Extended Special Linear group and square root in matrix groups
$SL_{2}(\mathbb{F}_{p}),ESL_{2}(\mathbb{F}_{p})$, $SL_{2}(\mathbb{Z})$ and
$GL_{2}(\mathbb{F}_{p})$

Skuratovskii Ruslan [0000-0002-5692-6123] 1

##  1 Abstract

First time, we introduce Extended special linear group $ESL_{2}(\mathbb{F})$,
which is generalization of the matrix group $SL_{2}(\mathbb{F})$, where
$\mathbb{F}$ is arbitrary perfect field. Extended special linear group
$ESL_{2}(\mathbb{F})$ is storage of all square matrix roots from
$ESL_{2}(\mathbb{F})$.

The analytical formula of square roots of 2-nd and 4-th power in
$SL_{2}(\mathbb{F}_{p})$, $ESL_{2}({\mathbb{F}_{p}})$, for any prime $p$, as
well as in $SL_{2}(\mathbb{Z})$, $ESL_{2}(\mathbb{Z})$ and in $SL_{2}({k}),\\\
ESL_{2}({k})$, over perfect field $k$, is found by us.

New linear group which is storage of square roots from
$SL_{2}({\mathbb{F}_{p}})$ is found and investigated by us.

The criterion of roots existing for different classes of matrix — simple and
semisimple matrixes from $SL_{2}({\mathbb{F}_{p}})$, $SL_{2}({\mathbb{Z}})$
are established.

The analytic formula of cubic square from matrix is founded.

The problems of square root from group element existing in $SL_{2}(F_{p})$,
$SL_{2}(F_{p})$ and $GL_{2}(F_{p})$ for arbitrary prime $p$ are solved in this
paper. The similar goal of root finding was reached in the GM algorithm
adjoining an $n$-th root of a generator [34] results in a discrete group for
group $SL(2,R)$, but we consider this question over finite field $F_{p}$. Well
known the Cayley-Hamilton method [37] for computing the square roots of the
matrix $M^{n}$ can give answer of square roots existing over a finite field
only after computation of $detM^{n}$ and some real Pell-Lucas numbers by using
Bine formula. Over method gives answer about existing $\sqrt{M^{n}}$ without
exponenting $M$ to $n$-th power. We only use the trace of $M$ or only
eigenvalues of $M$.

In [4] only the Anisotropic case of group $SL_{1}(\mathbb{Q})$, where
$\mathbb{Q}$ is a quaternion division algebra over $k$ was considered. The
authors of [4] considered criterion to be square only for the case $F_{p}$ is
a field of characteristics not equal 2. We solve this problem even for fields
$F_{2}$ and $\mathbb{F}_{2^{n}}$. The criterion to $g\in
SL_{2}(\mathbb{F}_{2})$ be square in $SL_{2}(\mathbb{F}_{2})$ was not found by
them what was declared in a separate sentence in [4]. Also in [4] the split
case of $SL_{2}(k)$ and its powers was considered, where under group splitting
authors mean Bruhat decomposition is the double coset decomposition of the
group $SL_{2}(k)$ with respect to the subgroup Borel $B$, consisting of upper
triangualar matrix from $SL_{2}(k)$. Moreover authors of [4] find recursive
formula of matrix root in $SL_{2}(k)$ only for case of field $char(k)\neq 2$ ,
but our formula solves this problem for a $\mathbb{F}_{p}$ with arbitrary
characteristic. Futhermore, our formula is analytical and does not require
sequential recursive calculation.

We consider a more general case [1] consisting in the whole group
$G=SL_{2}(F_{q})$. We do not provide additional condition of splitting. Also
the authors considered separetely conjugacy classes in
$SL_{2}(\mathbb{F}_{q})$ [4] such as: central classes, split regular
semisimple classes, non-semisimple classes, anisotropic regular semisimple
classes. For each case the criterion of solvability of equation is provided.
In the last two cases Bruhat decomposition is applied. Our criterion oriented
on more general class of matrix depending of the form of minimal and
characteristic polynomials.

The necessary and sufficient conditions when an element of one the following
matrix groups $GL_{2}(F_{p})$, $SL_{2}(F_{p})$ as well as
$ESL_{2}(\mathbb{F}_{p})$ can be presented as a squares of one element from
initial group are also found by us.

We investigate a condition of a matrix quadraticity depending on its Jordan
structure and spectrum.

Key words: extended special linear group, equation in matrix group, splittable
extension, formula of square roots in special linear group, set of squares in
matrix group, criterion of square root existing in $SL_{2}(\mathbb{F}_{p})$.
2000 AMS subject classifications: 20B27, 20E08, 20B22, 20B35, 20F65, 20B07.

##  2 Introduction

We research the conditions of matrix equation solvability $X^{2}=A$ in
$SL_{2}(\mathbb{F}_{p}),\\\ GL_{2}(\mathbb{F}_{p})$ and one of splitting
extension of $SL_{2}(\mathbb{F}_{p})$ that is $ESL_{2}(\mathbb{F}_{p})$ [1,
2].

Our statements can be easy reformulated for these groups over the field
$\mathbb{R}$ so it lead us to arguments of solving of discreteness problem
[34, 35] in some subgroups of $SL(2,\mathbb{R})$.

One method of computing square roots of two-by-two matrices was presented in
[38] but under unsolved by him condition which $\sqrt{A}$ exists. But we find
not only all such conditions but the method of square root computation for
$n\times n$ matrix. Moreover we consider this question in groups over finite
fields. Also we indicate in which group $\sqrt{A}$ lies.

The action of subgroup of new group $ESL_{2}(\mathbb{F}_{p})$ introduced here
also arose without description of group structure and generators in the
topology. Namely, if $G$ is a Morse-Bott foliation on the solid Klein bottle K
into 2-dimensional Klein bottles parallel to the boundary and one singular
circle $S^{1}$ then such group appears as leaf preserving diffeomorphisms for
foliations $G$ [8].

In many geometrical groups there are automorphisms preserve hyperbolic
distance (hyperbolic metric) and hyperbolic angles, furthermore they may
change orientation of space as well as keep it permanent [6].

In hyperbolic geometry there are groups in particular group projective special
linear group preserves hyperbolic length [7] and orientation as well as change
orientation. As well known $S{{L}_{2}}(R)$ possess such a property too, but
action of $S{{L}_{2}}(R)$ is non-faithful because of $PSL_{2}(R)$ is a
homomorphic image of $SL_{2}(\mathbb{R})$.

Define the algebraic properties and structures in the following item.

One of interesting algorithmic problem of combinatorial group theory was
solved by Roman’kov [23]. It was problem of determining for any element $g\in
G$ is $g$ a commutator for free nilpotent group $N_{r}$ of arbitrary rank $r$
with class of nilpotency 2 [23]. The analogous problem can be formulated for
$SL_{n}(F_{q})$, $GL_{n}(F_{q})$ and $ESL_{n}(F_{q})$ over a set of squares.

The problem of the solvability of an equation over a group is well known. We
consider the same problem with additional constrains on the solvability of an
equation of the form $X^{2}=A$ in a group.

Question of root existing in different forms appears in the Purtzitsky-
Rosenberger trace minimizing algorithm [36, 34] it was considered roots and
rational powers of one or both generators of in non-elementary two generator
discrete subgroups of $PSL_{2}(R)$ found by the GM algorithm. But we solve
existing root problem for arbitrary element of $PSL_{2}(F_{p})$.

Also such criterion of root existing for $SL_{2}(F_{p})$, $PSL_{2}(F_{p})$ and
$GL_{2}(F_{p})$ are established. This criterion is a stricter version of the
formulated question for group extensions how large must an overgroup of a
given group be in order to contain a square root of any element of the initial
group $G$, which was considered in the paper of Anton A. Klyachko and D. V.
Baranov [41].

The main result of this paper about criterion of quadraticity can be extended
to larger dimension matrices having a Jordan structure constructed of blocks
of dimension 2 or 1. Also, our result for a semisimple matrix of dimension 2
can be generalized to a semisimple matrix of higher dimension.

Since $PSL_{2}(\mathbb{F}_{p})$, $p>3$ is simple and is quotient-subgroup of
$SL_{2}(\mathbb{F}_{p})$ then study of verbal subgroup generated by squares in
$SL_{2}(\mathbb{F}_{p})$ gives possibility to study Waring problem for
matrices over finite fields [5].

Thanks to Anton A. Klyachko, who inspired the author to continue this study.

##  3 Preliminaries

In hyperbolic geometry there are groups in particular projective special
linear group preserves hyperbolic length [7] and orientation as well as
changes orientation. As well known $S{{L}_{2}}(\mathbb{R})$ that possess such
a property too besides besides property of changing orientation, because of
action of $S{{L}_{2}}(\mathbb{R})$ is non-faithful because of $PSL_{2}(R)$ is
a homomorphic image of $SL_{2}(R)$. A proposed by us group
$ES{{L}_{2}}(\mathbb{R})$ also preserves hyperbolic length [7] and orientation
as well as changes space orientation.

An action of a subgroup of $ESL_{2}(R)$ appears in leaf preserving
diffeomorphism group which is called foliated leaf preserving in Morse-Bott
foliation on the solid torus [9] of simplest Morse-Bott foliations. But this
action was defined geometrically by symmetry with respect to meridian and
parallel of torus and infinite shift on torus.

Also Morse-Bott foliation on solid torus [9] $T=S^{-1}\times D^{2}$ into 2-
tori parallel to the boundary and one singular circle consists of elements
presented by matrices with determinant 1 and -1 by author who characterize it
as a subgroup of the whole $GL_{2}(R)$ group. But now we characterize it more
precisely as a subgroup of smaller group $ESL_{2}(R)<GL_{2}(R)$. The
diffeomorphisms group of this manifold posses the subgroup $\mathcal{G}$
described in the geometrical terms, where the actions of shifts, symmetries
relative to a parallel and symmetries relative to a meridian appear, where
shift is generated by $\left(\begin{array}[]{rr}1&0\\\ 1&1\end{array}\right)$
which is called by reflection. Symmetries relative to a parallel and a
meridian are defined by matrices $\left(\begin{array}[]{rr}1&0\\\
0&-1\end{array}\right),\ \ \left(\begin{array}[]{rr}-1&0\\\
0&1\end{array}\right)$ correspondingly. This actions generate group
$\mathcal{G}$ which is a proper subgroup of $ESL_{2}(Z)$.

Define the algebraic properties and structures of $ESL_{2}(\mathbb{F}_{p})$ in
the next item.

Definition 1. The set of matrices

$\left\\{{{M}_{i}}:Det({{M}_{i}})=\pm 1,{M}_{i}\in
GL_{2}(\mathbb{F}_{p})\right\\}$

forms extended special linear group in $GL_{2}(\mathbb{F}_{p})$ and is denoted
by $\texttt{ESL}_{2}(\mathbb{F}_{p})$.

As it is studied by us, $ESL_{2}(\mathbb{F}_{p})\cong
SL_{2}(\mathbb{F}_{p})\rtimes\,{{\mathcal{C}}_{2}}$, where
$\,{{\mathcal{C}}_{2}}$ is generated by reflection
$\left(\begin{array}[]{cc}-1&0\\\ 0&1\\\ \end{array}\right)$. The involution
from $\,{{\mathcal{C}}_{2}}\simeq\left\langle\left(\begin{array}[]{cc}-1&0\\\
0&1\\\ \end{array}\right)\right\rangle$ induces the sign of automorphism in
$Aut\left(SL_{2}(F_{p})\right)$. Matrices with determinant -1 correspond to
the elements changing Euclidean space orientation. As it was found in our
study of the roots of matrices from group matrices from defined above group
$ESL_{2}(F_{p})$ arise.  Justification of
$SL_{2}(\mathbb{F}_{p}),SL_{2}(\mathbb{Z}),SL_{2}(\mathbb{Q}),SL_{2}(\mathbb{R})$
extensions existence based on the description
$Aut\left(SL_{2}((\mathbb{F}_{p})\right),Aut\left(SL_{2}(\mathbb{Z})\right),\ldots,\\\
Aut\left(SL_{2}(\mathbb{R})\right)$ and its subgroups of order 2. We can
spread the definition of $ESL_{2}(\mathbb{F}_{p})$ on case of matrices over
the arbitrary field $\mathbb{F}$ as well as over the ring $\mathbb{Z}$. In
similar way we can extend $SL_{n}(\mathbb{F}_{p})$ to
$ESL_{n}(\mathbb{F}_{p})$.

$SL_{2}(\mathbb{F}_{p})$ is subgroup of index 2 in $ESL_{2}(\mathbb{F}_{p})$
so its normality is established.

