## Algebraic numbers and character tables

Let $\zeta$ be a primitive $p$th 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.