Existence of an inverse when verifying whether $\mathbb R^3$ forms a group under this operation 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? 
 A: 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.  
A: 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.
A: 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 $y_2=-y_1+x_1z_1$.

