# Element structure of general semilinear group of degree one over a finite field

## Contents

This article gives specific information, namely, element structure, about a family of groups, namely: general semilinear group of degree one.
View element structure of group families | View other specific information about general semilinear group of degree one

This article describes the element structure of the general semilinear group of degree one. Recall that, for a field $K$, the group is defined as:

$\Gamma L (1,K) = GL(1,K) \rtimes \operatorname{Gal}(K/k) = K^\ast \rtimes \operatorname{Gal}(K/k)$

where $GL(1,K) = K^\ast$ is the multiplicative group of $K$, $k$ is the prime subfield of $K$, and $\operatorname{Gal}(K/k)$ denotes the Galois group of $K$ over $k$.

We are interested specifically in the case where $K$ is a finite field of size $q$, in which case the group is written as $\Gamma L(1,q)$. Suppose $q$ is a prime power with underlying prime $p$, so that $q = p^r$ for a positive integer $r$. $p$ is the characteristic of $K$. In this case, $K^\ast$ is cyclic of order $q - 1$ (see multiplicative group of a finite field is cyclic) and $\operatorname{Gal}(K/k)$ is cyclic of order $r$ (generated by the Frobenius map $a \mapsto a^p$).

Thus, $\Gamma L(1,K)$ is a metacyclic group of order $r(q - 1)$ with presentation:

$\langle a,x \mid a^q = a, x^r = e, xax^{-1} = a^p \rangle$

(here $e$ denotes the identity element).

Note that if $r = 1$, the group is the same as $GL(1,q) = \mathbb{F}_q^\ast$ and is cyclic of order $q - 1$.

## Summary

Item Value
order $r(q - 1)$
number of conjugacy classes $-1 + \left(\sum_{d \mid r} \frac{1}{d} \sum_{m \mid d} \mu(m)p^{d/m}\right) + \left( \sum_{d \mid r, d < r} \varphi(r/d)(p^d - 1) \right)$ (see explicit polynomials for fixed values of $r$ below)

### Number of conjugacy classes formulas

For every fixed value of $r$, the number of conjugacy classes simply becomes a polynomial in $p$. The values of these polynomials for small $r$ are listed below:

$r$ Polynomial in $p$ giving number of conjugacy classes
1 $p - 1$
2 $(p- 1)(p + 4)/2 = (p^2 + 3p - 4)/2$
3 $(p - 1)(p^2 + p + 9)/3 = (p^3 + 8p - 9)/3$

## Particular cases

$q$ (field size) $p$ (field characteristic) $r = \log_pq$ general semilinear group $\Gamma L(1,q)$ order of the group (= $r(q - 1)$) number of conjugacy classes (polynomial in $p$ dependent on $r$: $p - 1$ for $r = 1$, $(p^2 + 3p - 4)/2$ for $r = 2$, $(p^3 + 8p - 9)/3$ for $r = 3$) element structure page
2 2 1 trivial group 1 1 --
3 3 1 cyclic group:Z2 2 2 element structure of cyclic group:Z2
4 2 2 symmetric group:S3 6 3 element structure of symmetric group:S3
5 5 1 cyclic group:Z4 4 4 element structure of cyclic group:Z4
7 7 1 cyclic group:Z6 6 6 element structure of cyclic group:Z6
8 2 3 general semilinear group:GammaL(1,8) 21 5 element structure of general semilinear group:GammaL(1,8)
9 3 2 semidihedral group:SD16 16 7 element structure of semidihedral group:SD16

## Conjugacy class structure

### For generic r

This table is complicated! We consider conjugacy classes in the multiplicative group and conjugacy classes outside the multiplicative group. Note that $q = p^r$.

Nature of conjugacy class Number of rows of this type Size of conjugacy class for each row Number of such conjugacy classes for each row Total number of elements for each row Total number of conjugacy classes Total number of elements
in the multiplicative group and in the prime subfield 1 1 $p - 1$ $p - 1$ $p - 1$ $p - 1$
(one such row for every positive divisor $d$ of $r$ with $d > 1$) generates precisely the subfield of size $p^d$ for $d$ dividing $r$, $d > 1$ $\sigma_0(r) - 1$ where $\sigma_0$ is the divisor count function $d$ $\frac{1}{d} \sum_{m \mid d} \mu(m)p^{d/m}$ $\sum_{m \mid d} \mu(m)p^{d/m}$ $\sum_{d \mid r, d > 1} \frac{1}{d} \sum_{m \mid d} \mu(m)p^{d/m}$ $p^r - p = q - p$
(one such row for every positive divisor $d$ of $r$ with $d < r$) outside the multiplicative group, and induces an automorphism of raising to the power of $p^{ud}$ where $u$ is relatively prime to $r$ $\sigma_0(r) - 1$ $\frac{p^r - 1}{p^d - 1}$ $\varphi(r/d)(p^d - 1)$ $\varphi(r/d)(p^r - 1) = q - 1$ $\sum_{d \mid r, d < r} \varphi(r/d)(p^d - 1)$ $(p^r - 1)(r - 1) = (q - 1)(r - 1)$
Total -- -- -- -- $-1 + \left(\sum_{d \mid r} \frac{1}{d} \sum_{m \mid d} \mu(m)p^{d/m}\right) + \left( \sum_{d \mid r, d < r} \varphi(r/d)(p^d - 1) \right)$ (equal total number of conjugacy classes) $r(q - 1)$ (equals order of group)

### For r = 2

We take $q = p^2$:

Nature of conjugacy class Size of conjugacy class Number of such conjugacy classes Total number of elements
in the multiplicative group and in the prime subfield 1 $p - 1$ $p - 1$
outside the prime subfield 2 $p(p - 1)/2$ $p(p - 1)$
outside the multiplicative group $\frac{p^2 - 1}{p - 1} = p + 1$ $p - 1$ $p^2 - 1$
Total -- $(p - 1)(p + 4)/2 = (p^2 + 3p - 4)/2$ (equals number of conjugacy classes in the group) $2(p^2 - 1)$ (equals order of the whole group)

### For r = 3

We take $q = p^3$:

Nature of conjugacy class Size of conjugacy class Number of such conjugacy classes Total number of elements
in the multiplicative group and in the prime subfield 1 $p - 1$ $p - 1$
outside the prime subfield 3 $(p^3 - p)/3 = p(p + 1)(p - 1)/3$ $p^3 - p = p(p + 1)(p - 1)$
outside the multiplicative group $\frac{p^3 - 1}{p - 1} = p^2 + p + 1$ $2(p - 1)$ $2(p^3 - 1)$
Total -- $(p - 1)(p^2 + p + 9)/3 = (p^3 + 8p - 9)/3$ (equals number of conjugacy classes in the group) $3(p^3 - 1)$ (equals order of the whole group)