## walters Abstract Algebra help one year ago one year ago

1. walters

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

2. walters

0 is a binary operation not zero

3. walters

@experimentX

4. experimentX

use \\ in latex for new line

5. walters

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

6. experimentX

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

7. experimentX

hmm ... f1 is the identity element.

8. walters

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

9. experimentX

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

10. walters

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

11. walters

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

12. experimentX

huh?? how did you do that??

13. walters

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

14. experimentX

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

15. walters

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

16. experimentX

isn't |dw:1362683941681:dw|

17. walters

i got that but i am not sure what if we use f7 and f2 it seems we will have another different answer

18. walters

and we know that identity element is unique

19. experimentX

there isn't any other identity element on the group.

20. walters

if i am using f7 and f2 i have |dw:1362684306098:dw| how can i simplify it until i get x as my identity

21. experimentX

I think you are not getting the meaning|dw:1362684677725:dw| of identity

22. experimentX

|dw:1362684807109:dw|

23. experimentX

hell ... the page refreshed and my work is gone.

24. experimentX

|dw:1362685396384:dw|

25. experimentX

looks like i made some mistake.

26. walters

so if is like this how wat am i gonna say about the inverse

27. walters

actually i mean how to write them

28. experimentX

could you check those stuffs?

29. experimentX

if inverse exist, then it must be unique.

30. walters

the inverse not always is unique but the identity must always be unique

31. experimentX

the identity is unique on this case ... f(x) = x is the only identity element.

32. walters

the inverse is given by a*b=b*a=e

33. walters

where e is the identity we have found

34. walters

i think this is not a group

35. experimentX

why do you think it's not a group?

36. walters

the only way to get the inverse of a certain element u must multiply two elements and this does not clearly tell us the inverse of a function.As we know that for every function wen we use the operation combining the function and its inverse we must get the identity

37. experimentX

here, the operation is not multiplication ... but i think there is no unique inverser. But I am not sure ... let me do this seriously on copy.

38. walters

ok

39. experimentX

I am getting the same result as before have you verified associativity??

40. walters

yes it holds

41. walters

i think wat u did is similar wen trying to show that 0 is a binary operation ,because wat u have as the inverse i have them at my fisrt condition that one of checking indeed the our operation hold

42. experimentX

what a fool i've been ... f3, f5, f6 <-- they all are self inverse.

43. experimentX

this is a group.

44. walters

can u show them pls

45. experimentX

assuming associativity holds for all operations define your idenity by $$f_1(x) = x$$ ... in any case you won't get $$f_x f_y = f_x$$ if y $$\neq 1$$

46. experimentX

for every operation this is closed. earliser I showed, f2 and f4 are inverses of each other.|dw:1362690024527:dw|

47. experimentX

|dw:1362690149777:dw| these element are inverses of themselves.

48. walters

ok now i see it is a group

49. experimentX

most likely yes ... as always I am not sure.