The existence of a non-trivial homomorphism $\varphi:\,\,{{\mathbb{Z}}_{2}}\to
Aut\left(S{{L}_{2}}(\mathbb{Z})\right)$, as well as
$\phi:\,\,{{\mathbb{Z}}_{2}}\to
Aut\left(S{{L}_{2}}({{\mathbb{F}}_{p}})\right)$ and
$\phi:\,\,{{\mathbb{Z}}_{2}}\to
Aut\left(S{{L}_{2}}({{\mathbb{F}}_{p}})\right)$, can be proved by indicating
an element of order 2 in the automorphisms of base group that is the kernel of
the semidirect product we want to construct. There is countergradient
automorphism in $S{{L}_{2}}\left(\mathbb{Z}\right)$ that is
$\varphi:\,M\to{{\left({{M}^{T}}\right)}^{-1}}$ or alternating automorphism of
order 2 acting by conjugating $\varphi:\,M\to D^{-1}MD$, where
$D=\left(\begin{array}[]{rr}1&0\\\ 0&-1\\\ \end{array}\right)$ and is called
by diagonal automorphism [46].

$SL_{2}(\mathbb{Z})$ is a normal subgroup of $ESL_{2}(\mathbb{Z})$, as being
the kernel of the determinant, which is a group homomorphism whose image is
the multiplicative group $\\{-1,+1\\}$.

######  Remark 1.

It is obvious that orthogonal group $O_{2}(k)<ESL_{2}(k)$, where $k$ is a
field but $O_{2}(k)\ntriangleleft ESL_{2}(k)$ [32, 33].

In fact, the action by conjugation of elements from the $ESL_{2}(k)$ does not
preserve angles and does not fixe non-degenerate quadratic and Hermitian
forms.

We briefly introduce the minimal set of generators and new relations in
$ESL_{2}\left(\mathbb{Z}\right)$ [44] i.e. this group over integer ring. Each
relation of $SL_{2}\left(\,Z\right)$ holds. We denote
$\left(\begin{array}[]{cc}1&1\\\ 0&1\\\ \end{array}\right)$ by $s$ and
$\left(\begin{array}[]{cc}0&-1\\\ 1&0\\\ \end{array}\right)$ as $t$ they
generate $SL_{2}\left(\mathbb{Z}\right)$, new generator
$\left(\begin{array}[]{cc}-1&0\\\ 0&1\\\ \end{array}\right)$ is denoted by
$i$. Then new relation is ${{i}^{-1}}si=s$. Note that elements $i$ and $t$ are
orthogonal because of $ti=0$. Some interesting relation in this terms of the
kernel subgroup $SL(2,\mathbb{Z})$ are ${{t}^{2}}=-E$,
${{t}^{-2}}s{{t}^{2}}=s$.

Existence justification of such extension of $SL_{2}(\mathbb{Z})$ by $C_{2}$
is based on $Aut(SL_{2}(\mathbb{Z}))$ [47, 46, 48] structure which is
splitting extension $SL_{2}(\mathbb{Z})$ by $\mathbb{Z}$. As well known the
group of outer automorphisms of $S{{L}_{n}}(\mathbb{Z})$ is semidirect
products of the form $S{{L}_{n}}(\mathbb{Z}){{\rtimes}_{\varphi}}\mathbb{Z}$
and its isomorphism type depends only on $[\varphi]\in
Out(S{{L}_{n}}(\mathbb{Z}))$. Since $Aut(SL_{2}(\mathbb{Z}))$ contains an
element of order 2 that is $t^{2}$ therefore homomorphism from top group that
is cyclic group $C_{2}=<i>$ of order 2 in $Aut(SL_{2}(\mathbb{Z}))$ exists.

The action by right multiplication on $\left(\begin{array}[]{cc}-1&0\\\ 0&1\\\
\end{array}\right)$ of matrix from $S{{L}_{n}}(\mathbb{Z})$ inducing
automorphism changing sing (inverting sing) of first column of matrix $A$ from
$S{{L}_{n}}(\mathbb{Z})$. This automorphism invert sign of $det(A)$.

The concept of geometrical group $\mathcal{G}$ appears as subgroup in the
group ${{D}^{lp}}\left(F\right)$ of diffeomorphisms group of $T$ and
$\left[0;\text{ }1\right]$ on ${{C}^{\infty}}(T,\left[0;\text{ }1\right])$ and
now is characterized by us in more structural way and exact in algebraic
sense, because of the authors [9] consider the following subgroup
$\mathcal{G}$ as subgroup of very wide group $GL(2,\mathbb{Z})$ consisting of
matrices for which the vector $(0,1)$ is eigen with eigenvalue $±1$. Which was
defined as:

$\mathcal{G}=\,\left\\{\,\left(\begin{array}[]{rr}\varepsilon&0\\\
m&\delta\end{array}\right)\left|m\in\mathbb{Z},\,\,\,\varepsilon,\delta\in\left\\{\pm
1\right\\}\right.\right\\}.$

But $\mathcal{G}$ is a proper subgroup of $ESL_{2}(\mathbb{Z)}$ that is more
special whole $G{{L}_{2}}\left(\mathbb{Z}\right)$ Moreover
$ES{{L}_{2}}\left(\mathbb{Z}\right)$ has as kernel a proper subgroup of
$S{{L}_{2}}\left(\mathbb{Z}\right)$, because of
$det\left(\mathcal{G}\right)=\pm 1$. Furthermore the concept of new group
$ES{{L}_{2}}\left(\mathbb{Z}\right)$ admits us to obtain a structural
characterization and set of generators with relations for $\mathcal{G}$.
Furthermore $ES{{L}_{2}}\left(\mathbb{Z}\right)$ give us structural
characterization and set of generators with relations for $\mathcal{G}$. We
take in consideration involutions which generate symmetry of torus symmetries
with respect to the parallel are represented by matrix
$t=\left(\begin{array}[]{rr}1&0\\\ 0&-1\end{array}\right)$ and generator of
this subgroup of $ES{{L}_{2}}\left(\mathbb{Z}\right)$ top group
$i=\left(\begin{array}[]{rr}-1&0\\\ 0&1\end{array}\right)$ which is also is
symmetries with respect to the meridian of the torus in paper [9]. Now using
concept of new group $ESL_{2}(\mathbb{Z)}$ allows us to give exact and
structural characterization of group $\mathcal{G}$ which contains in
${{D}^{lp}}\left(F\right)$. For this goal we consider subgroup of
$ESL_{2}(\mathbb{Z)}$ with kernel
$K\simeq\left\langle\left(\begin{array}[]{rr}1&0\\\ 1&1\\\
\end{array}\right)\right\rangle$. Since
$K\simeq\left\langle\left(\begin{array}[]{rr}\,1&0\\\
\,1&1\end{array}\right)\right\rangle\simeq\mathbb{Z}$ then
$AutK\simeq{{\mathbb{Z}}_{2}}$ and therefore homomorphism from subgroup
$~{}\left\langle i\right\rangle$ as well as from subgroup $\left\langle
t\right\rangle$ to $AutK$ exist. One easy can check that

$i\left(\begin{array}[]{rr}1&0\\\
1&1\end{array}\right)i^{-1}=\left(\begin{array}[]{rr}1&0\\\
-1&1\end{array}\right)=\left(\begin{array}[]{rr}1&0\\\
1&1\end{array}\right)^{-1}$ and rest of conjugations remain $K$ invariant.
Thus, we find a structure of $\mathcal{G}$ which, up to a way to define a
semidirect product, is $\mathcal{G}\,\simeq K\ltimes\left\langle
t,i\right\rangle$. An important fact that $K\ltimes\left\langle
t,i\right\rangle$ is a subgroup in $ES{{L}_{2}}(\mathbb{Z})$. Top subgroup of
$\mathcal{G}$ has 2 generators but kernel subgroup $K$ is one generated,
unlike the kernel in $ES{{L}_{2}}(\mathbb{Z})$ having 2 generators. If we
denote $\left(\begin{array}[]{rr}\,1&0\\\ \,1&1\end{array}\right)$ then the
relations are following
$isi={{s}^{-1}},\,\,tst={{s}^{-1}},\,\,\,{{t}^{2}}={{s}^{2}}=e$.

We denote by e.v. — eigenvalues. Let $\mu_{A}$ be minimal polynomial of $A$.

A polynomial $P(X)$ over a given field $K$ is separable if its roots are
distinct in an algebraic closure of $K$, that is, the number of distinct roots
is equal to the degree of the polynomial. _Simple matrix_ is a matrix such
that characterstic polynomial is separable.

Recall that matrix $A$ is called semisimple if $\mu_{A}$ is a product of
distinct monic irreducible and separable polynomials; if moreover all these
irreducible polynomials have degree 1, then $A$ is called split semisimple or
diagonalizable.

We denote iff — necessary and sufficient condition, e,v. — eigenvalue.

###  3.1 Some possible applications in topology

Geometrical transformations corresponding to matrices that form the subgroup
of the introduced here $S{{L}_{2}}\left(R\right)\rtimes{{C}_{2}}$ group, occur
in leaf preserving diffeomorphism group in Morse-Bott foliation on the solid
Klein bottle [8], with the complementary circle, which are homeomorphisms.

Also an action defined in a geometrical form of subgroup over $\mathbb{R}$ of
our group $S{{L}_{2}}\left(\mathbb{R}\right)\rtimes{{C}_{2}}$ occurs in
continuous transformations of the Klein bottle (because of matrix $A$ with
$det(A)=-1$ change space orientation) with the complementary circle, which are
homeomorphisms. Action of the subgroup such a group appears in diffeomorphisms
groups of simplest Morse-Bott foliations which were presented in geometrical
and topological forms as shift and symmetries relatively meridian and parallel
by S. I. Maksymenko.

Group of continuous functions implementing rotation $D\left(y\right)$, which
is a linear isomorphism preserving concentric circles, simultaneously with a
shift as standing a second coordinate of tuple, is presented in [8]. Its
elements have a form of pair $(w{{e}^{2\pi i{{\lambda}_{h}}(s)}},s)$, where
${{\lambda}_{h}}(s)$ ensures sign inversion provided unit shift (on one). We
assume that this subgroup has structure of semidirect product and denote it by
$H$. Thus, from this group $H$ of diffeomorphisms with additional functions
${{\lambda}_{h}}\left(s+1\right)=-{{\lambda}_{h}}\left(s\right)$ making
changing of sign provided by action of shift on one described in [8]
homomorphism in subgroup of $ES{{L}_{2}}(\mathbb{R})$ can be constructed.
Homomorphic image can be realized by matrices of rotation with sign inversion
inducing by the top group of semidirect product $ES{{L}_{2}}(\mathbb{R})$ that
could be also generated by Frobenius normal form
$\left(\begin{array}[]{rr}0&-1\\\ 1&0\\\ \end{array}\right)$. Thus this
subgroup of $ES{{L}_{2}}(\mathbb{R})$ can be embedded in $H$ and this subgroup
is realized by matrices of rotation on with sign inversion due to the top
group of semidirect product $ES{{L}_{2}}(\mathbb{R})$. In $H$ one of subgroup
of our new group $ES{{L}_{2}}(\mathbb{R})$ is embedded. This subgroup is
$\left(SO(2)\right)\ltimes\left\langle\left(\begin{array}[]{rr}-1&0\\\ 0&1\\\
\end{array}\right)\right\rangle\simeq:O(2)$.

We denote this subgroup by $\left\langle\rho\right\rangle\ltimes\left\langle
i\right\rangle$. Note that group of the diffeomorphisms $h$ coinciding with
some vector bundle morphism also function
${{\lambda}_{h}}:\,\,\mathbb{R}\to\mathbb{R}$ is described in [8] in item 3),
there are subgroup $h^{\prime}(w,s)=\left({{e}^{2\pi
i{{\lambda}_{h}}(s)}},s\right),\,\,\,\,\,{{\lambda}_{h}}\left(s+1\right)=-{{\lambda}_{h}}\left(s\right)$
presented in form of functions. Additionally we can describe its structure as
semidirect product. We establish a homomorphism from this group to
$\left\langle\rho\right\rangle\ltimes\left\langle i\right\rangle$. Furthermore
the top group of $ES{{L}_{2}}(R)$ is the same matrix
$i=\left(\begin{array}[]{rr}-1&0\\\ 0&1\\\ \end{array}\right)$ coinciding with
a matrix $\Lambda$ presenting the meridian of torus respect to the parallel
[9].

Matrices of $G{{L}_{2}}\text{ }\left(\mathbb{Z}\right)$ are present in
subgroup $L_{p,q}$ of diffeomorphism denoted by $D\left({{L}_{p,q}}\right)$ in
[9], whence its group closure Is $ES{{L}_{2}}(\mathbb{Z})$ but algebraic
structure of set was not investigated before it was recalled as subset of
$G{{L}_{2}}\text{ }\left(\mathbb{Z}\right)$ with determinant -1 also there is
transformation $T$ in that item with $\det(T)=1$. Thus there is such subgroup
$ES{{L}_{2}}(\mathbb{Z})$ in nature but it was not defined and investigated as
algebraic group before.

