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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?

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You're done, yes. –  Andreas Caranti Feb 7 '13 at 13:34
    
Oh ok, thank you i assumed i did it wrong because it was easy :) –  Faust7 Feb 7 '13 at 13:38
    
When checking if a subset $H$ of a group $G$ is a subgroup, the same three things needs to be checked every time: 1) that the identity element is in $H$. 2) That the inverse of any element of $H$ is in $H$. 3) That the product of any two elements of $H$ is contained in $H$. Most of the time these three points are very easy to check. At least in exercises given in books. –  Arthur Feb 7 '13 at 13:43
    
Yes you are done. Just a comment: once you know $H$ is nonempty, it is enough just to show the last part, which is technically not closure, but for groups even stronger than closure as you indicate in your second last sentence. You have done more than you need to. –  Barbara Osofsky Feb 7 '13 at 13:46
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up vote 2 down vote accepted

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.

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