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

1. walters Group Title

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

0 is a binary operation not zero

3. walters Group Title

@experimentX

4. experimentX Group Title

use \\ in latex for new line

5. walters Group Title

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

6. experimentX Group Title

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

7. experimentX Group Title

hmm ... f1 is the identity element.

8. walters Group Title

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 Group Title

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

10. walters Group Title

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

11. walters Group Title

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

12. experimentX Group Title

huh?? how did you do that??

13. walters Group Title

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

14. experimentX Group Title

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

15. walters Group Title

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

16. experimentX Group Title

isn't |dw:1362683941681:dw|

17. walters Group Title

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 Group Title

and we know that identity element is unique

19. experimentX Group Title

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

20. walters Group Title

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 Group Title

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

22. experimentX Group Title

|dw:1362684807109:dw|

23. experimentX Group Title

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

24. experimentX Group Title

|dw:1362685396384:dw|

25. experimentX Group Title

looks like i made some mistake.

26. walters Group Title

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

27. walters Group Title

actually i mean how to write them

28. experimentX Group Title

could you check those stuffs?

29. experimentX Group Title

if inverse exist, then it must be unique.

30. walters Group Title

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

31. experimentX Group Title

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

32. walters Group Title

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

33. walters Group Title

where e is the identity we have found

34. walters Group Title

i think this is not a group

35. experimentX Group Title

why do you think it's not a group?

36. walters Group Title

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 Group Title

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 Group Title

ok

39. experimentX Group Title

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

40. walters Group Title

yes it holds

41. walters Group Title

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 Group Title

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

43. experimentX Group Title

this is a group.

44. walters Group Title

can u show them pls

45. experimentX Group Title

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 Group Title

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

47. experimentX Group Title

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

48. walters Group Title

ok now i see it is a group

49. experimentX Group Title

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