How do I approach finding the inverse of a group? The question is as follows:
$G= (x_1,y_1,z_1) \in \mathbb{R}^3$
Is $(G,*)$ a group? Where multiplication is defined as $(x_1,y_1,z_1)*(x_2,y_2,z_2) = (x_1 + x_2 ,y_1 + x_1z_2+y_2,z_1 + z_2)$
I was able to show non-empty, associativity, commutativity and the identity element, $(0,0,0)$. However I cannot come up with an inverse. Is there no inverse for this question?
Is this really a group? If $e=(a,b,c)$ identity element than we have $$x_1+a=x_1\implies a=0$$ $$y_1+x_1z_1 +b = y_1\implies b= -x_1z_1$$ $$ z_1+c = z_1\implies c=0$$
So $a$ and $c$ are defined, but $b$ is not unique, it depends on $x_1$ and $z_1$. This should not be.
We have a bijection between $\mathbb R^3$ and unipotent upper-triangular matrices by sending $$(x,y,z) \mapsto \begin{pmatrix}1&x&y\\0&1&z\\0&0&1\end{pmatrix}$$ You can verify that under this bijection, $*$ is just matrix multiplication. The inverse of an invertible upper-triangular matrix is upper-triangular(1), and the inverse of a unipotent matrix is unipotent.
(1) An invertible filtered homomorphism on a finite-dimensional filtered vector space is a filtered isomorphism, or see Inverse of an invertible upper triangular matrix of order 3 if you like computations.
For the inverse, you can solve directly $$\left\{\begin{align}&x_1+x_2=0\\ &y_1+x_1z_2+y_2=0\\ &z_1+z_2=0 \end{align}\right.$$ for $x_2$, then $z_2$, then $y_2$.
We find $x_2=-x_1$, $z_2=-z_1$ and finally $y_2=-y_1+x_1z_1$.