##  4 Square root of an element in $GL_{2}(\mathbb{F}_{p})$,
$SL_{2}(\mathbb{F}_{p})$ and $ESL_{2}(\mathbb{F}_{p})$

Let $SL_{2}(\mathbb{F}_{p})$ denotes the special linear group of degree 2 over
a finite field of order $p$. And a degree always means an irreducible
character degree in this paper.

We recall the well known relation between eigenvalues of $A$ and $f(A)$.

######  Lemma 2.

If $\beta$ is an eigenvalue for $B$ then ${{\beta}^{2}}$ is an eigenvalue for
${{B}^{2}}$.

Consider the criterion of squareness of elements in $SL_{2}(\mathbb{F}_{p})$
as well as in $GL_{2}(\mathbb{F}_{p})$ which can be presented by diagonal
matrix. As well known [33] a matrix can be presented in the diagonal form iff
the algebraic multiplicity of its eigenvalues are the same as the geometric
multiplicity.

######  Theorem 3.

Let $A$ be simple matrix and $A\in SL_{2}(\mathbb{F})$ [33], then for $A$
there is a solution $B\in SL_{2}(\mathbb{F})$ of the matrix equation

$X^{2}=A$ (1)

if and only if

${trA+2}$ (2)

is quadratic element in $\mathbb{F}$ or 0, where $\mathbb{F}$ is a field.

If $X\in ESL_{2}(\mathbb{F})$ then the matrix equation (1) has a solution in
$ESL_{2}(\mathbb{F})$ iff

${trA\pm 2}$ (3)

is quadratic element in $\mathbb{F}$ or 0.

This solution $X\in ESL_{2}(\mathbb{F})\setminus SL_{2}(\mathbb{F})$ iff
$\left({trA-2}{}\right)$ is quadratic element or 0 in $\mathbb{F}$ but
$\left({trA+2}{}\right)$ is not. Conversely $X\in SL_{2}(\mathbb{F}_{p})$ iff
$\left({trA+2}{}\right)$ is quadratic element. Solutions $X_{i}\in
ESL_{2}(\mathbb{F})$ and $SL_{2}(\mathbb{F})$ iff $\left({trA+2}{}\right)$ and
$\left({trA-2}{}\right)$ are quadratic elements.

In the case $A\in GL_{2}({{\mathbb{F}}})$ this condition (2) takes form:

${trA\pm 2\sqrt{detA}}$ (4)

is quadratic element in $\mathbb{F}$ or 0.

###### Proof.

Throughout the proof a quadraticity of element $x$ or $x=0$ in a field
$\mathbb{F}$ be denoted by $\left(\frac{x}{p}\right)\in\\{0,1\\}$. For
concretization, we provide a proof over $\mathbb{F}p$. But out prove can be
spread without changes on _arbitrary field_ $\mathbb{F}$ instead
$\mathbb{F}_{p}$.

We assume that matrices $A$ and $B$ have eigenvalues $\lambda_{1},\lambda_{2}$
and $\mu_{1},\mu_{2}$ respectively. Let a characteristic polynomial
${{\chi}_{B}}(x)$ of $B$ be the following:
${{\chi}_{B}}(x)=(x-\mu_{1})(x-\mu_{2})$. We denote $tr(A)$ by $a$.

Then according to Lemma 2 $a$ is the sum of the roots $\mu_{1}^{2}$,
$\mu_{2}^{2}$ of a polynomial
${{\chi}_{A}}(x)=(x-\mu_{1}^{2})(x-\mu_{2}^{2})$. Therefore
$tr(A)=a=\mu_{1}^{2}+\mu_{2}^{2}=(\mu_{1}+\mu_{2})^{2}-2\mu_{1}\mu_{2}=(tr(B))^{2}-2$.
Since $det(A),A\in PSL_{n}(F_{p})$ is 1, then eigenvalues of $A$ satisfy the
following equality: $\mu_{1}^{2}\mu_{2}^{2}=1$ that implies
$\mu_{1}\mu_{2}=\pm 1$. Therefore $a+2\mu_{1}\mu_{2}=a\pm
2=(\mu_{1}+\mu_{2})^{2}$. As is known $tr(B)=\mu_{1}+\mu_{2}\in F_{p}$ and
$det(B)=\mu_{1}\mu_{2}\in F_{p}$.

We show the existence of ${{\chi}_{B}}(x):={{x}^{2}}-cx+1$ having roots
${{\mu}_{1}},\,\,{{\mu}_{2}}$ which will be the e.v. of $B$. Let
${{\chi}_{{{B}^{2}}}}(x)={{\mu}^{2}}-a\mu+1$. Then
$\mu_{1}^{2},\,\,\,\mu_{2}^{2}$ are e.v. for $A$ and according to Viet’s
theorem, $\mu_{1}^{2}+\mu_{2}^{2}=a.$

Let us prove the sufficiency of the condition $(\frac{trA+2}{p})=1$. We take
into account Lemma 2 whence $\mu_{1}^{2},\,\,\,\mu_{2}^{2}$ are e.v. of $A$,
provided ${{\mu}_{1}},\,\,\,{{\mu}_{2}}$ are e.v. of $B$. Since
$\left(\frac{tr(A)+2}{p}\right)\in\\{0,1\\}$, then there exists such
$c\in{{F}_{p}}$ for which ${{c}^{2}}=tr(A)+2$. For such $c$ we get, that the
roots ${{\mu}_{1}},{{\mu}_{2}}$ of the polynomial ${{x}^{2}}-cx+1$ satisfy the
following property: $\mu_{1}^{2}$ and $\mu_{2}^{2}$ are equal to e.v. of $A$.
Hence any matrix $B$ with $tr(B)=c$ and $\det(B)=1$ satisfies the property:
${{B}^{2}}$ has the same e.v. as $A$.

Also $c=\mu_{1}+\mu_{2}=tr(B)$. Thus matrix $B$ is in $SL_{2}(F_{p})$.

The condition $(\frac{trA+2}{p})=1$ is necessary and sufficient for existing a
solution of ${{X}^{2}}=A$ in $S{{L}_{2}}({{\text{F}}_{p}})$. Let us construct
the solution of equation $X^{2}=A$ in $SL_{2}(F_{p})$. In general case we
obtain the solution

${{B}^{2}}=A^{\prime},$

where $A^{\prime}\sim A$ with eigenvalues
${{\lambda}_{1}}=\mu_{1}^{2},\,\,\,{{\lambda}_{2}}=\mu_{2}^{2}.$ Since
$c\in{{\text{F}}_{p}}$ then we can construct in normal Frobenius form a matrix

$\left(\begin{array}[]{rr}0&-1\\\ 1&c\\\ \end{array}\right)={{B}}$

therefore this matrix is over base field ${{\text{F}}_{p}}$ or Q or arbitrary
field $\mathbb{F}$. Since
${{\lambda}_{1}}+{{\lambda}_{2}}={{(\mu_{1}^{2}+\mu_{2}^{2})}^{2}}-2=trA$ and
that is why ${{\left(\mu_{1}+\,\mu_{2}\right)}^{2}}=trA+2$ this equality holds
iff $(\frac{trA+2}{p})=1$. Thus, the condition $(\frac{trA+2}{p})=1$ is
sufficient for existing of ${{\chi}_{b}}(x)$. But it remains to show that
these eigenvalues $\sqrt{\lambda_{1}}=\mu_{1}$, $\sqrt{\lambda_{2}}=\mu_{2}$
are the roots of the characteristic polynomial $\chi_{B}(x)$.

By the condition of theorem $trA+2$ is a quadratic residue or 0, there is
$\sqrt{tr(A)+2}=\sqrt{{{({{\mu}_{1}}+{{\mu}_{2}})}^{2}}}$ in
${{\text{F}}_{\text{p}}}$, whence
$tr(\text{B})\in{{\text{F}}_{\text{p}}},\,\,\det B\in{{\text{F}}_{\text{p}}}$
holds in view of well known theorems, therefore ${{\chi}_{B}}(x)$ has
coefficients $c=\sqrt{tr(A)+2}={\mu}_{1}+{\mu}_{2}$ in
${{\text{F}}_{\text{p}}}$, hence $B$ presented in the Frobenius normal form
belongs to $S{{L}_{2}}({{\text{F}}_{\text{p}}})$.

Furthermore $B$ having e.v. ${{\mu}_{1}},\,\,{{\mu}_{2}}$ is the matrix over
${{\text{F}}_{\text{p}}}$, but ${{\mu}_{1}},\,\,{{\mu}_{2}}$ can be from
${{\text{F}}_{{{\text{p}}^{2}}}}\backslash{{\text{F}}_{\text{p}}}$.

For analyzing distributions of roots of (1) by cosets of
$ES{{L}_{2}}\left(\mathbb{F}\right)$ consider
${{\left({{\mu}_{1}}+{{\mu}_{2}}\right)}^{2}}=\mu_{1}^{2}+\mu_{2}^{2}+2{{\mu}_{1}}{{\mu}_{2}}=tr(A)+2det(A)$.
If we have $\mu_{1}^{2}+\mu_{2}^{2}-2=tr(A)-2$ is square in $\mathbb{F}$, then
$\det(A)={{\mu}_{1}}{{\mu}_{2}}=-1$ and so these roots of (1) from coset
$ES{{L}_{2}}\left(\mathbb{F}\right)\backslash
S{{L}_{2}}\left(\mathbb{F}\right)$. If $\mu_{1}^{2}+\mu_{2}^{2}+2=trA+2$ is
square in $\mathbb{F}$ then the roots of equation (1) are from normal subgroup
$S{{L}_{2}}\left(\mathbb{F}\right)$ from $ES{{L}_{2}}\left(\mathbb{F}\right)$.

At the same time we have relation
$\mu_{1}^{2}+\mu_{2}^{2}={{\left(\mu_{1}+\mu_{2}\right)}^{2}}-2=tr(A)$.
Consider characteristic polynomial ${{x}^{2}}-cx+1=0$, where
$c={{\mu}_{1}}+{{\mu}_{2}}$. Show that it is exactly a characteristic
polynomial for $B=\sqrt{A}$. Indeed e.v. of $B$ have to be
${{\mu}_{1}},\,\,\,{{\mu}_{2}}$ and have to satisfy condition 2. Substitute
${{\mu}_{1}},\,\,\,{{\mu}_{2}}$ in ${{x}^{2}}-cx+1=0$ we can verify that they
are roots
${{x}^{2}}-({{\mu}_{1}}+{{\mu}_{2}})x+1={{\mu}_{1}}^{2}-({{\mu}_{1}}+{{\mu}_{2}}){{\mu}_{1}}+1=0-{{\mu}_{1}}{{\mu}_{2}}+1=0$
and the analogously for verification of ${{\mu}_{2}}$.

We show the necessity of condition (2) provided the rest of the theorem
conditions hold. We assume that there exists $B\in F_{p}$ such that $B^{2}=A$
and deduce the condition $\frac{trA+2}{p}=1$. It provides existing of
correspondent characteristic polynomial $\chi_{B}(x)=x^{2}+cx+1$, where by
Viet theorem $c=\mu_{1}+\mu_{2}$ which in accord to equalities
$\mu_{1}^{2}=\lambda_{1},\mu_{2}^{2}=\lambda_{2}$. In the same time
$trA+2={\mu_{1}+\mu_{2}}^{2}+2$. Since $B\in SL_{2}(F_{p})$ then
$({{\mu}_{1}}+{{\mu}_{2}})\in F_{p}$. Thus, $\sqrt{trA+2}\in F_{p}$ too. The
case $({{\mu}_{1}}+{{\mu}_{2}})\equiv 0(modp)$ lead us to ${trA+2}\equiv
0(modp)$.

Using condition (2) we show that matrix $B$ with trace
$tr(\text{B})={{\mu}_{1}}+{{\mu}_{2}}$ and $det(B)={{\mu}_{1}}{{\mu}_{2}}=1$
exists. For this goal we express
${{\left(\mu_{1}+\mu_{2}\right)}^{2}}=\mu_{1}^{2}+\mu_{2}^{2}+2=tr(A)+2$.
Since $(\frac{trA+2}{p})=1$ then the square root
$\sqrt{tr(A)+2}=\sqrt{{{({{\mu}_{1}}+{{\mu}_{2}})}^{2}}}$ exists, i.e. matrix
$tr(A)={{\mu}_{1}}+{{\mu}_{2}}\in F_{p}$.

If the matrix $B$ is an element from $SL_{2}(\mathbb{F}_{p})$, it yields to
$\mu_{1}\mu_{2}=1$. Therefore, the matrices with $\mu_{1}\mu_{2}=-1$ are not
valid. But if $B\in ESL_{2}(\mathbb{F}_{p})$ then $\mu_{1}\mu_{2}=\pm 1$.

$B=\left(\begin{array}[]{cc}\mu_{1}&0\\\ 0&\mu_{2}\\\ \end{array}\right).$

The case, where $\mu_{1}=\mu_{2}=\mu$, implies that $\mu=g^{2}$, where $g\in
F_{p}$, this condition demands to $\mu$ be a square in $F_{p}$. In this case
$\mu\in F_{p}$ where $F_{p}$ is base field, so according to the theorem about
Jordan basis Jordan matrix exists over this field. Considering the multiple
eigenvalues $\mu_{1}=\mu_{2}=\mu$ and Viet theorem for characteristic
polynomial, where $b\in F_{p}$ holds, we see that $b=\mu_{1}+\mu_{2}=2\mu$. So
transformation of matrix $A$ in Jordan form is possible. Note that in case
$p=2$ expression $\mu=b/2$ is impossible. But existing of such $\mu$ can be
proven in another way.

There are two possible matrix forms, either $A$ has 2 multiple eigenvalues or
$A$ has different eigenvalues. If $A$ has a form of a Jordan block then this
case satisfies the conditions of the theorem 15 as well as when $A$ is a
scalar matrix then it satisfies the condition of the theorem described below.

We note that the case when $\lambda=0$ is irrelevant for consideration,
because of belonging $A$ to $SL_{2}(\mathbb{F}_{p})$ demands $det(A)=1$.

Two different eigenvalues $\lambda_{1},\lambda_{2}$ occur only in the case of
a diagonal matrix. These eigenvalues are
$\lambda_{1}=\mu_{1}^{2},\lambda_{2}=\mu_{2}^{2}$. In the case of Jordan block
of size 2 by 2 only multiple eigenvalues are possible.

A sufficient condition in the subcase of the Jordan cell being obvious is the
same existence of a root $\sqrt{\lambda}$ in the finite field ${{F}_{p}}$,
$p>2$. It is this root that is the eigenvalue

$\mu=\sqrt{\lambda}$

of the matrix $B$.

Let us express the square by trace and determinant. Since
$\mu_{1}^{2}+\mu_{2}^{2}=(\mu_{1}+\mu_{2})^{2}-2\mu_{1}\mu_{2}$ then
$(\mu_{1}+\mu_{2})^{2}=\mu_{1}^{2}+\mu_{2}^{2}+2\mu_{1}\mu_{2}=tr(A)+2$. A
value in the left part is the quadratic residue so $tr(A)+2$ is too. Since in
$ESL_{2}(F_{p})$ and $det(A)$ admits as $1$ as well as $-1$ and in case action
of such $A$ changes space orientation, $SL_{2}(F_{p})$ then in case of
$ESL_{2}(F_{p})$ the both signs $"\pm"$ before the 2 in $tr(A)+2$ fit. Thus,
in the case of $ESL_{2}(F_{p})$ this conditions transforms in the form
$tr(A)\pm 2$ due to possible root $B$ with $det(B)=-1$. But such case $B\notin
SL_{2}(F_{p})$. If $tr(A)+2$ is multiple to $p$ then characteristic polynomial
$\chi_{A}(x)$ of $A$ takes form $x^{2}+1$, because of general formula is such
$\chi_{A}(x)$ has form $x^{2}-tr(A)x+detA$.

Define the polynomial squared if there is such ${{\chi}_{B}}(x)=x^{2}-cx+d$,
that $d=\mu_{1}\mu_{2},\,c=\mu_{1}+\mu_{2},\,c,d\in F_{p}$ and
$\mu^{2}_{1}=\lambda_{1}$, $\mu^{2}_{2}=\lambda_{2}$, where $\lambda_{1}$,
$\lambda_{2}$ are e. v. of another polynomial ${{\chi}_{A}}(x)$. We can
formulate the similar criterion for a polynomial. In particular characteristic
polynomials of $A$ and $B$ could be in such a role. Let
${{\chi}_{A}}(x)=x^{2}-ax+1$, $(a\in F_{p})$.

Let us investigate requirement of existence of squared polynomial. We take
into account that ${\chi}_{B}(x)$ for matrix for $SL_{2}(F_{p})$, so
$\mu_{1}\mu_{2}=1$. But roots of ${\chi}_{B}(x)$ and even ${\chi}_{A}(x)$ can
belong to $F_{p^{2}}$ but not to $F_{p}$. So we will consider symmetric
functions from $\lambda_{1}$ and $\lambda_{2}$.
$(\mu_{1}+\mu_{2})^{2}=\lambda_{1}+\lambda_{2}+2\sqrt{\lambda_{1}\lambda_{2}}=\lambda_{1}+\lambda_{2}+2$
because of $\lambda_{1}\lambda_{2}=1$. Therefore condition
$(\frac{\lambda_{1}+\lambda_{2}+2}{p})\in\\{0,1\\}$ is sufficient that squared
polynomial exists.

In the case $p=2$ this condition becomes to be
$(\frac{\lambda_{1}+\lambda_{2}}{p})\in\\{0,1\\}$ is sufficient for existing
of squared polynomial ${\chi}_{B}(x)$.

The eigenvalues $\mu_{1}$, $\mu_{2}$ are determined as
$\mu_{i}=\pm\sqrt{(\mu_{i})^{2}}$, where
$\lambda=\mu_{i}^{2}\,\,i\in\\{1,2\\}$. A combination of signs in terms
representing $\mu_{1}$ and $\mu_{2}$ have to satisfy condition of polynomial
squared: $d=\mu_{1}\mu_{2},\,c=\mu_{1}+\mu_{2},\in F_{p}$.

To prove for $A\in S{{L}_{2}}({{F}_{p}})$ that condition

$(\frac{trA+2}{p})\in\\{0,1\\}$

is sufficient for the existence of a solution of the equation $A={{X}^{2}}$
then we shall show that $\exists B:{{B}^{2}}=A$. Let us prove that this
condition $(\frac{trA+2}{p})\in\\{0,1\\}$ provides the existence of
characteristic polynomial $\chi_{B}(x)={{x}^{2}}-cx+1$ with the
$c\in{{F}_{p}}$ whose roots are the eigenvalues $\mu_{1},\,\,\mu_{2}$ of the
matrix $B$ and ${{\mu}_{1}}{{\mu}_{2}}=1$ provided that
$\mu_{1}^{2},\mu_{2}^{2}$ are the eigenvalues of matrix $A$.

The proof is very similar in case $A\in GL_{2}({{F}_{p}})$, then the condition
$(\frac{trA+2\sqrt{detA}}{p})=1$ is enough for solvability of the equation
$A={{X}^{2}}$.

Consider the e. v., denoted by $\lambda_{1}$ and $\lambda_{2}$, of matrix $A$.
Then we need to find such $\mu_{1},\,\,\mu_{2}$ that
${{\lambda}_{1}}=\mu_{1}^{2},\,\,{{\lambda}_{2}}=\mu_{2}^{2}$ and
${{\mu}_{1}}{{\mu}_{2}}=1$.

Note that existing of such $\mu_{1},\,\,\mu_{2}$ implies the following
condition
${{({{\mu}_{1}}+{{\mu}_{2}})}^{2}}=\mu_{1}^{2}+2{{\mu}_{1}}{{\mu}_{2}}+{{\mu}^{2}_{2}}=\mu_{1}^{2}+\mu_{2}^{2}+2={{\lambda}_{1}}+{{\lambda}_{2}}+2=trA+2.$
holds. Consider such $c$ that ${{c}^{2}}=trA+2,\,\,\,c\in{{F}_{p}}$. We prove
that roots $\mu_{1},\,\,\mu_{2}$ of ${{\mu}^{2}}-c\mu+1$ endowed the property
${{\mu}_{1}}{{\mu}_{2}}=1$ and $\mu_{1}^{2},\,\,\mu_{2}^{2}$ coincide with
${{\lambda}_{1}},\,\,{{\lambda}_{2}}$.

Indeed, the equality ${{\mu}_{1}}{{\mu}_{2}}=1$ is provided by Viet theorem.
It remains to show that
$\mu_{1}^{2}={{\lambda}_{1}},\,\,\mu_{2}^{2}={{\lambda}_{2}}$ this is
equivalent to we need to show the equality of polynomials
$\chi(A)=\left(x-{{\lambda}_{1}}\right)\left(x-{{\lambda}_{2}}\right)$ and
$\left(x-\mu_{1}^{2}\right)\left(x-\mu_{2}^{2}\right)$. We have
$\left(x-{{\lambda}_{1}}\right)\left(x-{{\lambda}_{2}}\right)={{x}^{2}}-\left({{\lambda}_{1}}+{{\lambda}_{2}}\right)x+{{\lambda}_{1}}{{\lambda}_{2}}={{x}^{2}}-\left(trA\right)x+\det
A={{x}^{2}}-\left(trA\right)x+1.$ In the similar way
$\left(x-\mu_{1}^{2}\right)\left(x-\mu_{2}^{2}\right)={{x}^{2}}-\left(\mu_{1}^{2}+\mu_{2}^{2}\right)x+\mu_{1}^{2}\mu_{2}^{2}=\\\
={{x}^{2}}-\left(\mu_{1}^{2}+\mu_{2}^{2}+2{{\mu}_{1}}{{\mu}_{2}}-2{{\mu}_{1}}{{\mu}_{2}}\right)x+\mu_{1}^{2}\mu_{2}^{2}.$
Since $\mu_{1},\,\,\mu_{2}$ are roots of ${{\mu}^{2}}-c\mu+1$ then according
to Viet theorem: ${{\mu}_{1}}{{\mu}_{2}}=1$ and ${{\mu}_{1}}+{{\mu}_{2}}=c$,
This implies
$\left(x-\mu_{1}^{2}\right)\left(x-\mu_{2}^{2}\right)={{x}^{2}}-\left({{c}^{2}}-2\right)x+{{1}^{2}}$.
Taking into account that ${{c}^{2}}=trA+2$ we derive
$\left(x-\mu_{1}^{2}\right)\left(x-\mu_{2}^{2}\right)={{x}^{2}}-\left(trA+2-2\right)x+1={{x}^{2}}-\left(trA\right)x+1$,
which coincides with the characteristic polynomial
$\chi(A)=\left(x-{{\lambda}_{1}}\right)\left(x-{{\lambda}_{2}}\right)$ for the
matrix $A$. This completes the proof.

In cases when $A\in G{{L}_{2}}({{F}_{p}})$ the proof is very similar. ∎

######  Example 4.

Consider Fibonacci matrix $F=\left(\begin{array}[]{rr}0&\,\,1\\\
1&\,\,1\end{array}\right)$ in $S{{L}_{2}}\left({{F}_{p}}\right)$ then
${{F}^{2}}=\left(\begin{array}[]{rr}1&\,\,1\\\ 1&\,\,2\end{array}\right)$
which confirms criterion 3 of existing roots in
$ES{{L}_{2}}\left({\mathbb{F}_{p}}\right)$ because $trA-2=1$ because of 1 is
square in each field ${\mathbb{F}_{p}}$ as well as in $\mathbb{Q}$ and
$\mathbb{R}$.

Next one is $R=\left(\begin{array}[]{rr}0&\,\,-2\\\ 2&\,\,0\end{array}\right)$
in $S{{L}_{2}}\left({\mathbb{F}_{3}}\right)$ then
$R^{2}=\left(\begin{array}[]{rr}-1&\,\,0\\\ 0&\,\,-1\end{array}\right)$.

In another hand we can justify the root existing by criterion for
$ES{{L}_{2}}\left({\mathbb{F}_{3}}\right)$ because $tr{{R}^{2}}-2=0$.

######  Example 5.

The case of roots contain in both cosets of quotient
${}^{ES{{L}_{2}}\left(\mathbb{F}\right)}/{}_{S{{L}_{2}}\left(\mathbb{F}\right)}$
appear in particular matrix $A$ with $tr(A)=3$ and
$\mathbb{F}=\mathbb{F}_{11}$. In fact, in this case $tr(A)-2=1$, $tr(A)+2=5$
one easily can verify that 5 is quadratic residue by $11$ because of
${{4}^{2}}\equiv 5\left(\bmod 11\right)$ and $1$ is always square.

