# Endomorphism structure of general linear group over a finite field

This article gives specific information, namely, endomorphism structure, about a family of groups, namely: general linear group. This article restricts attention to the case where the underlying ring is a finite field.
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This article describes the endomorphism structure of the general linear group $GL(n,q)$ of degree $n$ over a finite field $\mathbb{F}_q$ of size $q$, where $q$ is a prime power. We will denote by $p$ the characteristic of the field, i.e., the prime of which $q$ is a power, and by $r$ the value $\log_pq$.

## Particular cases

Value of $n$ Comments Endomorphism structure page
1 $GL(1,q) \cong \mathbb{F}_q^*$, which in turn is isomorphic to the cyclic group $\mathbb{Z}/(q - 1)\mathbb{Z}$. Infer from endomorphism structure of cyclic groups
2 The transpose-inverse map is in the subgroup generated by the inner automorphisms and radial automorphisms. Specifically, it is the composite of the map $A \mapsto A/(\det A)$ and conjugation by $\begin{pmatrix} 0 & -1 \\ 1 & 0 \\\end{pmatrix}$. endomorphism structure of general linear group of degree two over a finite field
3 -- endomorphism structure of general linear group of degree three over a finite field

## Endomorphism structure

### Automorphism structure

The automorphism group can be described as a semidirect product in these equivalent ways. This applies only in the case $n > 2$. In the case $n = 2$, the transpose-inverse map is already in the product of the inner automorphism group and the radial automorphism group, and need not be introduced separately. See endomorphism structure of general linear group of degree two over a finite field. The case $n= 1$ is also qualitatively different.

Format Description of base Description of acting group
(Inner automorphism group) $\rtimes$ (Radial automorphism group $\times$ Field automorphism group $\times$ Cyclic group of order two generated by transpose-inverse map) inner automorphism group (the projective general linear group) outer automorphism group
(Inner automorphism $\rtimes$ Cyclic group of order two generated by transpose-inverse map) $\rtimes$ (Radial automorphism group $\times$ Field automorphism group). projective outer linear group no specific name
(Inner automorphism $\rtimes$ Field automorphism group) $\rtimes$ (Radial automorphism group $\times$ Cyclic group of order two generated by transpose-inverse map) general semilinear group
(Inner automorphism group $\times$ Radial automorphism group) $\rtimes$ (Field automorphism group $\times$ Cyclic group of order two generated by transpose-inverse map) no specific name no specific name

The components are described below briefly and then in detail (by default, $n > 2$, see right columns for explanation of how $n = 2$ and $n = 1$ differ).

Construct Value Order Comment Case $n = 2$ Case $n = 1$
automorphism group (Inner automorphism group) $\rtimes$ (Radial automorphism group $\times$ Field automorphism group $\times$ Cyclic group of order two generated by transpose-inverse map) $\frac{2r\varphi(n(q - 1))}{\varphi(n)}q^{\binom{n}{2}} \prod_{i=0}^{n-2} (q^{n-i} - 1)$ The subgroup generated by the transpose-inverse map does not get included as a separate factor because it is a composite of an inner automorphism and a radial automorphism. Thus, the factor of 2 in the numerator must be removed. The factor of $2r$ in the numerator needs to be removed, because both the transpose-inverse and the field automorphisms get included in the radial automorphisms.
inner automorphism group projective general linear group $PGL(n,q)$ $q^{\binom{n}{2}} \prod_{i=0}^{n-2} (q^{n-i} - 1)$ See order formulas for linear groups. Applies as is. Applies as is, though the group becomes a trivial group.
radial automorphism group Kernel of the natural homomorphism $(\mathbb{Z}/n(q - 1)\mathbb{Z})^* \to (\mathbb{Z}/n\mathbb{Z})^*$ $\frac{\varphi(n(q - 1))}{\varphi(n)}$ Here, $\varphi$ denotes the Euler totient function. See description of radial automorphism group below. Applies as in, though the denominator becomes trivial, so it simplifies to $\varphi(2(q - 1))$ Applies as is, though the denominator becomes trivial, so it simplifies to $\varphi(q - 1)$
field automorphism group (Galois group) $\operatorname{Aut}(\mathbb{F}_q) \cong \mathbb{Z}/r\mathbb{Z}$, generated by Frobenius $x \mapsto x^p$ $r$ See description below. Applies as is. Applies as is, but already included in radial automorphism group, so should not count it separately.
cyclic group generated by transpose-inverse map $\mathbb{Z}/2\mathbb{Z}$ 2 Applies as is, but is in direct product of inner automorphism group and radial automorphism group, so should not count it separately. Applies as is, but already included in radial automorphism group, so should not count it separately. Exception: Case that $q = 2$, in which case the group is a trivial group.
semidirect product of inner automorphism group and field automorphism group projective semilinear group $P\Gamma L(n,q)$ $rq^{\binom{n}{2}} \prod_{i=0}^{n-2} (q^{n-i} - 1)$ Applies as is. Applies as is.
semidirect product of inner automorphism group and cyclic group generated by transpose-inverse map projective outer linear group $POL(n,q)$ $2q^{\binom{n}{2}} \prod_{i=0}^{n-2} (q^{n-i} - 1)$ Applies as is. Applies as is, except in the case that $q = 2$, in which case the group is trivial.
outer automorphism group direct product of radial automorphism group, Galois group, and cyclic group of order two generated by transpose-inverse map $\frac{2r\varphi(n(q - 1))}{\varphi(n)}$ The subgroup generated by the transpose-inverse map does not get included as a separate factor because it is a composite of an inner automorphism and a radial automorphism. Thus, the factor of 2 in the numerator must be removed. The factor of $2r$ in the numerator needs to be removed, because both the transpose-inverse and the field automorphisms get included in the radial automorphisms.

#### Inner automorphism group

The inner automorphism group is the quotient of $GL(n,q)$ by its center, which is isomorphic to $\mathbb{F}_q^*$ (explicitly, it is the scalar matrices). This quotient group is the projective general linear group $PGL(n,q)$.

#### Radial automorphism group

The radial automorphism group is a group of automorphisms of the form:

$A \mapsto A(\det A)^m$

where $m$ is considered modulo $q - 1$, and has the property that $mn + 1$ is invertible modulo $q - 1$. The latter condition is necessary for invertibility. Values of $m$ that fail the condition define endomorphisms that are not automorphisms.

The radial automorphism group is a group of automorphisms of the form:

$A \mapsto A (\det A)^m$

where $m$ is considered modulo $q - 1$, and has the property that $2m + 1$ is relatively prime to $q - 1$. The latter condition is necessary for invertibility. Values of $m$ that fail the condition define endomorphisms that are not automorphisms.

Since $m$ is defined modulo $q - 1$, $mn + 1$ is defined modulo $n(q - 1)$. The set of possible values for $mn + 1$ is the subgroup of the multiplicative group modulo $n(q - 1)$ comprising elements that are 1 modulo $n$. Further, the composition of radial automorphisms corresponds to multiplication of the corresponding $(mn + 1)$ values. Therefore, the radial automorphism group is isomorphic to the subgroup of the multiplicative group modulo $n(q - 1)$ comprising the elements that are 1 modulo $n$. We can see that this is the same as the kernel of the natural surjective homomorphism obtained by going modulo $n$:

$(\mathbb{Z}/n(q - 1)\mathbb{Z})^* \to (\mathbb{Z}/n\mathbb{Z})^*$

By Lagrange's theorem and the fundamental theorem of group homomorphisms, the order of the kernel is the quotient of the order of the group on the left by the group on the right, and hence, is given as follows, where $\varphi$ is the Euler totient function:

$\frac{\varphi(n(q - 1))}{\varphi(n)}$

Note that these calculuations are valid in the cases $n = 1$ and $n = 2$. In both cases, the denominator $\varphi(n)$in the expression for the order becomes 1.

### Field automorphism group (Galois group)

The group of field automorphisms of the field $\mathbb{F}_q$ is the same as its Galois group over its prime subfield $\mathbb{F}_p$, because any automorphism fixes the prime subfield pointwise by definition. This Galois group is a cyclic group of order $r = \log_p q$ (for instance, it is generated by the Frobenius $x \mapsto x^p$, an automorphism of order $r$). It is thus isomorphic to $\mathbb{Z}/r\mathbb{Z}$.

The Galois group acts on the direct product of the inner automorphism group and radial automorphism group by conjugation, but does not preserve inner automorphisms pointwise: a Galois automorphism $\sigma$ acting by conjugation on conjugation by a matrix $B$ gives conjugation by the matrix $\sigma(B)$. The Galois group does commute with the radial automorphism group.

This also applies in the cases $n = 1$ and $n= 2$. The case $n = 1$, however, is anomalous in the following respect: the Galois group is a subgroup of the radial automorphism group, i.e., field automorphisms are radial automorphisms. Note that for $n \ge 2$, the groups intersect trivially (and in fact, form an internal direct product inside the whole automorphism group) so $n = 1$ is very anomalous in this respect.

### Cyclic group of order two generated by the transpose-inverse map

The transpose-inverse map $A \mapsto (A^T)^{-1}$ is an automorphism of order two of the general linear group. This generates a cyclic subgroup of order two inside the automorphism group (the exception is the case $GL(1,2)$, where the transpose-inverse map is trivial).

The transpose-inverse map commutes with the radial automorphism group as well as with the field automorphism group. It acts nontrivially by conjugation on the inner automorphism group. Conjugating the inner automorphism by a matrix $A$ by the transpose-inverse map gives the inner automorphism by the matrix $(A^T)^{-1}$.

The cases $n = 1$ and $n= 2$ are anomalous:

• For $n = 1$, the transpose-inverse map is in the radial automorphism group.
• For $n = 2$, the transpose-inverse map is the composite of the radial automorphism $A \mapsto A/(\det A)$ and conjugation by $\begin{pmatrix} 0 & 1 \\ -1 & 0 \\\end{pmatrix}$. Therefore, it is in the subgroup generated by the inner automorphism group and the radial automorphism group.