Quiz:Splitting field

Rational and ambivalent
An ambivalent group is a group in which every element is conjugate to its inverse. For a finite group, this is equivalent to every character in characteristic zero being real-valued.

A rational group is a group in which any two elements generating the same cyclic subgroup are conjugate. For a finite group, this is equivalent to every character in characteristic zero being rational-valued.

{ There are only finitely many values $$n$$ such that the alternating group of degree $$n$$ is ambivalent. For which of the following values of $$n$$ is the alternating group $$A_n$$ of degree $$n$$ not ambivalent? - 5 - 6 + 8 - 9 - 10
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 * See classification of ambivalent alternating groups. The degrees $$n$$ for which $$A_n$$ is ambivalent are $$1,2,5,6,9,10,14$$. See also linear representation theory of alternating groups.

{What is the smallest possible order of an ambivalent group that is not a rational group? - 6 - 8 + 10 The group dihedral group:D10 (order 10, degree 5) is ambivalent but not rational. See also dihedral groups are ambivalent, classification of rational dihedral groups. - 12 - 16
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{Which of the following conditions on a natural number $$n > 1$$ is necessary and sufficient for there to be an ambivalent group of order $$n$$? - $$n$$ must be a prime power - $$n$$ must be divisible by at least two primes + $$n$$ must be even - $$n$$ must be divisible by one of the primes 2, 3, or 5 - $$n$$ must be divisible by the square of a prime
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 * See odd-order and ambivalent implies trivial to note that nontrivial odd-order groups fail to be ambivalent. On the other hand, dihedral groups are ambivalent, and there is a dihedral group of every even order.

{A rational-representation group is a group all of whose irreducible representations in characteristic zero can be realized over the rationals, i.e., the rational numbers are a splitting field for the group. All the groups below are rational groups. Which of them is not a rational-representation group? - symmetric group:S3 - dihedral group:D8 (see linear representation theory of dihedral group:D8) + quaternion group (see linear representation theory of quaternion group) - direct product of D8 and Z2 - symmetric group:S4
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 * See faithful irreducible representation of quaternion group. This representation, which has Schur index two, is the culprit.

Splitting fields in characteristic zero
{For a group of order 27, what is the maximum possible degree of a minimal splitting field in characteristic zero as an extension over the rationals? - 3 - 6 - 9 + 18 - 27
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 * The maximum occurs for cyclic group:Z27, where we need to adjoin a primitive 27th root of unity. The degree of this extension is the Euler totient function $$\varphi(27)$$, which takes the value $$3^2(3 - 1) = 18$$.

{Of the five groups of order 8 (see also linear representation theory of groups of order 8), there is exactly one group that does not have a unique minimal splitting field in characteristic zero, i.e., there are multiple non-isomorphic minimal splitting fields in characteristic zero. Which group is this? - cyclic group:Z8 - direct product of Z4 and Z2 - dihedral group:D8 + quaternion group - elementary abelian group:E8
 * type=""}
 * See linear representation theory of quaternion group. The culprit is the faithful irreducible representation of quaternion group, which has Schur index two. For comparison, see linear representation theory of groups of order 8.

{It's obvious that for a finite group, if the field generated by character values in characteristic zero is a splitting field, then it is the unique minimal splitting field. Further, it is true that if every irreducible representation has Schur index one, then the field generated by character values is a minimal splitting field. It may, however, be possible that the field generated by character values is a splitting field, despite the existence of irreducible representations with Schur index more than one. Which of the following groups is an example of this phenomenon? - dihedral group:D8 (order 8, see linear representation theory of dihedral group:D8) - quaternion group (order 8, see linear representation theory of quaternion group) - dihedral group:D16 (order 16, see linear representation theory of dihedral group:D16) - semidihedral group:SD16 (order 16, see linear representation theory of semidihedral group:SD16) + special linear group:SL(2,3) (order 24, see linear representation theory of special linear group:SL(2,3))
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 * See quaternionic representation of special linear group:SL(2,3). Another example (of order 16) is nontrivial semidirect product of Z4 and Z4, which has a non-faithful irreducible representation that factors through the faithful irreducible representation of quaternion group.

{Which of the fields does not occur as the field generated by character values for a finite group of order 16? See also linear representation theory of groups of order 16. - $$\mathbb{Q}$$ -- the field of rational numbers - $$\mathbb{Q}(i) = \mathbb{Q}[t]/(t^2 + 1)$$ - $$\mathbb{Q}(\sqrt{2}) = \mathbb{Q}[t]/(t^2 - 2)$$ - $$\mathbb{Q}(\sqrt{-2}) = \mathbb{Q}[t]/(t^2 + 2)$$ + $$\mathbb{Q}(\sqrt{3}) = \mathbb{Q}[t]/(t^2 - 3)$$
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 * Wrong. This arises from elementary abelian group:E16.
 * Wrong. This arises from M16 and other groups.
 * Wrong. This arises from dihedral group:D16.
 * Wrong. This arises from semidihedral group:SD16.
 * The field generated by character values must be contained inside $$\mathbb{Q}(\zeta)$$ where $$\zeta$$ is a primitive sixteenth root of unity, and $$\mathbb{Q}(\sqrt{3})$$ is not so contained.

Rings of realization
{Consider the following integral domains $$R$$ of characteristic zero. For which of them is it not true that any linear representation realized over the field of fractions of $$R$$ can also, based on a suitable choice of basis, be realized with all matrix entries coming from $$R$$? - $$\mathbb{Z}$$ -- the ring of integers - $$\mathbb{Z}[i] = \mathbb{Z}[t]/(t^2 + 1)$$ -- the ring of Gaussian integers + $$\mathbb{Z}[\sqrt{5}] = \mathbb{Z}[t]/(t^2 - 5)$$ - $$\mathbb{Z}[(1 + \sqrt{5})/2]$$ - $$\mathbb{Z}[t]/(t^2 + t + 1)$$
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 * The key fact is that linear representation is realizable over principal ideal domain iff it is realizable over field of fractions. All the other options give principal ideal domains. This one doesn't. In fact, it isn't even a normal domain (normal meaning integrally closed in its field of fractions), and its closure is the next option $$\mathbb{Z}[(1 + \sqrt{5})/2]$$. In particular, there is a representation of dihedral group:D10 that can be realized over $$\mathbb{Z}[(1 + \sqrt{5})/2]$$ but not over $$\mathbb{Z}[\sqrt{5}]$$.