In group $ GL(2,R) $ I have subgroup H, generated by 2 elements: $ a=\begin{pmatrix} 1 & 2 \\ 0 & 1\\ \end{pmatrix} $ , $ b=\begin{pmatrix} 1 & 3 \\ 0 & 1\\ \end{pmatrix} $
I have to find group of automorphisms $ Aut H $
So, I have found out that multiplication of this matrix always gives:
$ \begin{pmatrix} 1 & n \\ 0 & 1\\ \end{pmatrix} $ $ \begin{pmatrix} 1 & m \\ 0 & 1\\ \end{pmatrix} $ = $ \begin{pmatrix} 1 & n+m \\ 0 & 1\\ \end{pmatrix} $
Order of matrix gives $ \begin{pmatrix} 1 & n*ord \\ 0 & 1\\ \end{pmatrix} $
Am I right that subgroup H is infinite cyclic group? And the only one automorphims is $Z2$ ? I dont get this stuff about automorhisms, to be honest.
$H$ is infinite cyclic, so isomorphic to $\Bbb Z$. There are only two automorphisms of $\Bbb Z$.
Note that $\Bbb Z$ has two generators, $1$ and $-1$. How about sending $1$ to $-1$?
So, in $H$, that would mean $\begin{pmatrix}1&1\\0&1\end{pmatrix}\to\begin{pmatrix}1&-1\\0&1\end{pmatrix}$.