######  Example 6.

Consider matrix equation $X^{2}=A$ with e.v. in
${{\mathbb{F}}_{9}}\backslash{{\mathbb{F}}_{3}}$ and having root in
$ES{{L}_{2}}\left({{\mathbb{F}}_{3}}\right)\backslash
S{{L}_{2}}\left({{\mathbb{F}}_{3}}\right)$ and check our new formula from
Proposition 1 (which is directly below) for root expression. Let
$A=\left(\begin{array}[]{rr}0&-1\\\ 1&0\\\ \end{array}\right)$

since $trA+2=2$ that is non-square residue in ${{\mathbb{F}}_{3}}$ but
$trA-2=-(-1)=1$ in ${{\mathbb{F}}_{3}}$, then according to our formula we use
branch of expression with minus in $trA\pm 2$ i.e.
$\sqrt{A}=\frac{A-E}{\sqrt[]{trA-2}}=\frac{1}{\sqrt[]{trA-2}}\left(\begin{array}[]{rr}-0-1,&-1\\\
\,1,&-0-1\end{array}\right)=\frac{1}{\sqrt[]{trA-2}}\left(\begin{array}[]{rr}-1,&-1\\\
\,1,&-1\\\ \end{array}\right)=\left(\begin{array}[]{rr}-1,&-1\\\ \,1,&-1\\\
\end{array}\right)=\left(\begin{array}[]{rr}\,2,&-1\\\ \,1,&2\\\
\end{array}\right)=B.$ Another branch with "=" before the root $trA\pm 2$ lead
us to second root:
$\sqrt{A}=\frac{A-E}{-\sqrt[]{trA-2}}=\frac{1}{\sqrt[]{trA-2}}\left(\begin{array}[]{rr}-0+1,&1\\\
\,=1,&-0+1\end{array}\right)=\frac{1}{\sqrt[]{trA-2}}\left(\begin{array}[]{rr}1,&1\\\
\,-1,&1\\\ \end{array}\right)=-B.$

Its ${{\chi}_{A}}(x)={{x}^{2}}+1=0$ therefore its roots are $\pm
i\in{{\mathbb{F}}_{9}}\backslash{{\mathbb{F}}_{3}}$ and $\pm i$ are square in
${{\mathbb{F}}_{9}}$ that confirms our criterion 3.

For case $\mathbb{F}=\mathbb{F}_{p}$ our criterion can be formulated in terms
of Legendre symbol.

######  Corollary 7.

Let $A$ be simple matrix and $A\in{SL_{2}(\mathbb{F}_{p})}$ [33], then for
matrix $A\in SL_{2}(\mathbb{F}_{p})$ there is a solution $B\in
SL_{2}(\mathbb{F}_{p})$ of the matrix equation

$X^{2}=A$ (5)

if and only if

$\left(\frac{trA+2}{p}\right)\in\\{0,1\\}.$ (6)

If $X\in ESL_{2}(\mathbb{F}_{p})$ then the matrix equation (5) has a solution
iff

$\left(\frac{trA\pm 2}{p}\right)\in\\{0,1\\}.$ (7)

This solution $X\in ESL_{2}(\mathbb{F}_{p})\setminus SL_{2}(\mathbb{F}_{p})$
iff $\left(\frac{trA-2}{p}\right)=1$ or 0, but
$\left(\frac{trA+2}{p}\right)=-1$. Conversely $X\in SL_{2}(\mathbb{F}_{p})$
iff $\left(\frac{trA+2}{p}\right)=1$. Solutions $X_{i}\in ESL_{2}(\mathbb{F})$
and $SL_{2}(\mathbb{F})$ iff $\left(\frac{trA+2}{p}\right)=1$ and
$\left({trA-2}{}\right)=1$.

In the case $A\in GL_{2}({{\mathbb{F}}_{p}})$ this condition (2) takes form:

$\left(\frac{trA\pm 2\sqrt{detA}}{p}\right)\in\\{0,1\\}.$ (8)

The proof is the same but instead of $\mathbb{F}$ we put $\mathbb{F}_{p}$. But
we emphasize that theorems of such a kind were for algebraic closed field
before this paper.

######  Corollary 8.

If $A\in GL(F_{2})$ the condition 2 takes form:

$\left(\frac{trA}{p}\right)\in\\{0,1\\}.$

######  Remark 9.

The formulated criterion for a diagonizable matrix is also true over fields
$\mathbb{Q}$ and $\mathbb{R}$.

###### Proof.

The proof is the same only with the change of quadraticity criterion over the
new field. ∎

######  Corollary 10.

If matrix $A$ admits diagonal form over ${{\text{F}}_{2}}$ then $A$ is square
over ${{\text{F}}_{2}}$.

###### Proof.

Since in $F_{2^{n}}$ all elements $g_{i}\in F_{2^{n}}$ are quadratic squares,
therefore a diagonal matrix $A$ is always square of the mentioned above $B$
over $F_{2^{n}}$. ∎

######  Proposition 1.

If a simple matrix $A\in S{{L}_{2}}({\mathbb{F}_{p}})$ and
$\left(\frac{Tr(A)+2}{p}\right)=1$, then

$\sqrt{A}=\frac{1}{\pm\sqrt{trA\pm 2}}\left(A\pm E\right),$

where $E$ is identity element of $S{{L}_{2}}({\mathbb{F}_{p}})$, in case of
sign ’-’ in $\left(A\pm E\right)$ roots $\sqrt{A}\in
ESL_{2}({\mathbb{F}_{p}})$. Namely for $A=\left(\begin{array}[]{cc}a&b\\\
c&d\\\ \end{array}\right)$ in a coordinate form in case $\sqrt{A}\in
SL_{2}({\mathbb{F}_{p}})$ we have
${\sqrt{A}}=\frac{1}{\pm\sqrt{tr(A)+2}}\left(\begin{array}[]{cc}a+1&b\\\
c&d+1\\\ \end{array}\right)$.

###### Proof.

Consider the characteristic equation for $A$: ${{x}^{2}}-tr(A)x+\det(A)=0$.
According to Cayley Hamilton theorem we have

$\displaystyle{{A}^{2}}-tr(A)A+E=0,$ $\displaystyle{{A}^{2}}+E=tr(A)A.$

We add $2A$ to the both sides of this equation

$\displaystyle{{A}^{2}}+2A+E=-tr(A)A+2A,$
$\displaystyle{{\left(A+E\right)}^{2}}=A\left(trA+2\right).$

And finally we express the root:

$\sqrt{A}=\frac{1}{\pm\sqrt{trA+2}}\left(A+E\right).$ (9)

In a coordinate it takes the form:
${\sqrt{A}}=\frac{1}{\pm\sqrt{trA+2}}\left(\begin{array}[]{cc}a+1&b\\\
c&d+1\\\ \end{array}\right).$

The similar proof for the case ${trA-2}$ is square give us roots in
$ESL_{2}(\mathbb{F_{p}})$.

$\sqrt{A}=\frac{1}{\pm\sqrt{trA-2}}\left(A-E\right).$ (10)

In a coordinate form it presents by matrix:

${\sqrt{A}}=\frac{1}{\pm\sqrt{trA-2}}\left(\begin{array}[]{cc}a-1&b\\\
c&d-1\\\ \end{array}\right).$ ∎

The similar proof yields the formula of roots in $ESL_{2}(\mathbb{Z})$ and in
$ESL_{2}({k})$, where $k$ is arbitrary perfect field.

######  Example 11.

For instance, $tr(E)=2$ and as a result
${\sqrt{E}}=\frac{1}{\pm\sqrt{2+2}}\left(\begin{array}[]{cc}2&0\\\ 0&2\\\
\end{array}\right)=\pm E.$

Note the analitical formula of root in $GL_{2}(\mathbb{R})$ was found in [43]
but for finite field $\mathbb{F}_{p}$ there did not found such formula in
linear group of 2 dimension.

######  Corollary 12.

The formula of 4-th power root is the following

$\sqrt[4]{A}=\frac{A\pm E\pm\sqrt{trA\pm 2}}{\pm\sqrt{\pm\sqrt{trA\pm 2}\pm
2}}.$

###### Proof.

We construct the formula to 4-th power root in recursive way where base of
recursion in formula (9). Taking into account $tr(A+E)=trA+2$ we obtain for
case $det(A)=det(\sqrt{A})=1$ that
$\sqrt[4]{A}=\frac{\sqrt{A}+E}{\pm\sqrt{tr\sqrt{A}+2}}=\frac{A+E\pm\sqrt{trA+2}}{\pm\sqrt{\sqrt{trA+2}+2}}.$
∎

######  Remark 13.

Extended special linear group $ESL_{2}(k)$, where $k$ is arbitrary perfect
field, is storage of all square matrix roots from $ESL_{2}(k)$.

######  Proposition 2.

If matrix $A$ do not admits diagonal form over ${{\text{F}}_{2}}$ then $A$ is
not square in $G{{L}_{2}}\left({{\text{F}}_{2}}\right)$ over
${{\text{F}}_{2}}$.

###### Proof.

If matrix $A$ do not admits diagonal form over ${{\text{F}}_{2}}$ then $A$ is
not square in $G{{L}_{2}}\left({{\text{F}}_{2}}\right)$ over
${{\text{F}}_{2}}$. We consider equation of form ${{X}^{2}}=A$ and show that
it has not solutions over ${{\text{F}}_{2}}$ a
$S{{L}_{2}}\left({{\text{F}}_{2}}\right)$ in case
${{\chi}_{A}}\left(x\right)\neq{{\mu}_{A}}\left(x\right)$. The conditions of
theorem implies that geometrical dimension of e.v. is 1 but algebraic
multiplicity of e.v. $\lambda$ is 2. We make prof by the contradiction,
assuming that is true then

${\left(\begin{array}[]{cc}\lambda&1\\\ 0&\lambda\\\
\end{array}\right)}^{2}=\left(\begin{array}[]{cc}{{\lambda}^{2}}&2\lambda\\\
0&{{\lambda}^{2}}\\\ \end{array}\right)$ but
$\left(\begin{array}[]{cc}{{\lambda}^{2}}&2\lambda\\\ 0&{{\lambda}^{2}}\\\
\end{array}\right)=\left(\begin{array}[]{cc}{{\lambda}^{2}}&0\\\
0&{{\lambda}^{2}}\\\ \end{array}\right)$ over ${{\text{F}}_{2}}$. That
contradicts to condition of this Theorem. ∎

Let $B$ has characteristic polynomial ${{x}^{2}}+bx+c$. It is well known that
trace of $B$ is stable under choosing of vector space base.

We denote Jordan form of matrix $A$ as ${{J}_{A}}$.

######  Lemma 14.

If a matrix $A\in S{{L}_{2}}({{F}_{p}})$ has multiple eigenvalues
${{\beta}_{1}}={{\beta}_{2}}=\beta$ and non-trivial Jourdan block of size
$2\times 2$ then $\beta\in{{\text{F}}_{p}}.$

Proof. Since in this case eigenvalues are presented as elements of matrix $B$
standing on diagonal, then this matrix can be in form:
$B=\left(\begin{array}[]{cc}\beta&1\\\ 0&\beta\\\ \end{array}\right)$ or
$B=\left(\begin{array}[]{cc}\beta&0\\\ 0&\beta\\\ \end{array}\right).$

But the eigenvalues of the matrix are multiples, therefore
$\beta+\beta=tr(B)\in{{\text{F}}_{p}}$. This implies $2\beta=b$, therefore in
a field of characteristic non equal 2 we express this eigenvalue as
$\beta=\frac{b}{2}$. Hence $\beta\in{{\text{F}}_{p}}$. The proof is complete.

Our study of quadratic elements in $SL_{2}(\mathbb{F})$ gives an instrument to
solve the problem when a finite group G contains a conjugacy class $K$ whose
square $K^{2}$ is again a conjugacy class [45].

######  Theorem 15.

Under conditions $(\frac{\lambda}{p})=1$ in ${{F}_{p}}$ and matrix $A$ is
similar to a Jordan block of the form

${{J}_{A}}=\left(\begin{array}[]{cc}\lambda&1\\\ 0&\lambda\\\
\end{array}\right)$ (11)

a square root $B$ of $A$ exists in $S{{L}_{2}}({{F}_{p}})$.

