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Is this the only subgroup of $GL_2(\Bbb R)$ isomorphic to $\Bbb R$? $$ \left\{\begin{bmatrix}1&a \\ 0&1\end{bmatrix} : a \in \Bbb R\right\} $$ If not, can we describe all such subgroups?

(motivated by this question)

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    $\begingroup$ Hint: Conjugation. $\endgroup$
    – Lee Mosher
    Commented Apr 24, 2020 at 15:03
  • $\begingroup$ Up to conjugation, they are all upper triangular, not necessarily strictly. $\endgroup$ Commented Apr 25, 2020 at 1:16
  • $\begingroup$ You should clarify if you want these subgroups to be isomorphic to ${\mathbb R}$ as topological groups (with subspace topology) or as abstract groups. The answers are different since there are many more subgroups isomorphic to ${\mathbb R}$ as abstract groups. $\endgroup$ Commented Apr 25, 2020 at 4:13

1 Answer 1

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Well, at the very least there is $$ \left\{\begin{bmatrix}1&0 \\ a&1\end{bmatrix} : a \in \Bbb R\right\} $$

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    $\begingroup$ Sorry, I thought that the question is: Find all subgroup of $GL(2)$. $\endgroup$
    – C.F.G
    Commented Apr 24, 2020 at 16:00
  • $\begingroup$ What about $\left\{\begin{bmatrix}1+a^2&0 \\ a&2+a^2\end{bmatrix} : a \in \Bbb R\right\}$? $\endgroup$
    – C.F.G
    Commented Apr 24, 2020 at 16:03
  • $\begingroup$ Also $$\left\{\pmatrix{a&0\\0&1}:a>0\right\}?$$ $\endgroup$ Commented Apr 24, 2020 at 16:08
  • $\begingroup$ Those are all also valid suggestions. $\endgroup$
    – Arthur
    Commented Apr 24, 2020 at 17:02

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