I have this group presentation $G = \langle a,b | ba = a^{-1}b\rangle$. I'm wondering if this group is solvable or not. I tried to show that this group is simple, because if it is, then it can not be solvable as it is an infinite group. I'm not sure if it even is simple, as I do not have much experience with group presentations and proving that (infinite) groups are simple. Is there anyone who can tell me if this group is solvable and if so, if this strategy is will work.
Thank you in advance.
Consider the infinite cyclic group $G=\langle a\rangle$, this admits an automorphism that sends $a\to a^{-1}$ and this is in fact the only nontivial automorphism it admits. Let $H= \langle b\rangle$ be another copy. There is a group morphism $\eta:H\to {\rm Aut}\,G$ that sends $b$ to the automorphism $a\mapsto a^{-1}$. This is just the projection map $\Bbb Z\to\Bbb Z_2={\rm Aut}\,\Bbb Z$. Then your group is the semidirect product $K=G\rtimes H$ under this action, and in particular $G$ is normal, and $K/G\simeq H$, so there is a solvable series $1\lhd G\lhd K$, for $G$ and $K/G$ are both infinite cyclic, in particular abelian. In general if $A,B$ are abelian groups then any semidirect product of $A$ with $B$ is solvable by the same reasoning.