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I have the matrix A = $$\begin{pmatrix} a & b\\ c & d \end{pmatrix} \qquad a \equiv d \equiv 1 \pmod{7}, \quad b \equiv c \equiv 0 \pmod{7} \, . $$ I need to show it is a subgroup of $SL_2(\mathbb{Z})$

Am I correct in thinking that the only values for A =$$\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} . $$

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    $\begingroup$ That is not correct. $a,b,c,d\in \Bbb Z$, they need not be restrained to $\{0,1,2,3,4,5,6\}$. $\endgroup$ – Fimpellizieri Nov 8 at 4:53
  • $\begingroup$ To show it is a subgroup, you need to show its elements belong to the group, that it contains the group's identity, and that it is closed under the group operation (in this case, matrix multiplication). In other words, you need to show that if you take two matrices with the properties you described, their product also has these properties. $\endgroup$ – Fimpellizieri Nov 8 at 4:56
  • $\begingroup$ So, I am supposed to solve this for the general case rather than looking for values? I am a beginner learner btw $\endgroup$ – user975 Nov 8 at 4:58
  • $\begingroup$ Yes. You can sometimes 'look for values', when you are dealing with things that are finite (I'm not saying that in those cases this is the best approach, only that it may be viable). But in situations like this one, there are infinitely many matrices that satisfy the properties you described, so you must handle the problem in generality. $\endgroup$ – Fimpellizieri Nov 8 at 5:00
  • $\begingroup$ Thank you for the response. I was looking for all possible values that would give me a determinant of 1. $\endgroup$ – user975 Nov 8 at 5:04
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The natural projection $SL_2(\mathbb{Z}) \to SL_2(\mathbb{Z_7})$ is a group homomorphism.

The set in question is its kernel, and so is a (normal) subgroup.

Moreover, since the domain is infinite and the codomain is finite, the kernel cannot be trivial.

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  • $\begingroup$ This is the "right" answer, in my opinion. $\endgroup$ – Dietrich Burde Nov 8 at 10:22
  • $\begingroup$ @DietrichBurde This is, as often in algebra, the most proper answer with the least amount of information gathered from it :-) $\endgroup$ – yo' Nov 8 at 15:47
  • $\begingroup$ @yo', will testing the subgroup conditions explicitly give you any useful information? Besides practice, of course. $\endgroup$ – lhf Nov 8 at 15:51
  • $\begingroup$ @lhf Ah well, it's closer to seeing what actually falls into the subgroup, by testing $(7A+1)(7D+1)-(7B)(7C)=1$ $\endgroup$ – yo' Nov 8 at 16:01
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You don't need to know this. More details near the bottom of this answer.

What about $\displaystyle \begin{pmatrix} 8 & 7 \\ 14 & -6 \end{pmatrix}$? Well, that matrix satisfies the congruences, but doesn't have determinant $1$.

Are there any matrices satisfying the congruences that also satisfy the determinant condition? Any matrix satisfying the congruences is $$ M(a,b,c,d) = \begin{pmatrix} 7a + 1 & 7b \\ 7c & 7d + 1 \end{pmatrix} $$ for some choice of $(a,b,c,d)$. Then \begin{align*} 1 &= \det M(a,b,c,d) \\ &= (7a+1)(7d+1) - (7b)(7c) \\ &= 49(ad-bc) + 7a + 7d + 1 \text{,} \end{align*} so $$ d = \frac{7 b c - a}{7a + 1} \text{.} $$

Therefore, $$ M\left(a,b,c,\frac{7 b c - a}{7a + 1}\right) = \begin{pmatrix} 7a + 1 & 7b \\ 7c & 7\frac{7 b c - a}{7a + 1} + 1 \end{pmatrix} $$ is the set of matrices you are interested in.

But you don't need to know this. You only have to check that the group axioms are satisfied, and that doesn't require knowing all the elements of the set in detail, only being able to multiply generic pairs of such elements...

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  • $\begingroup$ Thank you very much Eric and all other responses. Much appreciated. $\endgroup$ – user975 Nov 8 at 5:59
  • $\begingroup$ But when does $7a + 1$ divide $7 b c - a$? $\endgroup$ – lhf Nov 8 at 9:35
  • $\begingroup$ It's better to solve for $b$ and get $b = (7 a d + a + d)/(7 c)$; the condition is then simply that $7$ divides $a+d$. $\endgroup$ – lhf Nov 8 at 9:42
  • $\begingroup$ @lhf : Since my main point is that asking for a details about the set is the wrong way to go about showing that it forms a subgroup, optimizing the expression is furtherance of the wrong application of effort. $\endgroup$ – Eric Towers Nov 8 at 15:04
  • $\begingroup$ @EricTowers, good point. $\endgroup$ – lhf Nov 8 at 15:04
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No. What about $\begin{pmatrix}1 & 7\\7 & 50\end{pmatrix}$?

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