Element structure of general linear group of degree two over a field

This article gives specific information, namely, element structure, about a family of groups, namely: general linear group of degree two.
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Let $K$ be a field. Consider the general linear group of degree two $GL(2,K)$. The goal of this article is to describe the element structure of $GL(2,K)$.

Related descriptions

• Element structure of general linear group of degree two over a finite field
• Element structure of general linear group of degree two over a finite discrete valuation ring
• Element structure of special linear group of degree two over a field
• Element structure of special linear group of degree two over a finite field

Conjugacy class structure

Broad description

The very broad description of conjugacy class types given here is valid for all fields. However, in order to be more specific, we need to split into cases, which we do later in the article.

Nature of conjugacy class	Eigenvalues	Characteristic polynomial	Minimal polynomial	What set can each conjugacy class be identified with? (rough measure of size of conjugacy class)	What can the set of conjugacy classes be identified with (rough measure of number of conjugacy classes)	What can the union of conjugacy classes be identified with?	Semisimple?	Diagonalizable over $K$?
Diagonalizable over $K$ with equal diagonal entries, hence a scalar	$\{a,a\}$ where $a\in K^{\ast }$	$(x-a)^{2}$ where $a\in K^{\ast }$	$x-a$ where $a\in K^{\ast }$	one-point set	$K^{\ast }$	$K^{\ast }$	Yes	Yes
Diagonalizable over $K$ with distinct diagonal entries	$\lambda ,\mu$ (interchangeable) distinct elements of $K^{\ast }$	$x^{2}-(\lambda +\mu )x+\lambda \mu$	Same as characteristic polynomial	set of decompositions of a fixed two-dimensional vector space over $K$ as a direct sum of one-dimensional subspaces	the set ${\binom {K^{\ast }}{2}}$	?	Yes	Yes
Diagonalizable over a quadratic extension of $K$ but not over $K$ itself. Must necessarily have no repeated eigenvalues.	Pair of conjugate elements of some separable quadratic extension of $K$	$x^{2}-ax+b$, irreducible	Same as characteristic polynomial	?	?	?	Yes	No
Not diagonal, has Jordan block of size two with eigenvalue in $K$	$a$ (multiplicity two) where $a\in K^{\ast }$	$(x-a)^{2}$ where $a\in K^{\ast }$	Same as characteristic polynomial	?	?	?	No	No
Not diagonal, has Jordan block of size two with eigenvalue not in $K$	$a$ (multiplicity two) where $a$ is in a purely inseparable quadratic extension of $K$, so $a^{2}\in K$. This case arises only when $K$ has characteristic two and is not a perfect field	$x^{2}-t$, $t$ a non-square in $K^{\ast }$, $K$ has characteristic two	Same as characteristic polynomial	?	?	?	No	No

Identification between conjugacy classes and monic quadratic polynomials

The "characteristic polynomial" is a mapping:

Conjugacy classes in $GL(2,K)$ $\to$ Monic quadratic polynomials over $K$ with nonzero constant term

The characteristic polynomial always has nonzero constant term because the constant term is the determinant, which must be invertible.

This mapping is surjective. Further, it is almost injective, with the following exception: for those monic quadratics that are of the form $(x-a)^{2}$ with $a\in K^{\ast }$, there are two conjugacy classes mapping to such a polynomial: the diagonalizable conjugacy class, and the non-diagonalizable conjugacy class, which corresponds to a Jordan block of size two.

Another way of putting this is that we have a bijection:

Conjugacy classes of non-central elements in $GL(2,K)$ $\leftrightarrow$ Monic quadratic polynomials over $K$ with nonzero constant term

We know that the set of monic quadratics over $K$ with nonzero constant term can be identified with $K\times K^{\ast }$, with $K$ denoting the possibilities for the linear coefficient (negative of the trace) and $K^{\ast }$ the possibilities for the constant term (the determinant). Thus:

Conjugacy classes of non-central elements in $GL(2,K)$ $\leftrightarrow$ $K\times K^{\ast }$

In total we can identify:

Conjugacy classes in $GL(2,K)$ $\leftrightarrow$ $(K\times K^{\ast })\sqcup K^{\ast }$

In particular, when $K$ is a finite field of size $q$, this allows us to compute the total number of conjugacy classes to be $q(q-1)+(q-1)=q^{2}-1$.

In fact, with a little work, we can make the mapping nicer, and get:

Conjugacy classes in $GL(2,K)$ $\leftrightarrow$ $\mathbb {P} ^{1}(K)\times K^{\ast }$