Algebra Prelim Solutions, August 2013

1. We use without further comment the property that a p-subgroup of a group is a Sylow p-subgroup if and only if it has index in the group prime to p.

   First suppose P is a Sylow p-subgroup of G. Then P∩H is a subgroup of P, so P∩H is a p-subgroup of H. Also P/P∩H≌PH/H, hence | H|/| P∩H| = | PH|/| P|. Furthermore | G|/| P| = | G|/| PH|·| PH|/| P| and we deduce that | H|/| P∩H| divides | G|/| P|. Since | G|/| P| is prime to p, it follows that | H|/| P∩H| is also prime to p, which proves that P∩H is a Sylow p-subgroup of H.

   Next, PH/H≌P/P∩H, so PH/H is a p-subgroup of G/H. Furthermore | G|/| P| = | G|/| PH|·| PH|/| P|, and we see that | G|/| PH| is prime to p. Therefore | G/H|/| PH/H| is also prime to p and it follows that PH/H is a Sylow p-subgroup of G/H.

   Now suppose P∩H and PH/H are Sylow p-subgroups. Since P/P∩H≌PH/H and | P| = | P/P∩H|·| P∩H|, we see that P is a p-group. Finally if | H| = p^ax and | G/H| = p^by, where x and y are prime to p, then | G| = p^a+bxy and xy is prime to p. This means that a Sylow p-subgroup of G has order p^a+b. But since | P∩H| = p^a and | PH/H| = p^b, we see that | P| = p^a+b and hence P is a Sylow p-subgroup of G, as required.

2. Suppose G is simple group of order 576. The number of Sylow 2-subgroups is congruent to 1 mod 2 and divides 9, so has to be 1, 3 or 9. It cannot be 1, because then G would have a normal Sylow 2-subgroup. Nor can it be 3, otherwise G would be isomorphic to a subgroup of A₃. Finally suppose it is 9. Then G is isomorphic to a subgroup of A₉; unfortunately at first sight this seems possible, because 576 divides | A₉|. However the isomorphism is induced by the representation of G on the 9 left cosets of a Sylow 2-subgroup P in G. Thus g∈G gives the permutation xP↦gxP. Then g stabilizes some xP if and only if g is in some Sylow 2-subgroup. So if g has order 6, it can be considered as an element of A₉ which fixes no points; a quick check shows that this is not possible and therefore G has no element of order 6.

   Now consider the Sylow 3-subgroups. If P and Q are distinct Sylow 3-subgroups and 1≠x∈P∩Q, then C_G(x) contains P and Q and hence contains an element of order 2. It follows that G has an element of order 6, which is not possible by the previous paragraph, so distinct Sylow 3-subgroups intersect trivially.

   Next the number of Sylow 3-subgroups is 16 or 64. If it is 16, we consider the representation of G on the left cosets of a Sylow 3-subgroup P. If Q is another Sylow 3-subgroup, then there is an orbit under Q which has order 3, which shows that there exists 1≠q∈Q such that q∈P∩Q, contradicting the previous paragraph. Finally if there are 64 Sylow 3-subgroups, then since two distinct Sylow 3-subgroups intersect in the identity, we can count elements to show that G has a normal Sylow 2-subgroup.

3. Consider p^m + qⁿ. If this is not a unit, then there exists some prime which divides it, which without loss of generality we may assume is p. Thus p divides p^m + qⁿ, hence p divides qⁿ, which is not possible.

   Now let I⊲R. We want to prove I is a principal ideal, and since 0 is clearly a principal ideal, we may assume that I≠ 0. Each nonzero element of I has a factorization upⁱq^j, where u is a unit and i, j are nonnegative integers. Choose 0≠x∈I such that x = pⁱq^j, with i as small as possible, and then choose 0≠y∈I such that y = p^kq^l, with l as small as possible. We will show that I = (pⁱq^l). Clearly I⊆(pⁱq^l). On the other hand p^k-i + q^j-l is a unit by the above, hence pⁱq^l is an associate of y + x and we see that pⁱq^l∈I. This proves that I = (pⁱq^l).

4. Note that k[x] is a PID. By the structure theorem for finitely generated modules over a PID, we may write M≌k[x]^d⊕k[x]/(f₁)⊕...⊕k[x]/(f_n), where the f_i are monic polynomials, say of degree a_i, and f_i | f_i+1 for all i. Suppose d = 0. Then dim_kM = ∑_i=1ⁿa_i, and then it is clear that if N is a proper k[x]-submodule of M, then N≇M, because dim_kN < dim_kM. Therefore d > 0, in particular there is an epimorphism M↠k[x]. Since C is a cyclic k[x]-module, there exists an epimorphism k[x]↠C. By composing these two epimorphisms, we obtain a k[x]-module epimorphism M↠C.

5. Let p denote the characteristic of K. Then we may write | K| = pⁿ where n∈ℕ. Set M = K⁺⊗_ℤL^×. Then | K⁺| M = 0 and | L^×| M = 0, so if (| K|,| L|-- 1) = 1, we see that | M| = 0. Now suppose (| K|,| L|-- 1)≠1. Then p divides | L|-- 1. Also K⁺≌(ℤ/pℤ)ⁿ and L^×≌ℤ/(| L|-- 1)ℤ.

   Now we have well defined homomorphisms θ : ℤ/pℤ⊗_ℤℤ/(| L|-- 1)ℤ and φ : ℤ/pℤ→ℤ/pℤ⊗_ℤℤ/(| L|-- 1)ℤ determined by θ($\bar{{x}}$⊗$\bar{{y}}$) = $\overline{{xy}}$ and φ($\bar{{x}}$) = $\bar{{x}}$⊗$\bar{{1}}$, and θφ and φθ are the identity maps. This shows that ℤ/pℤ⊗_ℤℤ/(| L|-- 1)ℤ≌ℤ/pℤ and we deduce that M≌(ℤ/pℤ)ⁿ. Therefore if (| K|,| L|-- 1)≠1, it follows that | M| = | K|.

6. Write K∩L = ℚ(α₁,...,α_n), let f_i denote the minimum polynomial of α_i over ℚ, and set f = f₁...f_n. Let F denote the splitting field of f over ℚ, a subfield of ℂ. Since K and L are Galois extensions of ℚ, all the roots of all f_i lie in both K and L and hence the splitting field of f is contained in K∩L. Therefore K∩L is the splitting field of f and it follows that K∩L is a Galois extension of ℚ.

7. We can split up the given exact sequence into two short exact sequences, namely 0→ℤ→P→Y→ 0 and 0→Y→Q→ℤ→ 0. Then using the long exact sequence for Ext in the first variable, we obtain
   H¹(G, X) = Ext_ℤG¹(ℤ, X)≌Ext_ℤG²(Y, X)≌Ext_ℤG³(ℤ, X) = H³(G, X),
   as required.