I need to describe a one parameter group in $$ \text G= \biggl\lbrace \begin{pmatrix} x & y \\ 0 & 1 \\ \end{pmatrix} \in GL_2(\mathbb R)|x>0 \biggl\rbrace $$ There is a proposition that for every $A\in gl_n(\mathbb K)\ \gamma(t):=e^{tA}$ is a one parameter group, thus I need to find such A that $e^{tA} \in G $, but I have no idea how to find it.
A one parameter group in a matrix group G is a differentiable group-homomorphism $\gamma:(\mathbb R,+)\rightarrow G$.
$\forall A \in gl_n(\mathbb K) \ e^A=\sum_{n=0}^\infty \frac{A^n}{n!}$
Thank you.
Hint: If you write $e^{tA}=\begin{pmatrix} a(t) & b(t) \\ c(t) & d(t)\end{pmatrix}$ for smooth real valued functions $a(t)$,..., $d(t)$, then $A=\begin{pmatrix} a'(0) & b'(0) \\ c'(0) & d'(0)\end{pmatrix}$. This immediately shows that $e^{tA}\in G$ implies that the second row of $A$ consists of zeros only. But for such a matrix $A$, you can easily compute $e^{tA}$ explicitly to show that it has values in $G$.