###### Proof.

Assume that square from $A$ exists in $S{{L}_{2}}({{F}_{p}})$ or in
$ES{{L}_{2}}({{F}_{p}})$ correspondently. We denote matrix $B$ transformed to
upper triangular form by $U{{T}_{B}}$. Let us show that there that provided
condition above it always exists such $B:\,\,UT_{B}^{2}={{J}_{A}}$, where
$U{{T}_{B}}$ is $B$ transformed to UTM form. Then we show that it implies
existing of solution of

${{X}^{2}}=A.$

From the existence of the Jordan block for $A$ follows the existence of a
similarity transformation $U$ transforming ${{B}^{2}}$ to the Jordan normal
form $J_{B}$ because of $A={{B}^{2}}$ and A has non-trivial Jordan block
denoted by $J_{A}$. But square root from $B^{2}$ this operator $U$ transforms
in upper triangular form $UT_{B}$. Then if we find solution for

$UT_{B}^{2}=J_{A}$ (12)

we can obtain solution for ${{X}^{2}}=A$ because of the following:

$A=U\cdot(U{{T}_{B}})^{2}\cdot{{U}^{-1}}=(U\cdot
U{{T}_{B}}\cdot{{U}^{-1}})(U\cdot U{{T}_{B}}\cdot{{U}^{-1}})={{B}^{2}}.$ (13)

It means that such matrix $U{{T}_{B}}$ satisfying (13), exists and it can be
transformed by the same similarity transformation by conjugation in form
$U{{T}_{B}}={{U}^{-1}}BU$ by the same matrix that transforms $A$ in
${{J}_{A}}$ because of ${{B}^{2}}=A$. To show the existing of such solution of
(12) we acting by invers transformation
$A=U\cdot(U{{T}_{B}})^{2}\cdot{{U}^{-1}}=(U\cdot
U{{T}_{B}}\cdot{{U}^{-1}})(U\cdot U{{T}_{B}}\cdot{{U}^{-1}})={{B}^{2}}$, where
$U$ is similarity transformation $B$ to

$U{{T}_{B}}=\left(\begin{array}[]{cc}\beta&\gamma\\\ 0&\beta\\\
\end{array}\right).$

note that its diagonal elements ${{b}_{11}}={{b}_{22}}=\beta$ are the same.
Therefore according to Lemma 14 we have $\beta\in{{F}_{p}}$. Even more easier
we can deduce it without Lemma 19. We have ${{b}_{11}}={{b}_{22}}=\beta$, then
$\beta+\beta=\text{Tr}({{U}^{-1}}BU)$. Therefore $2\beta\in{{\text{F}}_{p}}$.
It implies that $\beta\in{{F}_{p}}$ if $p>2$ and

$(UT_{B})^{2}=\left(\begin{array}[]{cc}{{\beta}^{2}}&2\beta\gamma\\\
0&{{\beta}^{2}}\\\ \end{array}\right).$

Here the element $\gamma$ can be chosen $\gamma:\,\,\,2\beta\gamma=1$ so
$\gamma={2\beta}^{-1}$ taking into account that $\beta=\sqrt{\lambda}$ which
is already determined by $A$. Then $(UT_{B})^{2}:$

$(UT_{B})^{2}=\left(\begin{array}[]{cc}{{\beta}^{2}}&2\beta\gamma\\\
0&{{\beta}^{2}}\\\
\end{array}\right)=\left(\begin{array}[]{cc}{{\beta}^{2}}&1\\\
0&{{\beta}^{2}}\\\
\end{array}\right)=J_{A}=\left(\begin{array}[]{cc}\lambda&1\\\ 0&\lambda\\\
\end{array}\right)$.

Furthermore we show that these conditions is also necessary but not only
sufficient. It means if $(\frac{\lambda}{p})=-1$, then there are no matrix $B$
over $S{{L}_{2}}({{F}_{p}})$ such that ${{B}^{2}}=A$. By a reversal of theorem
condition and using the representation in the form of UTM for and for we see
that $B$ from $PS{{L}_{2}}({{F}_{p}})$ such that ${{B}^{2}}=A$. We see that
according to the Lemma 14 the eigenvalue of $B$ over lie in the main field —
${{\text{F}}_{p}}$. However, we assumed that $(\frac{\lambda}{p})=-1$. Thus we
obtain the desirable contradiction.

Let us show that condition of non-diagonalizability of matrix is necessary in
the conditions of this Theorem. By virtue of the well-known theorem stating
that if the algebraic multiplicity is equal to the geometric multiplicity for
each eigenvalue, then matrix is diagonalizable otherwise it is not
diagonalizable, we see that if the condition of similarity to
${{J}_{A}}=\left(\begin{array}[]{cc}\lambda&1\\\ 0&\lambda\\\
\end{array}\right)$ indicated in this Theorem 15 does not holds, then such $A$
satisfy the conditions of this Theorem 3, where algebraic multiplicity is
equal to geometrical. And since the condition 11 of this criterion is nature,
therefore, it is no longer necessary to prove the non-diagonalizability
condition in Theorem 15.

Proof of necessity. Furthermore we show that these conditions is also
necessary but not only sufficient. It means if $(\frac{\lambda}{p})=-1$, then
there are no matrix $B$ having non trivial Jordan block over
$PS{{L}_{2}}({{F}_{p}})$ and $S{{L}_{2}}({{F}_{p}})$ such that ${{B}^{2}}=A$.
By a reversal of theorem condition and using the representation in the form of
UTM for and for we see that $B$ from $PS{{L}_{2}}({{F}_{p}})$ such that
${{B}^{2}}=A$. We see that according to the Lemma the eigenvalue of $B\in
S{{L}_{2}}({{F}_{p}})$or $B\in PS{{L}_{2}}({{F}_{p}})$ correspondingly, lie in
the main field – ${{\text{F}}_{p}}$. Furthermore according to Lemma 2 if
$\beta$ is an eigenvalue for $B$ then ${{\beta}^{2}}$ is an eigenvalue for
${{B}^{2}}$, so we have ${{\beta}^{2}}=\lambda$. However, we assumed that
$(\frac{\lambda}{p})=-1$. Thus we obtain the desirable contradiction. The
eigenvalue $\beta$ has geometrical dimension 1, because of in oppositive case
geometrical $\dim\beta=2$ (dimension of eigenvector space of $\beta$), then we
get that $J_{B}^{2}$ is only scalar matrix $B$.

The proof is fully complete. ∎

######  Example 16.

A sufficiency of the condition $(\frac{\lambda}{p})=1$ in Theorem 15 for
$\exists$ $B:\,\,{{B}^{2}}=A$, where $A\sim{{J}_{A}}$ of size $2\times 2$ with
one eigenvalue corresponding to one eigenvector is given by following matrix
from $S{{L}_{2}}(R)$:

${{J}_{A}}=\left(\begin{array}[]{cc}1&1\\\ 0&1\\\ \end{array}\right)$ then
$B={{\left(\begin{array}[]{cc}\mu&1\\\ 0&\mu\\\
\end{array}\right)}^{2}}=\left(\begin{array}[]{cc}{{\mu}^{2}}&2\mu\\\
0&{{\mu}^{2}}\\\ \end{array}\right),\,\,\,\,\mu=\pm\sqrt{1}$.

This confirms Theorem 3. Choosing the base for $B$ to $A$ be in Jordan form
(in Jordan base): $UB{{U}^{-1}}$ we obtain

$\left(\begin{array}[]{cc}\frac{\mu}{2}&1\\\ 0&\frac{\mu}{2}\\\
\end{array}\right)={{J}_{B}}.$

The last matrix is expressed by conjugating of $B$ by a diagonal matrix.

######  Example 17.

Consider an example confirming Theorem 3 . Let
$A=\left(\begin{array}[]{rr}-1&0\\\ 0&-1\end{array}\right)={{\rho}_{180}}$.
This is a $180$ degree rotation matrix. The $Tr(A)+2=0$ then root has to exist
in $S{{L}_{2}}(R)$. Then its square root $B\in S{{L}_{2}}(R)$ has form
${{\rho}_{90}}=\left(\begin{array}[]{cc}\,0&1\\\ -1&0\\\
\end{array}\right)=B\in S{{L}_{2}}(R)$. Note that $A$ is presented in the
diagonal form. There are also roots $B_{1}=\left(\begin{array}[]{cc}i&a\\\
0&-i\\\ \end{array}\right)$ from $S{{L}_{2}}(C)$ as well as
$B_{2}=\left(\begin{array}[]{cc}-i&a\\\ \,0&i\\\ \end{array}\right)$.

Consider the commutative matrix algebra $A\lg\left[A\right]=\left\langle
E,A\right\rangle$ over ${{F}_{p}}$, where$A\in
S{{L}_{2}}\left({{F}_{p}}\right)$, A is semisimple without multiple e.g. and
$E$ is identity matrix. Let us consider such formulation of Weederbern Artin
theorem (Theorem W. A.). Each simple algebra is isomorphic to full matrix ring
over algebra with division.

We refer to such formulation of theorem Wederburn Artin from Vanderwarden’s
book [1, page 372, 1979 year]. Theorem W. A.

######  Theorem 18.

Every artinian simple algebra with 1 is isomorphic to a complete matrix ring
${{M}_{{{n}_{i}}}}[{{D}_{i}}]$ over a division algebra ${{D}_{i}}$.

Recall well known statement.

######  Proposition 3.

Every finite commutative division algebra ${{D}_{i}}$ is finite field.

The following Lemma 19 can be proved with using of Weederbern Artin theorem in
the following way. As well known that every finite commutative division
algebra is isomorphic to finite field. (Note that the field is a division
commutative algebra and complete matrix ${{M}_{2\times 2}}[{{F}_{p}}]$ ring
over ${{F}_{p}}$ is algebra.) But we need more specialized isomorphism making
connection between $A$ and its e.g. as element of ${{F}_{p^{2}}}$ so we will
construct it now.

Since ${{F}_{{{p}^{2}}}}$ is simple commutative algebra and elements of group
$A\lg\left[A\right]=\left\langle E,A\right\rangle$ with additional zero matrix
$2\times 2$ forms an commutative matrix ring then according to Weederbern
Artin theorem we have $A\left[M\right]\simeq{{F}_{{{p}^{2}}}}$. Kelly Hamilton
theorem yields that dimension of such algebra $A\lg\left[A\right]$ is 2
because of characteristic polynomial $\chi_{A}(x)$ of $A$ has degree 2.

To give a self-contained proof, we propose the following Lemma.

Firstly we consider case $A\in S{{L}_{2}}({{F}_{p}})$ but after our reasoning
can be easy extended on the group $G{{L}_{2}}({{F}_{p}})$ because the property
property of the $\det A=1$ will not be involved in our reasoning.

######  Lemma 19.

The matrix algebra $Alg[A]=\left\langle
E,\,\,A\right\rangle\simeq{{F}_{{{p}^{2}}}}$.

###### Proof.

We show that algebra $Alg\left[M\right]=\left\langle E,A\right\rangle$ is
isomorphic to finite field ${{F}_{{{p}^{2}}}}$. As well-known from Galois
theory, a quadratic extension of $F_{p}$ can be constructed by involving of
any external element. As well-known from Galois theory, a quadratic extension
of ${{F}_{p}}$ can be constructed by involving of any external element
$g\in{{F}_{{{p}^{2}}}}\backslash{{F}_{p}}$ relatively to ${{F}_{p}}$. We
denote this element by $i$, in particular, for $p=4m+3$ it may be an element
satisfying the relation ${{i}^{2}}=-1$. Note that the matrix of the rotation
by 90 degrees, that is a matrix

$I:=\left(\begin{array}[]{cc}0&1\\\ -1&0\\\ \end{array}\right)={{\rho}_{90}}$

satisfies this relation and can used as an example of matrix $A$. In case when
$p=4m+3$ such matrix $J:\,\,\varphi(J)=j$, ${{j}^{2}}=-1$ exists too.

