TITLE: Simple subgroup proof, would love some advice
QUESTION [5 upvotes]: Let $\mathrm{GL}_2(\mathbb R)$ be the group of invertible $2\times 2$ matrices with real entries.
Consider $H$ which is the subset of matrices of the form
$$A=\begin{bmatrix}1 & b \\ 0 &1\end{bmatrix}$$
with $b$ an integer. Is $H$ a subgroup of $\mathrm{GL}_2(\mathbb R)$? Explain your answer.
well when $b = 0$ we have the identity, and $\det |A| = 1$ so it is invertible and the inverse is simply  $$\begin{bmatrix}1 & -b \\ 0 &1\end{bmatrix}$$ for all $b$ in the integers. I seem a little lost on closure. Clearly $$\begin{bmatrix}1 & -b \\ 0 &1\end{bmatrix}$$ multiplied by $$\begin{bmatrix}1 & a \\ 0 &1\end{bmatrix}$$ where $a$ is in the integers yields
$$\begin{bmatrix}1 & -b+a \\ 0 &1\end{bmatrix}$$ and $a - b$ where $a$ and $b$ are integers is an integer. So we got non empty and we got $A^{-1} B$.
So we're done?

REPLY [2 votes]: Pretty much. Use any $b \in \mathbb{R}$, not $-b$ to underline it's a generic real number, and as in your calculation with $-b$, you get a product matrix $$\begin{pmatrix} 1 & a+b \\ 0 & 1 \\ \end{pmatrix} := C.$$ As $a+b \in \mathbb{R}$, $C \in H$, so $H$ is closed.