Is this lie algebra isomorphic to $\mathfrak{sl}(2,\mathbb R)$? I have a lie group $G$, its (real? it was never mentioned in the text/context) lie algebra $\mathfrak {g}$ and three linear independent (over $\mathbb {R}$, but also over $\mathbb {C}$) elements $x,y,z\in \mathfrak{g}$ such that $[x,y]=az, [z,x]=bx, [z,y]=-by$ for constants $a<0<b$.
Is the lie subalgebra generated by $x,y,z$ isomorphic to $\mathfrak{sl}(2,\mathbb R)$?
It seems very similar. I want to take $u=\lambda x, v=\lambda y, w=\mu z$ and say that for a suitable $\lambda, \mu$ those elements satisfy the relations of $\mathfrak{sl}(2,\mathbb R)$. Indeed, if we demand $[u,v]=w, [w,u]=2u, [w,v]=-2v$ we get the equations $\lambda^2a=\mu, \lambda\mu b=2\lambda$ which turn as $\mu=\frac{2}{b}, \lambda^2=\frac{2}{ab}<0$. So if I would allow imaginary constants, I'm done. Problem is I'm not certain that this is allowed (or to be accurate, I'm pretty sure I am not). Yet, even if I'm not allowed, this doesn't prove that they are not isomorphic.
Any comments or different approaches are welcome. 
 A: Responding to the continuation of the question in the comments: up to isomorphism, there are two real Lie algebras whose complexifications are (isomorphic to) $\mathfrak{sl}(2,\mathbb R)$ and $\mathfrak{su}(2)$, attached to real Lie groups $SL_2(\mathbb R)$ and $SU(2)$. On the former, the Killing form has indefinite signature, but on the latter the Killing form is definite.
The problem (in your comment and in @Ennar's answer) of needing to take a square root of a negative number to put your Lie algebra in isomorphism with $\mathfrak{sl}(2,\mathbb R)$ strongly suggests to me (without computing anything) that your Lie algebra is isomorphic to $\mathfrak{su}(2)$, not $\mathfrak{sl}(2,\mathbb R)$. Yes, their complexifications are isomorphic, but this certainly does not imply that the real Lie groups are locally isomorphic.
A: If you change $v = -\lambda y$, then you get the following:
$$[u,v]=[\lambda x,-\lambda y] = -\lambda^2[x,y] = -\lambda^2az = -\frac{\lambda^2 a}{\mu}w$$
$$[w,u]=[\mu z,\lambda x] = \mu\lambda[z,x] = \mu\lambda bx = \mu b u$$
$$[w,v]=[\mu z,-\lambda y] = -\mu\lambda[z,y] = -\mu\lambda (-by) = -\mu b v$$
and thus, let $\mu = \frac 2b$, $\lambda = \sqrt{-\frac 2{ab}}$.
A: I think that your line of argument is slightly too complicated. If you first define $\tilde z=(2/b)z$, then you have $[\tilde z,x]=2x$ and $[\tilde z,y]=-2y$. These just mean that $x$ and $y$ are eigenvectors for $ad(z)$, so they remain the same if you multiply $x$ and/or $y$ by any non-zero factor. The third relation is $[x,y]=az=(ab/2)\tilde z$, so if you put $\tilde x=(2/ab)x$, then $\tilde x$, $y$, and $\tilde z$ form a standard $\mathfrak{sl}_2$-basis. 
