Is this group index infinite? Let $G=\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$ where $a,b,c,d\in\mathbb{Z}$ and determinant = 1, be a group.
And let $H=\left(\begin{matrix}1&n\\0&1\end{matrix}\right)$ where ($n\in\mathbb{Z}$) be a subgroup of $G$.
What is the index $[G:H]$?

$H$ is clearly infinite so $G$ must be infinite so the index must be infinite. Does this make sense?
 A: Notice that $$\left(\begin{matrix}a&b\\c&d\end{matrix} \right) \left(\begin{matrix}1&n\\0&1\end{matrix}\right )= \left(\begin{matrix}a&an+b\\c&cn+d\end{matrix}\right).$$ Given that calculation we can see all the elements in the coset $gH$ have the same first column as $g$.
So to show that $H$ is infinite index we only have to show there are infinitely many possible first columns. Well $$l_n= \begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix}$$ give infinitely many different first columns and $l_n \in G$, so each $l_n H$ is a distinct coset. There are infinitely many, so $H$ has infinite index.
A: Each $k \in \mathbb{Z}$ yields a distinct coset
$$\begin{pmatrix}k & k-1 \\ 1 & 1\end{pmatrix}H = \left\{ \begin{pmatrix}k & nk + k - 1 \\ 1 & n+1\end{pmatrix} : n \in \mathbb{Z}\right\},$$
so $[G:H]$ is infinite.
A: Consider the "opposite" subgroup $K = \{ \left( \begin{array}{clcr} 1 & 0\\m & 1 \end{array} \right) : m \in \mathbb{Z} \}$. If $a,b \in K$ are different, then $a^{-1}b$ must lie outside $H$, since $H \cap K = \{I \}$ and we certainly have $a^{-1}b \in K$. Thus $\{ aH : a \in K \}$ is an infinite set of distinct cosets of $H$ in $G$ ( note that $K$ is infinite, just as $H$ is).