Obviously $\det A=1$, that’s why $A\in S{{L}_{2}}({{F}_{p}})$ and
${{\mu}_{A}}(x)$ is irreducible. We define mapping
$\varphi:\,\,{{y}_{1}}A+{{x}_{1}}E\,\to
ae+b\lambda;\,\,\,\,a,b\in{{\text{F}}_{p}}$. The mapping $\varphi$ can be more
broadly described, in $S{{L}_{2}}[{{F}_{p}}]$ such a way that a matrix $A$ is
found such that ${{A}^{2}}=-E$, then its e.g. $\lambda$ is assigned to it in
the field ${{F}_{{{p}^{2}}}}$, while
$\lambda\in{{F}_{{{p}^{2}}}}\backslash{{F}_{p}}$.
$\varphi:\,\,{{y}_{1}}A+{{x}_{1}}E\,\to
ae+b\lambda;\,\,\,\,a,b\in{{\text{F}}_{p}}$. According to assumption of Lemma
the matrix $A$ is semisimple and has not multiple eigenvalues (e.g.) which are
not squares in ${{\text{F}}_{p}}$, so ${{\chi}_{A}}\left(x\right)$ is
irreducible because of definition of semisimple matrix and condition
${{\lambda}_{1}}\neq{{\lambda}_{2}}$. According to Lemma about Frobenius
automorphism its eigenvalues are conjugated in ${{\text{F}}_{{{p}^{2}}}}$. The
method of constructing of $\sqrt{A}$ is the following. Having isomorphism
$A\lg\left[A\right]=\left\langle
E,A\right\rangle\simeq{{\text{F}}_{{{p}^{2}}}}$ we set a correspondence
$\lambda\leftrightarrow A$ and correspondence between groups operations in
${{\text{F}}_{{{p}^{2}}}}$ and $A\lg\left[A\right]$. Therefore solving
equation ${{\left(x+\lambda y\right)}^{2}}=\lambda$ relatively coefficients
$x,\,\,y\in{{\text{F}}_{p}}$ we obtain coefficients for expression for
$\sqrt{A}$ i.e. $\sqrt{A}=x+Ay$. To prove the isomorphism, we establish a
bijection between the generators of the algebra
$A\lg\left[A\right]=\left\langle E,A\right\rangle$ and the field
${{F}_{{{p}^{2}}}}$. It is necessary to establish in more detail that
$A\leftrightarrow\lambda$ and $E\leftrightarrow e$ also the correspondence
between the neutral elements of both structures, i.e.
$\varphi\left({\bar{0}}\right)=0$ where $0$ is the zero matrix. To complete
proof, it remains to show that the kernel of this homomorphism $\varphi$ is
trivial. To do this, we show that among the elements of the algebra there are
no identical ones. The surjectivity of $\varphi$ is obvious. From the
opposite, we assume ${{y}_{1}}A+{{x}_{1}}E={{y}_{2}}A+{{x}_{2}}E$,
${{x}_{i}},\,{{y}_{i}}\in{{F}_{p}}$. Then
${{y}_{1}}A+{{x}_{1}}E={{y}_{2}}A+{{x}_{2}}E$ it yields that
$\left({{y}_{1}}-{{y}_{2}}\right)E=\left({{x}_{1}}-{{x}_{2}}\right)A$, which
is impossible since the characteristic polynomial of the matrix $A$ is
irreducible but the characteristic polynomial of the identity matrix is
reducible. Therefore, our algebra $A\lg\left[A\right]$ is isomorphic to the
completely linear space of linear polynomials from $E$ and $A$. In the similar
way we prove that polynomial of form $xe+y\lambda$ where $x,\,\,y\in{{F}_{p}}$
do not repeat. The proof is based on oppositive assumption about coinciding
${{x}_{1}}e+{{y}_{1}}\lambda={{x}_{2}}e+{{y}_{2}}\lambda$ of polynomial with
different coefficients. Then equality
${{x}_{1}}e+{{y}_{1}}\lambda={{x}_{2}}e+{{y}_{2}}\lambda$ implies that
$\left({{y}_{1}}-{{y}_{2}}\right)\lambda=\left({{x}_{1}}-{{x}_{2}}\right)e$
i.e. ${{y}_{1}}={{y}_{2}}$ and ${{x}_{1}}={{x}_{2}}$ that contradicts to
assumption. ∎

######  Theorem 20.

If a matrix $A\in G{{L}_{2}}({{F}_{p}})$ is semisimple with different
eigenvalues and at least one an eigenvalue ${{\lambda}_{i}}\in
F_{p^{2}}\setminus F_{p}$, $i\in\\{1,2\\},\,p>2$, then $\sqrt{A}\in
G{{L}_{2}}({{F}_{p}})$ iff of $A$ satisfies:

$(\frac{{{\lambda}_{i}}}{p})=1\,\,in\,\,the\,\,square\,\,extention\,\,that\,\,is\,\,{{F}_{p^{2}}}.$

###### Proof.

Firstly, we consider most complex and interesting case when $A$ is not
diagonalizable, then ${{\chi}_{A}}\left(x\right)$ is irreducible over
${{F}_{p}}$. By assumption, the matrix is semisimple and its characteristic
polynomial is irreducible. So root $\lambda$ of ${{\chi}_{A}}(x)$ belongs to
the quadratic extension of the field ${{F}_{p}}$. Since each element of
${{F}_{{{p}^{2}}}}$ can be presented in form
$a+b\lambda,\,\,\,a,b\in{{F}_{p}}$, then we can construct mapping of matrix
algebra generators $E$ and $A$ in generators of ${{F}_{{{p}^{2}}}}$ and apply
the aforementioned Lemma 19 about isomorphism establish correspondence between
property be square in ${{\text{F}}_{{{p}^{2}}}}$ and in
$Alg\left[A\right]=\left\langle E,A\right\rangle$. If one e.v.
${{\lambda}_{i}}$ is square in $F_{p^{2}}$ then so is second e.v. because of
they are conjugated as roots of characteristic polynomial $\chi_{A}(x)$ by
theorem about Frobenius automorphism (Frobenius endomorphism in perfect field
became to be automorphism). ∎

######  Example 21.

Consider the matrix $A=-E$, where $E$ is identity matrix in
$S{{L}_{2}}({{F}_{3}})$ satisfying conditions of Theorem 20 because of
$(\frac{-1}{9})=1$ in ${{\text{F}}_{9}}$. And there exists the matrix
$\left(\begin{array}[]{cc}0&2\\\ -2&0\end{array}\right)\in
S{{L}_{2}}({{F}_{3}})$ is square root for $A$. Indeed ${{I}^{2}}=-E$.

Another root of this equation ${{X}^{2}}=A$, where $A$ is matrix of elliptic
type realizing rotation on 90 degrees
${{\rho}_{90}}=\left(\begin{array}[]{cc}\,\,\,0\,\,\,\,1\\\ -1\,\,0\\\
\end{array}\right)=I$ because of ${{I}^{2}}=-E$, is matrix of parabolic type.

The matrix $2I$ is the square in $G{{L}_{2}}\left({{F}_{3}}\right)$ because of
existing such an element ${{\left(\begin{array}[]{cc}\,\,\,1\,\,\,1\\\
-1\,\,\,\,1\\\ \end{array}\right)^{2}}}=2\left(\begin{array}[]{cc}\,\,\
0\,\,\,\,1\\\ -1\,\,0\\\ \end{array}\right)=2I.$

Hypothesis. If we consider vector space over the same perfect field $k$ over
which we consider $S{{L}_{2}}(k)$ then we have $\sqrt[3]{A}\in
Span\\{A,\,\,E\\}$ over $k$, where Span is linear span.

For proof we take into account Kelly Hamilton’s equation and apply a reduction
transformation of the second degree, then we get an expression similar to
$\sqrt[3]{A}=\frac{A+tr\left(\sqrt[3]{A}\right)\det(\sqrt[3]{A})}{t{{r}^{2}}\left(\sqrt[3]{A}\right)-\sqrt[3]{\det\left(A\right)}}$
which after transformations yields expression of root $\sqrt[3]{A}=\lambda
A+\beta$, where $\lambda,\beta\in k$.

Proposition. If $B\in S{{L}_{2}}({{\mathbb{F}}_{p}})$ is root of equation
${{X}^{3}}=A$, then

$B=\frac{A+tr(\sqrt[3]{A})\sqrt[3]{\det(A)}}{tr{{\left(\sqrt{A}\right)}^{2}}-\sqrt[3]{\det(A)}},$

where $A\in S{{L}_{2}}({{\mathbb{F}}_{p}})$.

###### Proof.

Proof. If $\sqrt[3]{A}\in S{{L}_{2}}({{\mathbb{F}}_{p}})$ then we consider
Kelly Hamilton equation
(K.H.E.)${{A}^{3}}-tr\left(A\right){{A}^{2}}+\left({{\lambda}_{1}}{{\lambda}_{2}}+{{\lambda}_{1}}{{\lambda}_{3}}+{{\lambda}_{2}}{{\lambda}_{3}}\right)A-\det\left(A\right)=0$.
Note, that
$tr{{\left(A\right)}^{2}}={{\left({{\lambda}_{1}}+{{\lambda}_{2}}+{{\lambda}_{3}}\right)}^{2}}=\lambda_{1}^{2}+\lambda_{2}^{2}+\lambda_{3}^{2}-\left({{\lambda}_{1}}{{\lambda}_{2}}+{{\lambda}_{1}}{{\lambda}_{3}}+{{\lambda}_{2}}{{\lambda}_{3}}\right)$.

Consider K.H.E. for $A:\,dimA=2$, ${{A}^{2}}-tr\left(A\right)\cdot
A+\det\left(A\right)\cdot 1=0$. Multiplying last equation on $A$ admit us
obtain the chain of transformation:

$\displaystyle{{A}^{3}}$
$\displaystyle=\left(tr\left(A\right)A-\det\left(A\right)\right)A=tr\left(A\right){{A}^{2}}-\det\left(A\right)A=$
$\displaystyle=tr\left(A\right)\left(tr\left(A\right)A-\det\left(A\right)\right)-\det\left(A\right)A=$
(14)
$\displaystyle=tr{{\left(A\right)}^{2}}A-tr\left(A\right)\det\left(A\right)-\det\left(A\right)A=$
$\displaystyle=\left(tr{{\left(A\right)}^{2}}-\det\left(A\right)\right)A-tr\left(A\right)\det\left(A\right).$

By applying substitute matrix $\sqrt[3]{A}$ instead of $A$ we express

$\sqrt[3]{A}=\frac{A+tr\left(\sqrt[3]{A}\right)\sqrt[3]{\det
A}}{tr^{2}{{\left(\sqrt[3]{A}\right)}}-\sqrt[3]{\det\left(A\right)}}.$ (15)

Thus,
$\sqrt[3]{A}=\frac{A\,+tr\left(\sqrt[3]{A}\right)\sqrt[3]{\det\left(A\right)}}{\,\left(t{{r}^{2}}\left(\sqrt[3]{A}\right)-\sqrt[3]{\det\left(A\right)}\right)}$.

Note that $\det\left(\sqrt[3]{A}\right)=\sqrt[3]{\det\left(A\right)}$ because
of determinant is homomorphism.

But $tr\left(\sqrt[3]{A}\right)$ is still not computed. From (4) we conclude
${A}^{3}=\left(tr{{\left(A\right)}^{2}}-\det\left(A\right)\right)A-tr\left(A\right)\det\left(A\right).$
Computing a trace from both sides we obtain
$tr\left({{A}^{3}}\right)=tr{{\left(A\right)}^{3}}-3\det\left(A\right)tr\left(A\right)$.

We put $\sqrt[3]{A}$ instead of $A$, then we get
$tr\left(A\right)=tr{{\left(\sqrt[3]{A}\right)}^{3}}-3\sqrt[3]{\det
A}tr\left(\sqrt[3]{A}\right).$

We need to solve
$tr\left(A\right)=tr{{\left(\sqrt[3]{A}\right)}^{3}}-3\sqrt[3]{\det
A}tr\left(\sqrt[3]{A}\right)$.

We denote $\sqrt[3]{A}$ by $X$ and obtain the equation

${{X}^{3}}-3\sqrt[3]{\det\left(A\right)}X-tr\left(A\right)=0.$

The _solvability_ of this equation over base field $\mathbb{F}_{p}$ is
equivalent to the _existence_ of a trace $\sqrt[3]{A}$ in the base field. ∎

######  Example 22.

Consider the diagonal matrix $A\in S{{L}_{2}}({{F}_{3}})$ emphasizing the need
for the condition $F_{p^{2}}\setminus F_{p}$, $i\in\\{1,2\\}$ in Theorem 20
for semisimple matrix and confirming Remark 24.

It is easy to verify the absence of root from
$A=\left(\begin{array}[]{cc}1\,\,\,\,\,0\\\ \,0\,-1\\\ \end{array}\right)$ in
$S{{L}_{2}}({{F}_{3}})$.

######  Theorem 23.

If a matrix $A\in S{{L}_{2}}({{F}_{p}})$ ($A\in GL({F}_{p})$) possesses
diagonal Jordan form over ${{\text{F}}_{p}}$, then $\sqrt{A}\in
S{{L}_{2}}({{F}_{p}})$ ($GL({F}_{p})$) if and only if
$(\frac{{{\lambda}_{1}}}{p})=1$ and $(\frac{{{\lambda}_{2}}}{p})=1$ over
${{\text{F}}_{p}}$.

###### Proof.

From condition $(\frac{{{\lambda}_{1}}}{p})=1$ and
$(\frac{{{\lambda}_{2}}}{p})=1$ it is followed, that
${{\mu}_{A}}\left(x\right)$ is reduced over ${{\text{F}}_{p}}$. Therefore why
${{\mu}_{1}},\,\,{{\mu}_{2}}\in{{\text{F}}_{p}}$ exist
${{\mu}_{B}}\left(x\right)$ over ${{\text{F}}_{p}}$ exists for matrix
$B:\,\,{{B}^{2}}=A$. Assume that
$(\frac{{{\lambda}_{1}}}{p})=-1,\,\,(\frac{{{\lambda}_{2}}}{p})=-1$ prove,
that while $\sqrt{A}\notin S{{L}_{2}}({{F}_{p}})$. We use proof by
contradiction. Let
$(\frac{{{\lambda}_{1}}}{p})=-1,\,\,(\frac{{{\lambda}_{2}}}{p})=-1$ therefore
roots from eigenvalues ${{\lambda}_{1}},\,\,{{\lambda}_{2}}$ in general
belongs to ${{\text{F}}_{{{p}^{2}}}}$ while its roots
${{\mu}_{1}},\,\,{{\mu}_{2}}$ is not conjugated as roots from different values
of ${{\lambda}_{1}},\,\,{{\lambda}_{2}}$.

