MATH 701 SPRING 2007 HOMEWORK 6. Due Tuesday, March 20 at the beginning of class.

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1 MATH 70 SPRING 2007 HOMEWORK 6 Due Tuesday, March 20 at the beginning of class. 4. (5 points) Classify all groups of der 2p where p is an odd prime integer. (This instruction means state and prove a result which says, If G is a group of der 2p, where p is an odd prime integer, then G is isomphic to exactly one of the following groups:.... ) Theem. If G is a group of der 2p, where p is an odd prime integer, then G is a cyclic group a dihedral group. Proof. Let G be a group of der 2p. The Sylow Theems guarantee that G has an element b of der p and an element a of der 2. The subgroup <b> has index 2 in G; hence, <b> is a nmal subgroup of G. We also know that G = <a, b>. We first suppose that the subgroup <a> of G is also nmal in G. It follows that gag is an element of der 2 in <a> f all g G; and therefe, gag = a f all g G, and G is abelian. There is no difficulty in showing that G is generated by ab. Hencefth, we assume that <a> is not a nmal subgroup of G. The Sylow Theems guarantee that G has me than one sugroup of der 2 and the number of subgroups of der 2 is odd and divides 2p. Thus, G must have p subgroups of der 2. In other wds, every element of G that is not in <b> has der 2. It follows that ab has der two. Thus, G is a group of der 2p with G generated by a and b with a 2 = (ab) 2 = b p = id. It follows that G is a dihedral group. 6. (5 points) Classify the groups of der 8. (8) (6) Theem. If G is a group of der 8, then G is isomphic to D 9 S 3 θ, where θ : Aut sends from to the automphism of which sends each element to its inverse. Proof. Let G be a group of der 8. If G is abelian, then G is isomphic to either (8) (6). (Either make a direct proof appeal to the structure theem.)

2 2 MATH 70 SPRING 2007 HOMEWORK 6 Hencefth, we assume that G is not abelian. The Sylow Theem guarantees that G has a subgroup N of der 9. This subgroup has index 2 in G; so it is a nmal subgroup of G. Every group of der p 2 is abelian, so N is abelian. Let be the center of G. Claim. The der of is 3. Proof. If b is any element of G not in N, then G = N bn. If b were in, then G would be abelian. Thus, b / and N. We know that can not be all of N; else, once again, G would be abelian. We conclude that is a proper subgroup of N. Claim is established. Claim 2. If = 3, then G = S 3. Proof. In this case G is isomphic to (6) S 3. However, if G were isomphic to, then G would be abelian and we have ruled this case out of consideration. So, (6) G must be isomphic to S 3. Thus, there exist a and b in G with a 3, b 2, abab, and G is generated by a, b, and. I now claim there exist A and B in G so that A 3 = id, B 2 = id, ABAB = id, and G is generated by A, B, and. If b 2 = c, then take B = bc. Observe that B 2 = bcbc = bbcc = c 3 = id. If (ab) 2 = c, then take A = c a. Observe that ABAB = c abc ab = c c (ab) 2 = c c c = id. Observe that A 3 must equal id. Indeed, let us suppose A 3 = z. We already showed that BAB = A ; hence, z = BzB = BA 3 B = A 3 = z. The only element of which is its own inverse is id. We now have an isomphism and Claim 2 is established. <a, b> G, Claim 3. If = {id} and x is any element of G with x / N, then x 2 = id. Proof. The group G N is cyclic of der 2; so, x2 N. Thus, x 2 commutes with x and also with every element of N. It follows that x 2 = {id}. The proof of claim 3 is complete. Claim 4. If = and N is cyclic, then G = D 9. Proof. Fix a generat a f N and any element b G \ N. We have that G is a group of 8 elements generated by a, b with a 9 = id, b 2 = id (take x = b in Claim 3) and (ab) 2 = id (take x = ab in Claim 3). Thus, G is D 9 and the proof of Claim 4 is complete.

