Abstract Algebra help

- walters

Abstract Algebra help

- schrodinger

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- 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
i've shown that 0 is associative on T
can u pls help on showing the identity and the inverse

- walters

0 is a binary operation not zero

- walters

@experimentX

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- experimentX

use \\ in latex for new line

- walters

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

- experimentX

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

- experimentX

hmm ... f1 is the identity element.

- walters

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

- experimentX

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

- walters

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

- walters

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

- experimentX

huh?? how did you do that??

- walters

\[(f _{7}0f _{1})(x)=f _{1}(x)\]
then \[f _{7}(f _{1}(x))=f _{1}(x)\]
\[f _{7}(x)=x\]

- experimentX

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

- walters

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

- experimentX

isn't |dw:1362683941681:dw|

- walters

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

- walters

and we know that identity element is unique

- experimentX

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

- 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

- experimentX

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

- experimentX

|dw:1362684807109:dw|

- experimentX

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

- experimentX

|dw:1362685396384:dw|

- experimentX

looks like i made some mistake.

- walters

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

- walters

actually i mean how to write them

- experimentX

could you check those stuffs?

- experimentX

if inverse exist, then it must be unique.

- walters

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

- experimentX

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

- walters

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

- walters

where e is the identity we have found

- walters

i think this is not a group

- experimentX

why do you think it's not a group?

- 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

- 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.

- walters

ok

- experimentX

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

- walters

yes it holds

- 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

- experimentX

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

- experimentX

this is a group.

- walters

can u show them pls

- 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 \( \neq 1\)

- experimentX

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

- experimentX

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

- walters

ok now i see it is a group

- experimentX

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

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