# Math Help - Group - How to transform

1. ## Group - How to transform

Hello

I have problems forming a term.

The exercise is:

$\text{Let G be a group and } x,y,z,u \in G$

$\mathrm{Z\kern-.3em\raise-0.5ex\hbox{Z}}: ~ \left(x \left( \left( \left( y^{-1} \left( x^{-1} \cdot z \right) \right) ^{-1} \cdot u \right) \cdot \left( y \cdot u \right)^{-1} \right) ^{-1} \right) = z$

I kwon that I have to show that:

i.) Associative Law

ii.) Identity Element

iii.) Inverse Element

If I look the term it is clear that I have to form it such that I only have z=z and the other elements x,y,u should be transformed into the identity because of their inverse elements.

I don't know how to form this concretely

Thanks for help
Greetings

2. That's how you use your law's properties:

$\left(x \left(\left( \left( y^{-1} \left( x^{-1} \cdot z \right) \right) ^{-1} \cdot u \right) \cdot \left( y \cdot u \right)^{-1} \right) ^{-1} \right)$
$= \left(x \left(\left( \left( y^{-1} \cdot x^{-1} \cdot z \right) ^{-1} \cdot u \right) \cdot \left( y \cdot u \right)^{-1} \right) ^{-1} \right)$ associativity
$= \left(x \left(\left( \left( z^{-1} \cdot x \cdot y \right) \cdot u \right) \cdot \left( y \cdot u \right)^{-1} \right) ^{-1} \right)$ inverse formula
$= \left(x \left(\left( z^{-1} \cdot x \cdot y \cdot u \right) \cdot \left( y \cdot u \right)^{-1} \right) ^{-1} \right)$ associativity
$= \left(x \left(\left( z^{-1} \cdot x \cdot y \cdot u \right) \cdot \left( u^{-1} \cdot y^{-1} \right) \right) ^{-1} \right)$ inverse formula
$= \left(x \left( z^{-1} \cdot x \cdot y \cdot u \cdot u^{-1} \cdot y^{-1} \right) ^{-1} \right)$ associativity
$= \left(x \left( z^{-1} \cdot x \cdot y \cdot y^{-1} \right) ^{-1} \right)$ associativity and inverse
$= \left(x \left( z^{-1} \cdot x \right) ^{-1} \right)$ associativity and inverse
$= \left(x \left( x^{-1} \cdot z \right) \right)$ inverse formula
$= \left(x \cdot x^{-1} \cdot z \right)$ associativity
$= z$ associativity and inverse

where "inverse formula" is $(a_{1}...a_{n})^{-1}=a_{n}^{-1}...a_{1}^{-1}$

3. Thank you very much

That helps really a lot. Now I know how to show it in the future.