Algebraic numbers and character tables

Let \zeta be a primitive pth root of unity in \mathbb{C}. Let a be a primitive root modulo p, so the field automorphism \sigma \in \mathrm{Gal}(\mathbb{Q}(\zeta):\mathbb{Q}) defined by \zeta^\sigma = \zeta^a generates the Galois group. Fix t dividing p-1, let b = a^t, and let \alpha = \zeta + \zeta^b + \cdots + \alpha^{b^{s-1}} where st = p-1. Thus \mathbb{Q}(\alpha) is the fixed field of \tau = \sigma^{t}, the unique element of the Galois group of order s. We have [\mathbb{Q}(\zeta) : \mathbb{Q}(\alpha)] = t and [\mathbb{Q}(\alpha) : \mathbb{Q}] = s.

A few days ago, while reading Burnside’s proof (in Section 252 of his wonderful book, Theory of groups of finite order) that a permutation group of prime degree p is either 2-transitive or a Frobenius group contained in an affine general linear group \mathrm{AGL}_2(\mathbb{F}_p), I needed the identity

\alpha \alpha^{-\sigma^j} + \alpha^{\sigma} \alpha^{-\sigma^{j+1}} + \cdots + \alpha^{\sigma^{t-1}}\alpha^{-\sigma^{j+(t-1)}} = -s

where 0 < j < t. One rather sneaky proof is to look at the intended conclusion of Burnside's argument, namely that G is Frobenius of order ps, and work back. I've put in details of the construction of the character table below, but taking this as read, the proof is essentially one line.

Character tables of affine Frobenius groups

Let G = \langle g, h | g^p = h^s = 1, g^h = g^b \rangle. Setting N = \langle g \rangle, we have G = N \rtimes \langle h \rangle. In the action of G on the cosets of \langle h\rangle, the non-identity elements of N act as p-cycles, and all elements of G \backslash N have a unique fixed point. Therefore G is a Frobenius group. The conjugacy class of g is \{g, g^h, \ldots, g^{h^{s-1}}\} = \{g,g^b, \ldots, g^{b^{s-1}} \}, and the non-identity conjugacy classes have representatives

g, g^a, \ldots, g^{a^{t-1}}, h, h^2, \ldots, h^{s-1}

and are of sizes s, s, \ldots, s, p, p, \ldots, p respectively.

There are s linear characters of G inflated from G / \langle g \rangle \cong C_{s-1}. Let \phi : N \rightarrow \mathbb{C} be defined by \phi(g) = \zeta. Let \theta = \phi\uparrow^G. Then

\begin{aligned} \theta(g) &= \phi(g) + \phi(g^h) + \cdots + \phi(g^{h^{s-1}}) \\&= \phi(g) + \phi(g^b) + \cdots + \phi(g^{b^{s-1}}) \\ &= \zeta + \zeta^b + \cdots + \zeta^{b^{s-1}} \\ &= \alpha, \end{aligned}

and more generally, \theta(g^{a^j}) = \alpha^{\sigma^j} for all j. It follows that \theta, \theta^\sigma, \ldots, \theta^{\sigma^{s-1}} are s distinct characters of G. By Frobenius reciprocity

\begin{aligned} \langle \theta, \theta \rangle &= \langle \theta\downarrow_N, \phi \rangle \\ &= \langle \phi, \phi + \phi^{\tau} + \cdots + \phi^{\tau^{s-1}} \rangle \\ &= 1. \end{aligned}

The sum of the squares of the degrees of the characters constructed so far is 1^2 \times s + s^2 \times t = s + s^2 t = s (1+st) = sp, so we have found all irreducible characters.

Conclusion

Let 0 < j < t. From

|G| \langle \theta, \theta^{\sigma^j} \rangle = s^2 + s(\alpha \alpha^{-\sigma^j} + \alpha^{\sigma} \alpha^{-\sigma^{j+1}} + \cdots + \alpha^{\sigma^{t-1}} \alpha^{-\sigma^{j+(t-1)}})

we get

\alpha \alpha^{-\sigma^j} + \alpha^{\sigma} \alpha^{-\sigma^{j+1}} + \cdots + \alpha^{\sigma^{t-1}}\alpha^{-\sigma^{j+(t-1)}} = \frac{0-s^2}{s} = -s

for j\not= 0. If j=0, the left-hand side is |G| and we get instead p-s.

Two related well known tricks

In Wielandt’s well-known proof of Sylow’s Theorem, one needs to know that if m is coprime to p then \binom{p^a m}{p^a} \not\equiv 0 mod p. This follows immediately from Lucas’ Theorem, or one can follow Wielandt and use a short direct argument, but it is in the spirit of his proof to fix a set \Omega of size p^a m and let g \in \mathrm{Sym}(\Omega) have m disjoint p^a cycles. Then \langle g \rangle clearly has m fixed points, in its action on the set of subsets of \Omega of size p^a and all other orbits have size divisible by p. So \binom{p^a m}{p^a} \equiv m modulo p.

Another nice argument, applying linear algebra to algebraic number theory, shows that if \alpha and \beta are algebraic numbers then \alpha \beta is also algebraic. Indeed, \alpha \beta is an eigenvalue of X_\alpha \otimes X_\beta where X_\alpha and X_\beta are irreducible integer matrices having \alpha and \beta as eigenvalues. (For example, take the companion matrices of the minimal polynomials.) This trick also shows that \alpha \beta^{-1} is an eigenvalue of an integer matrix, Y say, and so 1 + \alpha\beta^{-1} is an eigenvalue of I + Y. Therefore 1 + \alpha\beta^{-1} is algebraic, and multiplying through by \beta (using the result just proved), we get that \alpha + \beta is algebraic.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: