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

    Next, PH/HP/PH, 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 PH and PH/H are Sylow p-subgroups. Since P/PHPH/H and | P| = | P/PH|·| PH|, we see that P is a p-group. Finally if | H| = pax and | G/H| = pby, where x and y are prime to p, then | G| = pa+bxy and xy is prime to p. This means that a Sylow p-subgroup of G has order pa+b. But since | PH| = pa and | PH/H| = pb, we see that | P| = pa+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 gG gives the permutation xPgxP. 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≠xPQ, then CG(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≠qQ such that qPQ, 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 pm + 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 pm + qn, hence p divides qn, which is not possible.

    Now let IR. 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 upiqj, where u is a unit and i, j are nonnegative integers. Choose 0≠xI such that x = piqj, with i as small as possible, and then choose 0≠yI such that y = pkql, with l as small as possible. We will show that I = (piql). Clearly I⊆(piql). On the other hand pk-i + qj-l is a unit by the above, hence piql is an associate of y + x and we see that piqlI. This proves that I = (piql).

  4. Note that k[x] is a PID. By the structure theorem for finitely generated modules over a PID, we may write Mk[x]dk[x]/(f1)⊕...⊕k[x]/(fn), where the fi are monic polynomials, say of degree ai, and fi | fi+1 for all i. Suppose d = 0. Then dimkM = ∑i=1nai, and then it is clear that if N is a proper k[x]-submodule of M, then NM, because dimkN < dimkM. Therefore d > 0, in particular there is an epimorphism Mk[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 MC.

  5. Let p denote the characteristic of K. Then we may write | K| = pn where n∈ℕ. Set M = K+L×. 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⊗ℤ/(| L|-- 1)ℤ and φ : ℤ/pℤ→ℤ/pℤ⊗ℤ/(| 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⊗ℤ/(| 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 KL = ℚ(α1,...,αn), let fi denote the minimum polynomial of αi over ℚ, and set f = f1...fn. 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 fi lie in both K and L and hence the splitting field of f is contained in KL. Therefore KL is the splitting field of f and it follows that KL is a Galois extension of ℚ.

  7. We can split up the given exact sequence into two short exact sequences, namely 0→ℤ→PY→ 0 and 0→YQ→ℤ→ 0. Then using the long exact sequence for Ext in the first variable, we obtain

    H1(G, X) = ExtℤG1(ℤ, X)≌ExtℤG2(Y, X)≌ExtℤG3(ℤ, X) = H3(G, X),

    as required.

Peter Linnell 2013-08-23