Showing a subset of $\text{SL}(2,Z)$ forms a group Show that
$$\left\{\pmatrix{a & b \\ c & d}\in\text{SL}(2,\mathbb{Z})\ \bigg|\ c\equiv 0\bmod{n} \right\}$$
forms group under multiplication.
As a beginner in group theory how to deal with this question?
 A: Hint: To prove matrix in $SL(2,\mathbb Z)$ with $c=kn$ for $k\in \mathbb Z$ form a subgroup of $SL(2,\mathbb Z)$,  you need to show that multiplication is closed, and the inverse is closed. That is:
(1)  $\pmatrix{ a& b\\\ kn& d}$  $\pmatrix{ a'& b'\\\ k'n& d'}$ = $\pmatrix{ a''& b''\\\ k''n& d''}$ for some $a'', b''. k'', d''\in \mathbb Z$
(2) $\pmatrix{ a& b\\\ kn& d}^{-1}=$$\pmatrix{ *& *\\\ -kn& *}$ $\in SL(2,\mathbb Z)$
A: Let 
$$G = \left\{\pmatrix{a & b \\ c & d} \in\text{SL}_2(\mathbb{Z})\ \bigg|\ c\equiv 0\bmod{n}\right\}.$$
Then, any $A,B\in G$ will look like this:
$$A = \pmatrix{a & b \\ k_1n & d},\quad\quad B = \pmatrix{x & y \\ k_2n & z}.$$
Their product is:
$$AB=\pmatrix{a & b \\ k_1n & d}\pmatrix{x & y \\ k_2n & z} = \pmatrix{ax+bk_1n & ay+bz \\ xk_1n+dk_2n & yk_1n+dz} = \pmatrix{ax+bk_1n & ay+bz \\ (xk_1+dk_2)n & yk_1n+dz} \in G,$$
since $(xk_1+dk_2)n\equiv 0\bmod{n}.$ Since $A\in\text{SL}_2(\mathbb{Z})$, it is invertible, and its inverse is equal to:
$$A^{-1} = \frac{1}{\det(A)}\pmatrix{d & -b \\ -k_1n & a} = \pmatrix{d & -b \\ -k_1n & a},$$
since $A\in\text{SL}_2(\mathbb{Z})$ implies $\det(A)=1$. This is just the standard inverse of an invertible $2\times 2$ matrix. Since $-k_1n\equiv 0\bmod{n}$, it follows that $A^{-1}\in G$ also. Associativity is inherited from $\text{SL}_2(\mathbb{Z})$, as is the identity. So $G$ is a subgroup of $\text{SL}_2(\mathbb{Z})$.
Edit: For good measure:
$$A^{-1} = \frac{1}{ad-bk_1n}\pmatrix{d & -b \\ -k_1n & a},$$
so:
$$AA^{-1} = \frac{1}{ad-bk_1n}\pmatrix{a & b \\k_1n & d}\pmatrix{d & -b \\ -k_1n & a} = \frac{1}{ad-bk_1n}\pmatrix{ad-bk_1n & -ba+ba \\ k_1nd-dk_1n & -bk_1n+ad} = \pmatrix{1 & 0 \\ 0 & 1}.$$
