Algebra Prelim Solutions, August 2014

  1. (a)
    If x∈X, then o(g) is a power of p, and since o(g) = o(xgx-1), we see that o(g·x)∈X. Also g·(h·x) = g·(hxh-1) = ghxh-1g-1 = (gh)·x for g, h∈P. Finally 1·x = x, and so we have an action.

    (b)
    {z} is an orbit of size 1 if and only if g·z = z for all g∈P, if and only if z is in the center of P.

    (c)
    The size of the orbits divide | P| and therefore are powers of p. Let Z denote the center of G. By (b), the number of orbits of size 1 is | Z|. Since p | | G|, we see that p | | P| and hence p | | Z|, because the center of a nontrivial p-group is nontrivial. The result follows.

  2. We prove the result by induction on | G|. We may assume that G≠1, because if G is the trivial group, then G has no maximal subgroups. Let Z denote the center of G and first suppose H⊇Z. By subgroup correspondence theorem, H/Z is a maximal subgroup of G/Z. By induction, H/Z⊲G/Z and |(G/Z)/(H/Z)| = p. Therefore H⊲G and | G/H| = p.

    Now assume that H⊉Z. Since HZ≤G and HZ≠H, we see that HZ = G. Since the normalizer of H in G contains H and Z, we see that H⊲G. Since G/H is a nontrivial p-group, its center Y/H is nontrivial and we see that Y = G, by maximality of H. Therefore G/H is abelian. But then G/H has a subgroup K/H of order p, and we must have K = G, again by maximality of H, and the result is proven.

  3. Let I⊲R. Since R is noetherian, there exist x1,..., xn∈R such that I = (x1,..., xn). Let g denote the greatest common divisor of {x1,..., xn}. Since g | xi for all i, there exist ri∈R such that xi = gri and we see that I⊆(g). Also xi/g∈R for all i and no prime divides all the xi. Therefore (x1/g,..., xn/g) = R, in particular there exist si∈R such that x1s1/g + ... + xnsn/g = 1 and hence g = x1s1 + ...xnsn. Therefore g∈I, consequently I = (g) and it follows that R is a PID, as required.

  4. Since M is an injective ℤ-module over the PID I and q≠ 0, we see that qM = M. Now let m⊗z be a simple tensor in M⊗ℤℤ. Since qM = M, there exists n∈M such that qn = m and the

    m⊗z = qn⊗z = n⊗qz = n⊗0 = 0.

    Since every tensor is a sum of simple tensors, it follows that M⊗ℤℤ/qℤ = 0.

  5. Since M is a finitely generated module over the PID ℂ[x], we may write M = F⊕T, where F is a free ℂ[x]-module of finite rank and T is a finitely generated torsion module. Furthermore we may write F = ⊕i=1nℂ[x]/(x -ai)bi for some integers n, bi and ai∈ℂ. First suppose F = 0. Note that dimℂT < ∞, so dimℂM < ∞, in particular no such N can exist ( dimℂM = dimℂN⇒M≌N as ℂ-modules).

    Therefore we may assume that F≠ 0. Now choose any c∈ℂ with c≠ai for all i. By the Chinese remainder theorem (x -c)T = T. Also (x -c)F≌F and hence (x -c)M≌M. Finally F = ℂ[x]m for some m∈ℕ, consequently (x -c)F = (x -c)ℂ[x]m and we deduce that (x -c)F≠F. Therefore (x -c)M≠M and the result follows (in fact there exist infinitely many such c).

  6. (a)
    A polynomial has a degree 1 factor if and only if it has a root. Therefore a degree 2 polynomial f∈𝔽2[x] is irreducible if and only if f (0) = f (1) = 1. There are only 4 degree 2 polynomials, and it is easy to see that only x2 + x + 1 satisfies this criterion.

