Algebra Prelim Fall 2014 Harini Chandramouli Problem 1. Exhibit a non-abelian group of order n3 for ever 2 ≤ n 2 Z. Solution. Consider the Heisenberg group: 80 1 9 1 a c < = H(n) = @0 1 bA a; b; c 2 Z=nZ : : 0 0 1 ; We can see that jH(n)j = n3 since there are n many choices for a; b, and c. The 3-by-3 identity matrix, I3, is clearly in H(n) and so we have an identity element. Inverses are given by: 01 a c1−1 01 −a −ab − c1 @0 1 bA = @0 1 −b A : 0 0 1 0 0 1 Now we want to show that H(n) is closed. Let a1; a2; b1; b2; c1; c2 2 Z=nZ, then 0 1 0 1 0 1 1 a1 c1 1 a2 c2 1 a1 + a2 c2 + a1b2 + c1 @0 1 b1A @0 1 b2A = @0 1 b1 + b2 A 2 H(n) 0 0 1 0 0 1 0 0 1 Thus, H(n) is a group. Now we want to show it is not abelian. Consider 01 1 01 01 0 01 X := @0 1 0A Y := @0 1 1A : 0 0 1 0 0 1 We see that X; Y 2 H(n). 01 1 01 01 0 01 01 1 11 XY = @0 1 0A @0 1 1A = @0 1 1A 0 0 1 0 0 1 0 0 1 01 0 01 01 1 01 01 1 01 YX = @0 1 1A @0 1 0A = @0 1 1A ; 0 0 1 0 0 1 0 0 1 so XY 6= YX and thus H(n) is a non-abelian group of order n3. 1 Problem 2. Let p; q be primes with p < q and q 6= 1 mod p. Show that every group of order pq is abelian. Proof. Let G be a group such that jGj = pq. By the Sylow Theorems, the number of p-Sylow subgroups, np, has the following two prop- erties: np = 1 mod p np q: So since q 6= 1 mod p, we know that np = 1. So there is one p-Sylow subgroup, call it Hp. By the Sylow Theorems, we know that Hp must be self-congruent and hence normal in G. We also know that jHpj = p and so by Lagrange's Theorem, we know that Hp is cyclic. By the Sylow Theorems, the number of q-Sylow subgroups, nq, has the following proper- ties: nq = 1 mod q nq p: So since p < q, we know that this means that nq = 1. Let Hq represent our q-Sylow sub- group. By the same logic as above, Hq is normal in G, cyclic, and jHqj = q. Since jHpj and jHqj are relatively prime, we know that Hp \ Hq = feg. Since Hp and Hq are normal subgroups of G, we know that HpHq is a normal subgroup of G such that jHpj · jHqj jHpHqj = = pq: jHp \ Hqj Thus, HpHq = G. So any element of G can be written as a combination of elements in Hp −1 and Hq. Let a 2 Hp and b 2 Hq. Since Hp is normal, b ab 2 Hp. Since Hq is normal, −1 a ba 2 Hq. Thus, −1 −1 −1 −1 b a ba 2 Hp \ Hq =) b a ba = e =) ab = ba: Thus, G is abelian. 2 Problem 3. Let S; T be linear operators on a finite-dimensional complex vector space V , and ST = TS. Show that S; T have a common eigenvector. Proof. Since C is algebraically closed, we know that the minimum polynomial of T on V has a root, λ 2 C, that is an eigenvalue of T with a corresponding, non-trivial eigenvector for T . Let Vλ represent the λ-eigenspace. That is, Vλ = v 2 V T v = λv : So 8 v 2 Vλ, since S and T commute, T (Sv) = T Sv = ST v = Sλv = λSv = λ(Sv): So S stabilizes Vλ. That is, S maps vectors in Vλ to vectors in Vλ. So S is a linear operator on Vλ. Let f(x) represent the minimum polynomial of S on Vλ. Since C is algebraically closed, we know that f(x) has a root, γ 2 C, that is an eigenvalue of S on Vλ with corresponding non-trivial eigenvector, u 2 Vλ. Notice that γ is also an eigenvalue for S on all of V since f(x) must divide the minimum polynomial of S on all of V and hence u is an eigenvector for S on all of V . Since u 2 Vλ, we know that T u = λu. But since u is an eigenvector for S, we know that Su = γu. Thus, S and T have a common eigenvector. 