The group to be determined is defined as follows:

$\{x\in\Bbb{Z^4}:x_1x_4=1+x_2x_3\}$ with $(x,y)\mapsto(x_1y_1+x_2y_3,x_1y_2+x_2y_4,x_3y_1+x_4y_3,x_3y_2+x_4y_4)$

$*$ denotes the operation. We have to prove

i) Associativity;

ii) Existence of identity;

iii) Existence of inverses.

Here is what I have so far:

Proof of associativity

We have to prove that

$(x_1,x_2,x_3,x_4)*((y_1,y_2,y_3,y_4)*(z_1,z_2,z_3,z_4)) = ((x_1,x_2,x_3,x_4)*(y_1,y_2,y_3,y_4))*(z_1,z_2,z_3,z_4)$

Then we have






Since they are equal, the property for associativity holds.

Proof of the existence of an identity

For the identity, we want


Then we have



The property of the existence of an identity holds also.

I have solved the problem until here. I am stuck with the proof of the existince of inverses. Could you please give me some hints?


1 Answer 1


This group is just the group of 2x2 matrices with integer coefficients and determinant $1$.

So the inverse of $(a,b,c,d)$ is $(d, -b, -c, a )$.

The easiest way to do compute this is to write down 4 linear equations with 4 indeterminates. So if you want to find the inverse element of $(a,b,c,d)$ write down the linear system $$ \left\{ \begin{matrix} ax_1 &&& +bx_3&& = 1 \\ &ax_2 & &&+bx_4 &= 0 \\ cx_1 && &+dx_3 &&= 0 \\ &cx_2 && &+dx_4& = 1 \end{matrix} \right. $$

and try to solve it.

  • $\begingroup$ Could you please give me some explanation on your solution? How can I get to $(d,-b,-c,a)$ when I have $(x_1,x_2,x_3,x_4)*(y_1,y_2,y_3,y_4)=(1,0,0,1)$? That is actually my problem. I do not know how to show that. $\endgroup$
    – Javiator
    Nov 19, 2014 at 17:30

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