    (b)
    If g ≔x5 + x3 + 1 is not irreducible, it has a factor of degree 1 or 2. But g(0) = g(1) = 1 and x2 + x + 1 does not divide g. Therefore g is irreducible and it follows that [𝔽2(α) : 𝔽] = 5. If h ≔x4 + x + 1 is not irreducible, it has a factor of degree 1 or 2. But h(0) = h(1) = 1 and x2 + x + 1 does not divide h. Therefore h is also irreducible and it follows that [𝔽2(β) : 𝔽2] = 4. Since 4 and 5 are coprime, we deduce that [𝔽2(α,β) : 𝔽2] = 4·5 = 20.

    (c)
    Since all field extensions involving finite fields are Galois extensions, K = 𝔽2(α,β). Also we know that the Galois group is cyclic with order the degree of the extension. Therefore Gal(K/𝔽2)≌ℤ/20ℤ.

  7. First we compute the character table for S3. There are three conjugacy classes in S3, and representatives are 1, (1 2) and (1 2 3). There is the trivial representation with character χ1 defined by χ1(x) = 1 for all x∈S3. Then there is the character χ2 which is defined by the sign of a permutation, so χ2(1) = χ2(1 2 3) = 1 and χ2(1 2) = -1. The number of irreducible characters equals the number of conjugacy classes, so there are exactly three irreducible characters. The final character χ3 can be determined by the orthogonality relations. We have χ3(1) = 2. Since χ an irreducible character if and only if $ \overline{{\chi}}$ (complex conjugate) is an irreducible character, we see that χ3(1 2) and χ3(1 2 3) are real numbers. Taking the inner product of the first two columns of the character table, we obtain χ3(1 2) = 0, and then it follows easily that χ3(1 2 3) = -1. Thus the character table of S3 is

    Class Size 1 3 2
    Class Rep 1 (1 2) (1 2 3)
    χ1 1 1 1
    χ2 1 -1 1
    χ3 1 0 -1

    Since ℤ/3ℤ is an abelian group, all its irreducible characters are of degree one and correspond to homomorphisms into the cube roots of 1 in ℂ, because |ℤ/3ℤ| = 3. Let ω = e2πi/3, a primitive cube root of 1. Let 0, 1, 2 represent the conjugacy classes $ \bar{{0}}$, $ \bar{{1}}$, $ \bar{{2}}$ respectively. Then the character table for ℤ/3ℤ is

    Class Size 1 1 1
    Class Rep 0 1 2
    ψ1 1 1 1
    ψ2 1 ω ω2
    ψ3 1 ω2 ω

    Now for a finite group of the form G×H, the conjugacy classes of G×H is C×D, where C is the set of conjugacy classes of G and D is the set of conjugacy classes of D, and then the irreducible characters are of the form χiψj ≔χi(c)ψj(d ), in particular there are | C|·| D| irreducible characters. Therefore the character table of S3×ℤ/3ℤ is

    Class Size 1 1 1 3 3 3 2 2 2
    Class Rep (1, 0) (1, 1) (1, 2) ((1 2), 0) ((1 2), 1) ((1 2), 2) ((1 2 3), 0) ((1 2 3), 1) ((1 2 3), 2)
    χ1ψ1 1 1 1 1 1 1 1 1 1
    χ1ψ2 1 ω ω2 1 ω ω2 1 ω ω2
    χ1ψ3 1 ω2 ω 1 ω2 ω 1 ω2 ω
    χ2ψ1 1 1 1 -1 -1 -1 1 1 1
    χ2ψ2 1 ω ω2 -1 - ω - ω2 1 ω ω2
    χ2ψ3 1 ω2 ω -1 - ω2 - ω 1 ω2 ω
    χ3ψ1 2 2 2 0 0 0 -1 -1 -1
    χ3ψ2 2 2ω 2ω2 0 0 0 -1 - ω - ω2
    χ3ψ3 2 2ω2 2ω 0 0 0 -1 - ω2 - ω



Peter Linnell 2014-08-12