3 MATH 70 SPRING 2007 HOMEWORK 6 3 Claim 5. If = and N ( =, then G is isomphic to as described above. ) θ Proof. Fix generats a, b f N. Pick c G \ N. We see that G is generated by a, b, c. We also know that a 3 = b 3 = id and ab = ba. Apply Claim 3 three times to see that c 2 = (ca) 2 = (cb) 2 = id. Consider the free group F = <X, Y, >. Let N be the smallest nmal subgroup of F which contains X 3, Y 3, 2, (X) 2, (Y ) 2, XY X 2 Y 2. It is clear that F N has at most 8. If F N has at least 8 elements then there is a surjective homphism F G which sends the class of X to a, the class N of Y to b and the class of to c and G = F N. I finish the proof by exhibiting a surjection from F N onto a group with 8 elements that is known to exist. This surjection shows that F N 8; and therefe, the calculation of the previous ( paragraph ) may be made. Of course, my honest group with 8 elements is θ. Take a = ((, 0), 0), b = ((0, ), 0), c = ((0, 0), ). There is no difficulty observing that a 3 = b 3 = c 2 = ((0, 0), 0); ca = ((0, 0), )((, 0), 0) = ((, 0), ), and ab = ba. cb = ((0, 0), )((0, ), 0) = ((0, ), ), (ca) 2 = ((, 0), )((, 0), ) = ((0, 0), 0) = (cb) 2 7. (2 points) Classify all groups of der less than 20, except the groups of der 6. (Actually, this is not a hard problem. F most of it you need only quote something that we have already done is otherwise obvious.) Theem. Let G be a group. () If G has der, 2, 3, 5, 7,, 3, 5, 7, 9 then G is a cyclic group. If G has der 4, then either G is cyclic G is isomphic to. If G has der 6, 0, 4, then either G is a cyclic group G is a Dihedral group. (4) If G has der 8, then G is isomphic to one of the groups: (8), (4),, D 4, the quaterion group <a, b; a 4 = id, b 2 = a 2, bab = a 3 >. (5) If G has der 9, then either G is cyclic G is isomphic to. (6) If G has der 2, then G is isomphic to (2), (6), D 6, A 4, C 3 θ C 4, where C n is the cyclic group of der n, and θ : C 4 Aut(C 3 ) is the group homomphism which sends a i to σ i, where C 4 = <a> and σ : C 3 C 3 is the automphism which sends x to x f all x C 3.

4 4 MATH 70 SPRING 2007 HOMEWORK 6 (7) If G has der 8, then G is isomphic to (8) (6) D 9 S 3 θ, where θ : Aut sends from to the automphism of which sends each element to its inverse. Proof. There are three cases involved in the proof of (). If G only has one element, then G isn t very interesting, but it is a cyclic group. If G has prime der, then it is a consequence of Lagrange s Theem that G is cyclic. If G has der 5, then it is a consequence of Sylows Theem that G is cyclic. (Count the number of subgroups of der 3 and the number of subgroups of der 5.) F and (5), we classified all groups of der p 2, where p is prime. Assertion is problem 4. Assertions (4) and (6) were done in class. Assertion (7) is problem (5 points) Find all subgroups of G = of der p2. I suppose that one can grub around until one has a list of all of the desired subgroups. I can think of two me interesting methods of attack. One could notice that G is a vect space over the field and that a subset of G is a subgroup of G if and only if it is a subspace of G. Furtherme, G is an inner product space, with inner product given by dot product. If V is a subspace of G, then the annhilat of V is a subspace of G of dimension 3 dim V, and ann(ann V ) = V. In particular every subspace of dimension 2 is the annhilat of a subspace of dimension. We found the subspaces of G of dimension on exam 2. (I don t think that I will outline any me of this approach. The calculation is equivalent to what I will do below.) Let K be a subgroup of G of der p 2. Then K is a nmal subgroup of G and G K is a group of der p. In other wds, G K = identically zero) group homomphism ϕ: G. In other wds, there exists a (not such that K = ker ϕ. Such a group homomphism is completely described by what it does to the elements 0, 0 0 0, and 0 0 of G. Furtherme, these elements of G may be sent to any element of. In other wds, there are integers a, a 2, a 3 with ϕ n n 2 = a n + a 2 n 2 + a 3 n 3 in n 3

5 MATH 70 SPRING 2007 HOMEWORK 6 5 f all n n 2 in G. Let i be the least index with a i not equal to zero in. We may n 3 write, which is the greatest common divis of a i and p, in the fm ra i + sp =, f some integers r and s. We see that the homomphism r ϕ: G has the same kernel as ϕ has. Thus, K is the kernel of [ b c ] [ 0 c ] [ 0 0 ] f some integers b, c. Thus, K is equal to one of the following subgroups of G: < b, c 0 > 0 < 0, 0 c > 0 < 0, 0 >, 0 0 f some elements b and c in. A few minutes wth of calculation shows that each of the listed subgroups are distinct. Therefe there are p 2 + p + subgroups of G of der p 2.