Surjectivity of SL_n(Z) to SL_n (Z/NZ)

The natural mod- ${N}$ projection map

$\displaystyle \pi: \mathrm{SL}_{2}(\mathbb{Z}) \rightarrow \mathrm{SL}_{2}(\mathbb{Z} / N)$

is surjective.

Proof: We want to show that for any integer matrix ${A}$ with ${\det(A) = 1 \mod N}$ , there is an matrix ${B \in \mathrm{SL}_{2}(\mathbb{Z})}$ such that ${B=A \mod N.}$

1) If ${\left(\begin{array}{ll}a & b \\ c & d\end{array}\right) \in \mathrm{SL}_{2}(\mathbb{Z} / N \mathbb{Z})}$ , we can choose integers ${c' = c \mod N, d'=d \mod N}$ such that ${c', d'}$ are coprime.

WLOG assume that ${c \neq 0}$ . Now if ${p}$ divides ${c}$ , then it cannot divide both ${N}$ and ${d}$ ; Because otherwise the determinant won’t be ${1 \mod N.}$
Choose ${k_p}$ such that ${d+k_pN}$ is not divisible by ${p}$ – you can take ${k_p=0}$ if ${p}$ doesn’t divide ${d}$ and ${k_p=1}$ if ${p}$ divides ${d.}$
Use Chinese remainder theorem to find ${k}$ such that ${k= k+p \mod p}$ for ${p}$ dividing ${c.}$
Now by construction ${d+kN}$ is not divisible by any prime ${p}$ dividing ${c}$ , and hence coprime to ${c.}$
${c'= c, d'= d+kN}$ is our choice.

2) Now consider

$\displaystyle B=\left(\begin{array}{cc}a+p N & b+q N \\ c' & d'\end{array}\right)$

We have $\displaystyle ad'-bc' =ad-bc=1 \mod N.$ Let $\displaystyle ad'-bc'=1+tN$

The determinant of the above matrix is $(a d'-b c')+N(p d'-qc').$

Because ${c',d'}$ are coprime we can choose ${p,q}$ such that ${p d'-qc' =-t.}$
Hence the determinant of ${B}$ equals $(a d'-b c')+N(p d'-qc') =1+tN -tN =1,$ and we are done.

Another proof: Given a matrix ${A}$ with ${\det(A) = 1 \mod N}$ , use elementary row and column operations to reduce the matrix to diagonal ${\left(\begin{array}{cc}a & 0 \\ 0 & a^{-1}\end{array}\right)}$ , where the inverse is over ${\mathbb{Z} / N.}$

Further using the following factorisation

$\displaystyle \left(\begin{array}{cc} a & 0 \\ 0 & a^{-1} \end{array}\right)=\left(\begin{array}{cc} 1 & a \\ 0 & 1 \end{array}\right)\left(\begin{array}{cc} 1 & 0 \\ -a^{-1} & 1 \end{array}\right)\left(\begin{array}{cc} 1 & a \\ 0 & 1 \end{array}\right)\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right)$

$\displaystyle \left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right) =\left(\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right)\left(\begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array}\right)\left(\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right)$

we see that ${A}$ can be lifted to a matrix in ${S L_{2}(\mathbb{Z}).}$ – the elementary row operators are in ${S L_{2}(\mathbb{Z}).}$ and the matrices in the above factorisation can be easily lifted.

We now move to the case of larger matrices.

$\displaystyle S L_{n}(\mathbb{Z})$

We also have surjectivity of the natural quotient map

$\displaystyle \pi: S L_{n}(\mathbb{Z}) \rightarrow S L_{n}(\mathbb{Z} / N)$

Proof: We reduce the case of diagonal matrices by elementary row and column operations.
Once we have the diagonal matrix, use ${\left(\begin{array}{cc} a & 0 \\ 0 & a^{-1} \end{array}\right)}$ matrices to set all the diagonal entries except one to ${1.}$ (Multiply with matrices of this form which by above identity is known to be generated by elementary matrices) But because the determinant is not changed, the last entry would be just ${1.}$ Thus we see that ${S L_{n}(\mathbb{Z} / N)}$ is generated by elementary matrices; we are now done because elementary matrices can be lifted to ${S L_{n}(\mathbb{Z}).}$

-These are special cases of much more general results about algebraic groups: See Strong Approximation.
– The above matrix identities and theory elementary matrices is studied in K-theory.

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