Let’s find minimal polynomial for $B=\sqrt{A}$. Minimal polynomial of matrix
$B$ is ${{\mu}_{B}}(x)={{x}^{2}}-bx+c$ and it has different roots
${{\mu}_{1}},\,\,{{\mu}_{2}}$, where ${{\mu}_{1}}+{{\mu}_{2}}=Tr(B)=b$ e $\det
B={{\mu}_{1}}{{\mu}_{2}}$. From the existence of diagonal representation for
$A$ reducibility of ${{\mu}_{A}}\left(x\right)$ follows. From the reducibility
of ${{\mu}_{A}}\left(x\right)$ over ${{\text{F}}_{p}}$ and the fact that
${{\lambda}_{1}}\neq{{\lambda}_{2}}$ follows ${{\mu}_{1}},\,\,{{\mu}_{2}}$ is
not conjugated as the roots of different values of
${{\lambda}_{1}},\,\,{{\lambda}_{2}}$ and it is obvious that
$\mu_{1}^{2}\,\neq\,\mu_{2}^{2}$. But the root ${{\mu}_{1}}$ is conjugated
with $-{{\mu}_{1}}$ and
${{\mu}_{1}}\in{{\text{F}}_{{{p}^{2}}}}\backslash{{\text{F}}_{p}}$. But
$-{{\mu}_{1}}$ is also a root, since
${{\left(\pm{{\mu}_{1}}\right)}^{2}}={{\lambda}_{1}}$ therefore it can be the
root for ${{\mu}_{B}}(x)$. Similar situation is with root ${{\mu}_{2}}$ and
$-{{\mu}_{2}}$. Therefore, we indicated as many as 4 roots for
${{\mu}_{B}}(x)$ but $B\in S{{L}_{2}}({{F}_{{{p}^{2}}}})$ therefore
$\deg\left({{\mu}_{B}}(x)\right)=2$. This contradiction arises from the
assumption that$\sqrt{A}\in S{{L}_{2}}({{F}_{p}})$ on condition
$(\frac{{{\lambda}_{1}}}{p})=-1,\,\,(\frac{{{\lambda}_{2}}}{p})=-1$. ∎

######  Remark 24.

If $A\in S{{L}_{2}}({{F}_{p}})$ possesses the capacity to be in diagonal
Jordan form over ${{\text{F}}_{p}}$ and
$(\frac{{{\lambda}_{1}}}{p})=-1,\,\,(\frac{{{\lambda}_{2}}}{p})=1$, then such
case does not give the existence of solution of ${{X}^{2}}=A$ in
$S{{L}_{2}}({{F}_{p}})$.

###### Proof.

The condition $(\frac{{{\lambda}_{1}}}{p})=-1$ means, that
$\sqrt{{{\lambda}_{1}}}={{\beta}_{1}}\in{{\text{F}}_{{{p}^{2}}}}\backslash{{\text{F}}_{p}}$
and simultaneously $\sqrt{{{\lambda}_{2}}}={{\beta}_{2}}\in{{\text{F}}_{p}}$,
therefore ${{\beta}_{1}}+{{\beta}_{2}}=Tr(B)\notin{{F}_{p}}$. This implies
non-existing of ${{\mu}_{B}}\left(x\right)$ over ${{\text{F}}_{p}}$. ∎

The following theorem it is true for $S{{L}_{2}}({k})$, even $k$ is arbitrary
perfect field. The following proof works for arbitrary perfect $k$ too.

######  Theorem 25.

If a matrix $A\in S{{L}_{2}}({{F}_{p}})$ is semisimple and diagonalizable over
${{\text{F}}_{p}}$ and
$(\frac{{{\lambda}_{1}}}{p})=(\frac{{{\lambda}_{2}}}{p})=-1$, then for the
existing $\sqrt{A}$, it is necessary and sufficient, that $A$ is similar to
scalar matrix $D$.

###### Proof.

From the facts that
$(\frac{{{\lambda}_{1}}}{p})=(\frac{{{\lambda}_{2}}}{p})=-1$ and the square of
diagonal matrix is again the diagonal matrix follows the existence of root
only in the off diagonal form, therefore we must find the solution $M$ among
the set of non-diagonalizable matrices
$D=\left(\begin{array}[]{cc}{{d}_{1}}&0\\\ 0&{{d}_{2}}\\\ \end{array}\right)$
is the diagonal representation of matrix $A$, and let

$D={{M}^{2}},$ (16)

where $M\in S{{L}_{2}}({{F}_{p}})$. Because of
$(\frac{{{d}_{1}}}{p})=(\frac{{{d}_{2}}}{p})=-1$ there is a root in non-
diagonal form. Also we note that there is a conjugation matrix $X$,

$X=\left(\begin{array}[]{cc}m_{11}^{-1}&0\\\ 0&m_{21}^{-1}\\\
\end{array}\right),$

transforming $M$ to $\tilde{M}$, where $\tilde{M}$ has following form

$\tilde{M}=~{}\left(\begin{array}[]{cc}{{m}_{11}}&1\\\
{{m}_{21}}&{{m}_{22}}\\\ \end{array}\right).$

Let’s transform the equality $D={{M}^{2}}$ into
$XD{{X}^{-1}}=XM{{X}^{-1}}XM{{X}^{-1}}$, where $XM{{X}^{-1}}=\tilde{M}$. Note
that $D$ and $XD{{X}^{-1}}$ have identical eigenvalues. Therefore we can solve
the equation (16) for $XD{{X}^{-1}}$. Let’s consider matrix equation
$D={{M}^{2}}$, let’s transform it by conjugation
$D=XD{{X}^{-1}}=XM{{X}^{-1}}\cdot
XM{{X}^{-1}}=\tilde{M}\tilde{M}={{\tilde{M}}^{2}}$, wherein
$M=\left(\begin{array}[]{cc}&{{m}_{11}}\,\,\,\,{{m}_{12}}\\\
&{{m}_{21}}\,\,{{m}_{22}}\\\ \end{array}\right)$,
$X=\left(\begin{array}[]{cc}&m_{11}^{-1}\,\,\,0\\\
&0\,\,\,\,\,\,m_{21}^{-1}\\\ \end{array}\right)$ and
$XM{{X}^{-1}}=\left(\begin{array}[]{ll}{{m}_{11}}&1\\\
{{m}_{21}}&{{m}_{22}}\\\ \end{array}\right).$

Since $D$ is a diagonal matrix, then it belongs to the commutative subgroup of
diagonal matrices from $S{{L}_{2}}({{F}_{p}})$, lets denote it as
$DS{{L}_{2}}({{F}_{p}})$. Therefore and $~{}XD{{X}^{-1}}$ is also a diagonal
matrix. Moreover, due to the commutativity of the field ${{F}_{p}}$ we have
$~{}XD{{X}^{-1}}=D.$ Now let’s solve the matrix equation for the reduced
$\tilde{M}$

$D=~{}XD{{X}^{-1}}=(XM{{X}^{-1}})(XM{{X}^{-1}})={{\tilde{M}}^{2}},$ (17)

Note that equations (17) and (16) are equivalent since they are obtained by
similarity transformations.

Note that equations (2) and (1) are equivalent since they are obtained by
similarity transformations. Let’s write down the equation

${{\tilde{M}}^{2}}={{\left(\begin{array}[]{cc}&{{m}_{11}}\,\,\,\,1\\\
&{{m}_{21}}\,\,\,{{m}_{22}}\\\
\end{array}\right)}^{2}}=\left(\begin{array}[]{cc}&{{d}_{1}}\,\,\,0\\\
&0\,\,\,{{d}_{2}}\\\ \end{array}\right).$

Thence we obtain the system of equations

$\left\\{\begin{array}[]{cc}{{m}_{21}}+m_{11}^{2}={{d}_{1}}\\\
{{m}_{21}}+m_{22}^{2}={{d}_{2}}\\\ {{m}_{11}}+{{m}_{22}}=0,\\\
\end{array}\right.$

by substitution $m_{11}$ from the equation 3) ${{m}_{22}}=-{{m}_{11}}$ into
equations 1) and 2) we express from

$2)\,\,{{m}_{21}}+m_{22}^{2}={{d}_{2}}\Rightarrow{{m}_{21}}+{{(-m_{11})}^{2}}={{d}_{2}}$

also we take into consideration equation 1) $m_{11}^{2}+{{m}_{21}}={{d}_{1}}$.
Thence ${{d}_{1}}={{d}_{2}}$ or more conveniently $d={{d}_{1}}={{d}_{2}}$.
Wherein $d$ doesn’t have to be a quadratic residue. Therefore the condition
$(\frac{d}{p})=-1$ of theorem is met. ∎

The previous theorem can be proven for group $SL_{2}(F_{p})$ also in following
way using finite field equations. The following proof works only for finite
field $F_{p}$ .

######  Remark 26.

If the matrix $A\in SL_{2}(F_{p})$ is semisimple and diagonalizable over
$F_{p}$ and $(\frac{{{\lambda}_{1}}}{p})=(\frac{{{\lambda}_{2}}}{p})=-1$, then
for $\sqrt{A}\in SL_{2}(F_{p})$, it is necessary and sufficient that $A$ be
similar to scalar matrix.

###### Proof.

Let according to condition both e.v.
$(\frac{{{\lambda}_{1}}}{p})=(\frac{{{\lambda}_{2}}}{p})=-1$ but
${\lambda}_{1}\neq{\lambda}_{2}$.

If we assume, that equation $X^{2}=A$ is solvable over $F_{p}$ then e.v.
$\beta_{1},\beta_{2}$ of $B$ are such that
$\beta_{1},\beta_{2}\in\\{F_{p^{2}}\backslash F_{p}\\}$ because of $\lambda$
is non-square over $F_{p}$. Since we assume existing of root from $A\in
SL_{2}(F_{p})$, then $tr(B)=\beta_{1}+\beta_{2}$. Consider expression
$\beta_{1}$ by trace $\beta_{1}=tr(B)-\beta_{2}$ then its square
$\beta_{1}^{2}=(tr(B)+\beta_{2})^{2}$ give us
$\beta_{1}^{2}=(tr(B))^{2}-2tr(B)\beta_{2}-\beta_{2}^{2}$. In the LHS of the
equality we have that $\beta_{1}^{2}\in F_{p}$ because $\beta_{1}$ is a root
from the element $\lambda_{1}\in F_{p}$. But in RHS
$(tr(B))^{2}-2t\beta_{2}+\beta^{2}_{2}\in F_{p}$ iff $tr(B)=0$ because of
$\beta_{1},\beta_{2}\in\\{F_{p}^{2}\backslash F_{p}\\}$ and $p>2$ as root from
$\alpha_{2}$ which is not square in $F_{p}$. But we assume that $B$ is an
arbitrary diagonal matrix over $F_{p}$ that means $tr(B)\neq 0$. The obtained
contradiction follows from our assumption that $A$ is not scalar matrix. ∎

##  5 Conclusion

The the analytical formula of square roots in $SL_{2}({\mathbb{F}_{p}})$,
$ESL_{2}({\mathbb{F}_{p}})$, for any prime $p$, as well as in $SL_{2}({Z})$
$ESL_{2}({Z})$ and in $SL_{2}({k}),ESL_{2}({k})$, where $k$ is arbitrary
perfect field, is found by us.

New linear group which is storage of square roots from
$SL_{2}({\mathbb{F}_{p}})$ is found and investigated by us.

The criterions of solvability of matrix equation $X^{2}=A$ over different
linear groups according of matrix classification according its $tr(A)$ and
type of space contracting is found and proved in this paper.

The criterion of roots existing for different classes of matrix — simple and
semisimple matrixes from $SL_{2}({\mathbb{F}_{p}})$, $SL_{2}({\mathbb{Z}})$
are established.

If a matrix $A\in G{{L}_{2}}({{F}_{p}})$ is semisimple with different
eigenvalues and at least one an eigenvalue ${{\lambda}_{i}}\in
F_{p^{2}}\setminus F_{p}$, $i\in\\{1,2\\}$, then $\sqrt{A}\in
G{{L}_{2}}({{F}_{p}})$ iff $A$ satisfies:

$(\frac{{{\lambda}_{i}}}{p})=1\,\,in\,\,the\,\,algebraic\,\,extention\,\,of\,\,degree\,\,2\,\,that\,\,is\,\,{{F}_{p^{2}}}.$

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