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Abstract Algebra help

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\[T=\left\{ f _{1} (x)=x,f _{2}(x)=\frac{ 1 }{ 1-x },f _{3}(x)\frac{ x }{ x-1 },f _{4}(x)=\frac{ x-1 }{ x },f _{5}(x)=1-x,f _{6}(x)=\frac{ 1 }{ x }\right\}\] 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
0 is a binary operation not zero

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\[f _{6}(x)=\frac{ 1 }{ x }\]} this is the lst element
\[ T=\left\{ f _{1} (x)=x,f _{2}(x)=\frac{ 1 }{ 1-x },f _{3}(x)\frac{ x }{ x-1 }, \\ f _{4}(x)=\frac{ x-1 }{ x },f _{5}(x)=1-x,f _{6}(x)=\frac{ 1 }{ x }\right\} \]
hmm ... f1 is the identity element.
let \[f _{7}(x) \in T\] be the identity with respect to 0,then \[(f _{7}0f _{1})(x)=(f _{1}0f _{7})(x)=f _{1}(x)\]
unfortunately, the last element ... has inverse as itself.
from there i did this \[(f _{7}0f _{1})(x)=f _{1}(x)\] and \[(f _{1} 0f _{7})(x)=f _{1}(x)\]
on the first part i got \[f _{7}(x)=x\] but the second part i don't know how to do it
huh?? how did you do that??
\[(f _{7}0f _{1})(x)=f _{1}(x)\] then \[f _{7}(f _{1}(x))=f _{1}(x)\] \[f _{7}(x)=x\]
hmm .. what is f7, where is f7 ??
i let it to be element of T because by the definition of identity we have e*a=a*e=a
isn't |dw:1362683941681:dw|
i got that but i am not sure what if we use f7 and f2 it seems we will have another different answer
and we know that identity element is unique
there isn't any other identity element on the group.
if i am using f7 and f2 i have |dw:1362684306098:dw| how can i simplify it until i get x as my identity
I think you are not getting the meaning|dw:1362684677725:dw| of identity
hell ... the page refreshed and my work is gone.
looks like i made some mistake.
so if is like this how wat am i gonna say about the inverse
actually i mean how to write them
could you check those stuffs?
if inverse exist, then it must be unique.
the inverse not always is unique but the identity must always be unique
the identity is unique on this case ... f(x) = x is the only identity element.
the inverse is given by a*b=b*a=e
where e is the identity we have found
i think this is not a group
why do you think it's not a group?
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
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.
I am getting the same result as before have you verified associativity??
yes it holds
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
what a fool i've been ... f3, f5, f6 <-- they all are self inverse.
this is a group.
can u show them pls
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 \( \neq 1\)
for every operation this is closed. earliser I showed, f2 and f4 are inverses of each other.|dw:1362690024527:dw|
|dw:1362690149777:dw| these element are inverses of themselves.
ok now i see it is a group
most likely yes ... as always I am not sure.

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