1
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ $(x_1,x_2,x_3)*(0,0,0)=(x_1,y_1+x_1z_1,z_1)$, so $(0,0,0)$ is not an identity element. $\endgroup$ – Dietrich Burde May 27 '18 at 16:23
  • 1
    $\begingroup$ "inherited from $\Bbb{R}^3$" ? $\endgroup$ – Bumblebee May 27 '18 at 16:28
  • $\begingroup$ @DietrichBurde I made a typo, it is actually $y_1+x_1z_2$ so it should hold. $\endgroup$ – Safder May 27 '18 at 17:07
1
$\begingroup$

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 $y_2=-y_1+x_1z_1$.

$\endgroup$
  • $\begingroup$ This makes so much sense now. I wasn't thinking about it this way. Thank you! $\endgroup$ – Safder May 27 '18 at 18:40
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Sorry I actually made a typo in the question. It should be $ (x_1 + x_2 ,y_1 + x_1z_2+y_2,z_1 + z_2)$ $\endgroup$ – Safder May 27 '18 at 17:08
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.