# Difference between revisions of "Special linear group:SL(2,Z)"

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## Definition

The group $SL(2,\mathbb{Z})$ is defined as the group, under matrix multiplication, of $2 \times 2$ matrices over $\mathbb{Z}$, the ring of integers, having determinant $1$.

In other words, it is the group with underlying set:

$\left \{ \begin{pmatrix} a & b \\ c & d \\\end{pmatrix} \mid a,b,c,d \in \mathbb{Z}, ad - bc = 1 \right \}$

This is the degree two case of a special linear group over integers and hence of a special linear group. It is also a special case of a special linear group of degree two.

The group also has the following equivalent descriptions:

### Definition by presentation

The group can be defined by any of the following presentations (here, $1$ denotes the identity element):

• From the amalgamated free product definition: $\langle x,y \mid x^4 = 1, x^2 = y^3 \rangle$

### Structures

Thinking of $SL(2,\mathbb{Z})$ as a group of matrices, we see that it is an example of an arithmetic group.

## Arithmetic functions

Function Value Explanation
order infinite (countable) As $SL(2,\mathbb{Z})$: The group is infinite because, for instance, it contains all matrices of the form $\begin{pmatrix} 1 & a \\ 0 & 1 \\\end{pmatrix}$ for $a \in \mathbb{Z}$.
As a set, the group is contained in the set of all $2 \times 2$ matrices over $\mathbb{Z}$. This can be identified with $\mathbb{Z}^4$, which is countable since $\mathbb{Z}$ is countable. Thus, $SL(2,\mathbb{Z})$ is also countable.
As an amalgamated free product: any amalgamated free product relative to a subgroup that is proper in both groups is infinite.
exponent infinite As $SL(2,\mathbb{Z})$: The group contains the element $\begin{pmatrix} 1 & 1 \\ 0 & 1 \\\end{pmatrix}$, which has infinite order.
minimum size of generating set 2 As $SL(2,\mathbb{Z})$: Follows from elementary matrices of the first kind generate the special linear group over a Euclidean ring, so $SL(2,\mathbb{Z})$ is generated by all matrices of the form $\begin{pmatrix} 1 & a \\ 0 & 1 \\\end{pmatrix}$ and $\begin{pmatrix} 1 & 0 \\ b & 1 \\\end{pmatrix}$ with $a,b$ varying over $\mathbb{Z}$. By the fact that the additive group of $\mathbb{Z}$ is cyclic, all these matrices are generated by $\begin{pmatrix} 1 & 1 \\ 0 & 1 \\\end{pmatrix}$ and $\begin{pmatrix} 1 & 0 \\ 1 & 1 \\\end{pmatrix}$.
As an amalgamated free product of two cyclic groups: follows that it is 2, just from the definition (note that the generators here are different from those used in the justification in matrix terms).
subgroup rank infinite (countable) $SL(2,\mathbb{Z})$ has a subgroup that is isomorphic to free group:F2 (see Sanov subgroup in SL(2,Z) is free of rank two). This in turn has free subgroups of countable rank.

## Group properties

Property Satisfied? Explanation Corollary properties satisfied/dissatisfied
2-generated group Yes See explanation for minimum size of generating set above satisfies: finitely generated group, countable group
Noetherian group No See explanation for subgroup rank above
finitely presented group Yes Any of the definitions ($SL$, B_amalgamated free product) gives a finite presentation
solvable group No contains subgroup isomorphic to free group:F2 -- see Sanov subgroup in SL(2,Z) is free of rank two dissatisfies: nilpotent group, abelian group
group satisfying no nontrivial identity Yes contains subgroup isomorphic to free group:F2 -- see Sanov subgroup in SL(2,Z) is free of rank two
SQ-universal group Yes contains subgroup isomorphic to free group:F2 -- see Sanov subgroup in SL(2,Z) is free of rank two
residually finite group Yes The kernels of the homomorphisms $SL(2,\mathbb{Z}) \to SL(2,\mathbb{Z}/n\mathbb{Z})$ for natural numbers $n$ are normal subgroups of finite index and their intersection is trivial. satisfies: finitely generated residually finite group
Hopfian group Yes Follows from finitely generated and residually finite implies Hopfian satisfies: finitely generated Hopfian group

## Elements

Further information: element structure of special linear group:SL(2,Z)

## Facts

• Sanov subgroup in SL(2,Z) is free of rank two
• The homomorphism $SL(2,\mathbb{Z}) \to SL(2,\mathbb{Z}/n\mathbb{Z})$ is surjective for any natural number $n$.

## GAP implementation

Description Functions used
SL(2,Integers) SL, Integers