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Consider the group homomorphism $\varphi : SL_2 (\Bbb Z) \longrightarrow SL_2 (\Bbb Z/ 3 \Bbb Z)$ defined by $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto \begin{pmatrix} \overline {a} & \overline {b} \\ \overline {c} & \overline {d} \end{pmatrix}.$$

What is the cardinality of the image of $\varphi$?

Since $\text {Im} (\varphi)$ is a subgroup of $SL_2 (\Bbb Z/ 3 \Bbb Z)$ so by Lagrange's theorem $\#\ \text {Im} (\varphi)\ \big |\ \#\ SL_2 (\Bbb Z/ 3 \Bbb Z) = 24.$ What I have observed is that $\#\ \text {Im} (\varphi) \geq 7.$ So $\#\ \text {Im} (\varphi) = 8,12\ \text {or}\ 24.$

I have just seen that the image contains at least $10$ elements. So the possibility for cardinality of $\text {Im} (\varphi)$ is $12$ or $24.$

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  • $\begingroup$ Perhaps I am missing something here, but it seems to me that any element of the target is clearly in the image of $\phi$. $\endgroup$
    – the_fox
    Commented Jan 27, 2020 at 14:25
  • $\begingroup$ Phi is surjective so #image must be 24. It is surjective because the homomorphism just sends each matrix component to its representative coset. $\endgroup$
    – Fomalhaut
    Commented Jan 28, 2020 at 9:57
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    $\begingroup$ @BalancedTryteOperators It is not completely trivial that the existence of an integer matirx with determinant $\equiv 1\pmod 3$ implies the existence of a matrix with determinant $1$ and congruent to the former $\pmod 3$. $\endgroup$ Commented May 13, 2020 at 18:20

2 Answers 2

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For any $n\geq 1$ and any prime number $p$, the canonical map $\pi:SL_n(\mathbb{Z})\to SL_n(\mathbb{F}_p)$ is surjective.

Indeed, it is a standard fact that if $R$ is an Euclidean ring (for example $R=\mathbb{Z}$ or $R=\mathbb{F}_p$), then $SL_n(R)$ is generated by the transvection matrices $$T_{i,j}(a)=I_n+aE_{ij}, i\neq j, a\in R$$ ($E_{ij}$ is the matrix whose entries are $0$, except the one at position $(i,j)$, which is $1$).

So, we just have to check that the image contains any transvection matrix. But this is clear, since $\pi (T_{i,j}(a))=T_{i,j}(\bar{a})$ for all $a\in\mathbb{Z}$.

Hence $f$ is surjective.

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The site didn't allow me to post this as a comment, so here is it as an answer.

See lemma 2.2 in this paper: https://arxiv.org/pdf/1306.2385.pdf I found that quite accessible and readable.

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  • $\begingroup$ Hmm. I see @jflipp. If you don't mind can I ask you a question? $\endgroup$ Commented Jan 27, 2020 at 10:03
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    $\begingroup$ Please consider writing at least the general idea of the lemma (and possibly the proof) in this answer post itself. As it stands now, if arxiv for some reason is down, or decides to move the paper to a different address on their servers, this answer becomes completely useless. Links can be terrific supplements to answers, but they are generally considered bad answers by themselves. $\endgroup$
    – Arthur
    Commented Jan 27, 2020 at 10:06

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