us that S4 {H – S3 This example is actually rare: there are no homomorphisms Sn Ñ Sn´1 when n ą 4 ! 3 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 17 More isomorphism theorems Last lecture we learned about the 1st Isomorphism Theorem. As its name suggests, it has a few sequels, whose names are just as imaginative! For subsets H and N of a group G we define the subset HN Ă G to be: HN “ tab ∶ a P H, b P N u § 2nd Isomorphism Theorem: Let H, N be subgroups a group G with N normal. Then HN is a subgroup of G, H X N is a normal subgroup of H, and H HN – H XN N Proof. Let us first show HN is a subgroup of G. Clearly e P HN , since e “ ee and e P H, e P N . Let ab, a1 b1 P HN where a, a1 P H, b, b1 P N . Then N a1 “ a1 N since N is normal, and this implies ba1 “ a1 b2 for some b2 P N . Thus pabqpa1 b1 q “ apba1 qb1 “ paa1 qpb2 b1 q P HN. Next, let ab P HN . Then N a´1 “ a´1 N implies b´1 a´1 “ a´1 b1 for some b1 P N . Thus pabq´1 “ b´1 a´1 “ a´1 b1 P HN . Thus HN is a subgroup of G. The argument that H X N is a normal subgroup of H is also straightforward, and is omitted. To prove the isomorphism, we construct a homomorphism φ ∶ H ÐÑ HN {N by setting, for each a P H, φpaq “ aN . Any coset in HN {N is of the form abN “ aN for ab P HN , and φpaq “ aN , so φ is onto. The map φ is a homomorphism: φpaa1 q “ aa1 N “ aN a1 N “ φpaqφpa1 q Now let us compute kerpφq. Suppose a P H and φpaq “ aN “ N . This means a P N . Thus kerpφq “ H X N . The 1st Isomorphism Theorem then gives HN {N – H{kerpφq “ H{H X N § Example: Let m, n be positive integers. Let G “ Z and H “ mZ, N “ nZ. Recall that gcdpm, nq is characterized as being the smallest positive integer contained in the set mZ ` nZ “ tam ` bn ∶ a, b P Zu This subset of Z is the subgroup “HN ” in this example. We see that mZ`nZ “ gcdpm, nqZ. On the other hand, the subset corresponding to H X N is given by mZ X nZ “ tk P Z ∶ k “ am “ an for some a, b P Zu 1 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 17 Noting that the smallest positive integer in this subset is the lcm of m and n, we obtain mZ X nZ “ lcmpm, nqZ. The 2nd Isomorphism Theorem gives us mZ H HN gcdpm, nqZ “ – “ lcmpm, nqZ H X N N nZ If k divides l, then |kZ{lZ| “ l{k. (This can be shown explicitly, but is also deduced as a consequence of the 3rd Isomorphism Theorem below.) By comparing orders of groups, lcmpm, nq{m “ n{gcdpm, nq Ôñ lcmpm, nqgcdpm, nq “ mn. § Example: Consider G “ S4 and the normal subgroup N “ te, p12qp34q, p13qp24q, p14qp23qu of order 4 inside of S4 . Then we saw from last lecture that S4 {N – S3 by constructing an onto homomorphism φ ∶ S4 Ñ S3 , recognizing its kernel to be N , and using the 1st Isomorphism Theorem. Let us give a different method for showing S4 {N – S3 . Take H “ S3 Ă S4 . This a (non-normal) subgroup. Then we first claim HN “ S4 . To see this, we proceed as follows. First, S3 “ S3 e Ă HN , so in particular p12q, p23q, p31q P HN . Next, as p12q, p23q, p31q P S3 and p12qp34q, p23qp14q, p31qp24q P N we get the products p12q pp23qp31qq “ p31q, p23q pp23qp14qq “ p14q, p31q pp31qp24qq “ p24q also as elements of HN . Thus every transposition p12q, p13q, p14q, p23q, p24q, p34q in S4 is in HN . Every permutation in S4 is a product of transpositions, and therefore S4 “ HN . Next note that H X N “ teu by direct inspection. Finally, the 3rd Isomorphism Theorem gives S4 {N “ HN {N – S3 {S3 X N “ S3 {teu – S3 § 3rd Isomorphism Theorem: Let G be a group and suppose H, N Ă G are normal subgroups with N Ă H. Then we have an isomorphism G{N G – H H{N Proof. We define a map φ ∶ G{N Ñ G{H by setting φpaN q “ aH. Note that if aN “ bN , then ab´1 P N Ă H, and thus aH “ bH. Thus φpaN q “ aH “ bH “ φpbN q, and φ is well-defined. Furthermore, φ is an onto homomorphism, by repeating earlier arguments. The kernel of φ is given by aN P G{N such that φpaN q “ aH “ H, which is true if and only if a P H. Thus kerpφq “ taN ∶ a P Hu “ H{N . The 1st Isomorphism Theorem gives G{H – G{N G{N “ kerpφq H{N 2 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 17 § Example: Consider Zn “ Z{nZ. Note that mnZ Ă nZ for any integer m. With G “ Z, H “ nZ and N “ mnZ we obtain from the 3rd Isomorphism Theorem: Zn “ Z{mnZ Z – nZ nZ{mnZ Taking orders, we get n “ |Zn | “ |Z{mnZ|{|nZ{mnZ| “ mn{|nZ{mnZ|. In particular, |nZ{mnZ| “ m. Setting l “ mn and k “ n we deduce in general that if k divides l then |kZ{lZ| “ l{k. 3
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