Basic question about finding a subgroup 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 =$$\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}
 .
$$
 A: 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.
A: 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...
A: No.  What about $\begin{pmatrix}1 & 7\\7 & 50\end{pmatrix}$?
