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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

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

$=(z_1x_1y_1+z_1x_2y_3+z_3x_1y_2+z_3x_2y_4,z_2x_1y_1+z_2x_2y_3+z_4x_1y_2+z_4x_2y_4,z_1x_3y_1+z_1x_4y_3+z_3x_3y_2+z_3x_4y_4,z_2x_3y_1+z_2x_4y_3+z_4x_3y_2+z_4x_4y_4)$

and

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

$=(z_1x_1y_1+z_1x_2y_3+z_3x_1y_2+z_3x_2y_4,z_2x_1y_1+z_2x_2y_3+z_4x_1y_2+z_4x_2y_4,z_1x_3y_1+z_1x_4y_3+z_3x_3y_2+z_3x_4y_4,z_2x_3y_1+z_2x_4y_3+z_4x_3y_2+z_4x_4y_4)$

Since they are equal, the property for associativity holds.

Proof of the existence of an identity

For the identity, we want

$(x_1,x_2,x_3,x_4)*(e_1,e_2,e_3,e_4)=(x_1,x_2,x_3,x_4)$

Then we have

$(x_11+x_20,x_10+x_21,x_31+x_40,x_30+x_41)$

$(e_1,e_2,e_3,e_4)=(1,0,0,1)$

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?

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1 Answer 1

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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.

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  • $\begingroup$ 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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