Algebra Prelim Solutions, August 2019

  1. Let G be a simple group of order 4860. The number of Sylow 3-subgroups is congruent to 1 mod 3 and divides 20, so if G does not have a normal Sylow 3-subgroup, it has 4 or 10 Sylow 3-subgroups. If there are 4 Sylow 3-subgroups, then G will be isomorphic to a subgroup of A4 which has order 12, which is clearly not possible because 12 < 4860. Therefore G must have 10 Sylow 3-subgroups and then G will be isomorphic to a subgroup of A10. This is not possible because 35 | 4860, but the largest power of 3 dividing | A10| = 10!/2 is 4. Therefore there is no such G, as required.

  2. Let f (x) = 2x3 +19x2 - 54x + 3. If f is not irreducible, then it must have a degree one factor, which we may assume is of the form ax + b where a, b∈ℤ and (a, b) = 1, a primitive polynomial in ℤ[x]. Write f (x) = (ax + b)g(x) where g∈ℚ[x]. Then by Gauss's lemma, g(x)∈ℤ[x]. Write g(x) = cx2 + dx + e where c, d, e∈ℤ. We now equate coefficients. We have ac = 1, so either a = ±1 or a = ±2, and be = 3. Suppose a = ±1. Then ±1 or ±3 is a root of f, which by inspection is not the case. On the other hand if a = ±2, then ±1/2 or ±3/2 is a root of f, which again by inspection is not the case. This proves that f is irreducible in ℚ[x].

  3. Define θ : S×S→S by θ(s, t) = st. Note that for s1, s2, t1, t2∈S and r∈R,

    θ(s1 + s2, t1) = (s1 + s2)t1 = s1t1 + s2t1 = θ(s1, t1) + θ(s2, t1),    
    θ(s1, t1 + t2) = s1(t1 + t2) = s1t1 + s1t2 = θ(s1, t1) + θ(s1, t2),    
    θ(s1r, t1) = s1rt1 = θ(s1, rt1).    

    This shows that θ is an R-balanced map. Therefore θ induces a group homomorphism φ : S⊗RS→S such that φ(s1, t1) = s1t1, in particular φ(1⊗1) = 1≠ 0. It follows that S⊗RS≠ 0.

  4. By the structure theorem for finitely generated modules over a PID, we may write I = T⊕F where T is the torsion submodule of I and F is a free R-module. First suppose F≠ 0. Then we may write F = E⊕R where E is a free module, so I = T⊕E⊕R. Since R is not a field, we may choose r∈R∖ 0 which is not a unit in R. Also I is an injective R-module, so sI = I and hence sR = R and we have a contradiction. Therefore F = 0 and hence I is a torsion module. It follows there exists s∈R∖ 0 such that sI = 0. But sI = I because I is injective and we conclude that I = 0 as required.

  5. Let A∈GL8(ℚ) be an element of order 7. Then A7 = I, which means that the minimal polynomial of A divides x7 - 1. Now x7 - 1 = f (x)(x - 1) where f (x) = x6 + x5 + x4 + x3 + x2 + x + 1, and f (x) is irreducible. Since the minimal polynomial of A is not x - 1, we see that the minimal polynomial of A is either f (x) or x7 - 1. Since the characteristic polynomial has degree 8, it must be f (x)(x - 1)2. It follows that there is one conjugacy class of matrices in GL8(ℚ) which consists of elements of order 7, namely the matrices with invariant factors {x7 -1, x - 1}.

  6. Write G = Gal(K/ℚ) and F = ℚ(e2πi/p). Since G≌S5, we know that [K : ℚ] = | S5| = 120.
    (a)
    If f is not irreducible, then we may write f = f1f2 where deg f1, deg f2≥1 and deg f1 + deg f2 = deg f = 5. We then have

    [K : ℚ]≤(deg f1)!(deg f2)! < 5! = 120

    and we have a contradiction. Therefore f is irreducible. Since an irreducible polynomial over a field of zero characteristic has distinct roots, it follows that f has 5 distinct roots.

    (b)
    If a is a root of f, then so is γa and it follows that γ permutes the roots of f. Also F is a cyclic extension of ℚ of degree p - 1. The subgroup corresponding to this extension in G will be a normal subgroup of index p - 1 in G with cyclic quotient. The only normal subgroups of G which have this property have index 1 or 2 and it follows that p = 3.

    (c)
    Since K⊈ℝ, we see that f must have at least one complex root. Since complex roots appear in pairs, this means that f has either 2 complex roots or 4 complex roots. Since A5 is the unique subgroup of index 2 in G, we see that F is the fixed field of A5. Now if f has 4 complex roots, then γ∈A5 and hence γ fixes F, which is not the case. It follows that f has 2 complex roots and therefore fixes 3 of the roots of f.

  7. Write ψ = IndHGχ. By Frobenius reciprocity, (ψ,ψ)G = (ψ|H,χ)H. Since {1, x, x2} is a transversal for H in G, we see that for h∈H,

    ψ(h) = χ(h) + χ(xhx-1) + χ(x2hx-2) = 3χ(h)

    and we deduce that (ψ|H,χ)H = 3. Therefore if we write ψ = a1χ1 + ... + anχn where ai∈ℕ and χi is an irreducible character for all i, then a12 + ... + an2 = 3 and we must have ai = 1 for all i and n = 3. This proves the result.