3 p p p Problem 4. Find all intermediate fields between Q and Q( 2; 3; 5). p p p p p p Proof. First observe that Q( 2; 3; 5) is a Galois extension of Q as Q( 2; 3; 5) is a separable, splitting field over Q. Notice h p p p i h p p p p p i h p p p i h p i Q( 2; 3; 5) : Q = Q( 2; 3; 5) : Q( 2; 3) Q( 2; 3) : Q( 2) Q( 2) : Q : p p p 2 First, it is easy to see that Q( 2) : Q = 2 as 2 is the root of x − 2 but 2 2= Q. p p p p p p Next, we will show that ( 2; 3) : ( 2) = 2. This is true iff 3 2= ( 2) as 3 Q Q p p Q is the root of x2 − 3. So suppose for contradiction that 3 2 Q( 2). Then, 9 a; b 2 Q such that p p 3 = a + b 2 p 3 = a2 + 2b2 + 2ab 2 ( a2 + 2b2 = 3 =) 2ab = 0 p p So either a = 0 or b = 0. If a = 0, then 2b2 = 3 =) b = p3 2= so we have a p 2 Q 2 contradiction. If b = 0, then a = 3 =) a = 3 2= Q, which is also a contradiction.p Notp as well that if a = b = 0, then we get that 0 = 3, which is a contradiction. Thus, 3 2= Q( 2) p p p and Q( 2; 3) : Q( 2) = 2. p p p p p Next, we want to show that ( 2; 3; 5) : ( 2; 3) = 2. This will be true iff p p p p Q Q p 2 5p2= pQ( 2; 3)p as p 5 is the root of x p− 5. So suppose for contradiction that 5 2 Q( 2; 3) = Q( 2)( 3). Then 9 a; b 2 Q( 2) such that p p 5 = a + b 3 p 5 = a2 + 3b2 + 2ab 3 ( a2 + 3b2 = 5 =) 2ab = 0 2 Now, sincep a = b = 0 cannotp hold, let us consider first if a = 0. Then 3b = 5 where b 2 Q( 2). Thus, b = α + β 2 where α; β 2 Q. So, p 3(α + β 2)2 = 5 p 3α2 + 6β2 + 6αβ 2 = 5 ( 3α2 + β2 = 5 =) 6αβ = 0 4 p So ifpα = 0, then we get that β = 5 2= Q, a contradiction. If β = 0, then we get that α = p5 2= , a contradiction. Thus a = 0 cannot hold. 3 Q If b = 0, then p p 2 a = 5 =) a = 5 2 Q( 2): p p p However,p 5 2= Q( 2) by an argument nearly identicalp to thep argumentp as to why 3 2= Q( 2) given above. Thus b = 0 cannot hold. Thus, 5 2= Q( 2; 3) and h p p p p p i Q( 2; 3; 5) : Q( 2; 3) = 2: So we see that h p p p i p p p Q( 2; 3; 5) : Q = 8 =) Gal(Q( 2; 3; 5=Q) = 8: p p p Now any automorphism in Gal(Q( 2; 3; 5=Q) must map a root of any polynomial to another root of that polynomial. Consider p p p p p p α : 2 7! − 2; 3 7! 3; 5 7! 5; Q fixed p p p p p p β : 2 7! 2; 3 7! − 3; 5 7! 5; Q fixed p p p p p p γ : 2 7! 2; 3 7! 3; 5 7! − 5; Q fixed Then the Galois group is as follows: p p p Gal(Q( 2; 3; 5)=Q) = fid; α; β; γ; α ◦ β; α ◦ γ; β ◦ γ; α ◦ β ◦ γg : We also notice that every element of the Galois group (other than the identity) is of order 2. By Lagrange's Theorem, any proper subgroup will be of order 2 or 4. The groups of order 2 are cyclic and hence we can see they are: hαi = fid; αg hβi = fid; βg hγi = fid; γg hα ◦ βi = fid; α ◦ βg hα ◦ γi = fid; α ◦ γg hβ ◦ γi = fid; β ◦ γg hα ◦ β ◦ γi = fid; α ◦ β ◦ γg It is easy to see thatp thesep p are all proper subgroups and all possible subgroups of order 2. Now since Gal(Q( 2; 3; 5)=Q) has no elements of order 4, the subgroups of order 4 will not be cyclic. Using a combinatorial argument, and facts about K4, one can show that there 5 are 7 subgroups of order 4. They are: hα; βi = fid; α; β; α ◦ βg hα; γi = fid; α; γ; α ◦ γg hβ; γi = fid; β; γ; β ◦ γg hα; β ◦ γi = fid; α; β ◦ γ; α ◦ β ◦ γg hβ; α ◦ γi = fid; β; α ◦ γ; α ◦ β ◦ γg hγ; α ◦ βi = fid; γ; α ◦ β; α ◦ β ◦ γg hα ◦ β; α ◦ γi = fid; α ◦ β; α ◦ γ; β ◦ γg : By the Fundamental Theorem of Galois Theory, there exists a one-to-one correspondence between the intermediate fields and the subgroups of the Galois group.