Abstract Algebra help

Question

walters · Answer

$$T=\left\{ f _{1} (x)=x,f _{2}(x)=\frac{ 1 }{ 1-x },f _{3}(x)\frac{ x }{ x-1 },f _{4}(x)=\frac{ x-1 }{ x },f _{5}(x)=1-x,f _{6}(x)=\frac{ 1 }{ x }\right\}$$
show  that T is a group or not  <T,0>
 
i've shown that 0 is associative on T
can u pls help on showing the identity and the inverse

walters · Answer

0 is a binary operation not zero

walters · Answer

@experimentX

experimentx · Answer

use \ in latex for new line

walters · Answer

$$f _{6}(x)=\frac{ 1 }{ x }$$} this is the lst element

experimentx · Answer

$$
T=\left\{ f _{1} (x)=x,f _{2}(x)=\frac{ 1 }{ 1-x },f _{3}(x)\frac{ x }{ x-1 }, \ f _{4}(x)=\frac{ x-1 }{ x },f _{5}(x)=1-x,f _{6}(x)=\frac{ 1 }{ x }\right\} $$

experimentx · Answer

hmm ... f1 is the identity element.

walters · Answer

let  $$f _{7}(x) \in T$$ be the identity with respect to 0,then
$$(f _{7}0f _{1})(x)=(f _{1}0f _{7})(x)=f _{1}(x)$$

experimentx · Answer

unfortunately, the last element ... has inverse as itself.

walters · Answer

from there i did this 
$$(f _{7}0f _{1})(x)=f _{1}(x)$$ and $$(f _{1} 0f _{7})(x)=f _{1}(x)$$

walters · Answer

on the  first part i got $$f _{7}(x)=x$$
but the second part i don't know how to do it

experimentx · Answer

huh?? how did you do that??

walters · Answer

$$(f _{7}0f _{1})(x)=f _{1}(x)$$
then $$f _{7}(f _{1}(x))=f _{1}(x)$$
$$f _{7}(x)=x$$

experimentx · Answer

hmm .. what is f7, where is f7 ??

walters · Answer

i  let it to be element of T because by the definition  of identity we have 

e*a=a*e=a

experimentx · Answer

isn't |dw:1362683941681:dw|