One parameter subgroups of group of isometries of plane I'm trying to work out what the one-parameter subgroups of the Lie group given by the set of isometries of the form $x \mapsto Ax + b$ are. I know that $A$ has to be orthogonal, but beyond there I'm pretty stuck.
I know that if $\phi : \mathbb{R} \to G$ is a one-parameter subgroup with $\phi(t)(x) = A(t)(x) + b(t)$ that then $A(t+s) = A(t)A(s) = A(s)A(t)$ and then $b(t+s) = A(t)b(s) + b(t) = A(s)b(t) + b(s)$, but I'm not sure what I should be doing after this.
Thanks for any help!
 A: I'm going to use complex numbers because it will make notation simpler.
First thing, note that $A(t) \in SO(2)$, because it must be in the same path component as the identity in $O(2)$. Thus we can write $A(t) = e^{i\theta(t)}$ for some function $\theta: \mathbb{R} \to \mathbb{R}$. Take your first equation and differentiate with respect to $s$ at $s = 0$.
$$A'(t) = A'(0)A(t) \implies \theta'(t) = \theta'(0) = k$$
for some constant $k \in \mathbb{R}$. Then, with the initial condition $A(0) = 1$, we have $A(t) = e^{ikt}$. Now we take a derivative of the second equation.
$$b'(t) = A'(0)b(t) + b'(0) = ikb(t) + b'(0)$$
That differential equation can be solved using an integrating factor. See http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx if you need help with this step. After solving with initial condition $b(0)=0$ (and renaming some constants), you should get
$$b(t) = z(1-e^{ikt})$$
for some constant $z \in \mathbb{C}$. For any values of $k$ and $z$ you get a different one parameter subgroup.
