Abstract Algebra: mapping composition

iceybloop

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The problem asked is in three parts:
For each element of a in the group G, definte Ta: G -> G such that Ta(x) = xa-1 for all x in G

1) Prove Ta is a permutation on G.
(I believe I did this part by showing Ta(x) = Ta(y) implies x = y (1-1). To show whether it is onto, letting y be in G and solve Ta(x) = y for x.

2) Prove that G' = {Ta for all a in G} is a group with respect to mapping composition
I wrote that if Ta is a permutation (I found that it was in the first part), then it is a subset of S, so I need to show it is a subgroup. Is this a fine approach?

3) Define Phi: G ->G' such that Phi(a) = Ta. Determine whether Phi is always an isomorphism.
For this part, I'm trying to show Phi is a homomorphism, 1-1, and onto. I know I could use the kernel to show it is 1-1 as well.

Any help would be appreciated.
 
What definition are you using for "permutation" operators? Are they really defined as "one-to-one and onto mappings", or is there something else to them?
 
I don't remember anything else, and I'm not seeing anything else in my notes. It just reads that to show a permutation, one needs to prove 1-1 and onto.
 
The problem asked is in three parts:
For each element of a in the group G, definte Ta: G -> G such that Ta(x) = xa-1 for all x in G
2) Prove that G' = {Ta for all a in G} is a group with respect to mapping composition
I wrote that if Ta is a permutation (I found that it was in the first part), then it is a subset of S, so I need to show it is a subgroup. Is this a fine approach?.
You are correct about the definition of permutation. But I have concerns about how you view the above.
You are asked to show that \(\displaystyle G'\) itself is a group. It is not a subgroup of anything, (btw, what the heck is S?)
The elements of \(\displaystyle G'\) are all the permutations on the group \(\displaystyle G\). The operation is function composition.
Show closure, associatitivey, identity, & inverse. Inverse is given by the elements being onto-to-one onto functions,

Again do you see that \(\displaystyle G~\&~G'\) are different groups with different operations? That is important for part (3).
 
You are correct about the definition of permutation. But I have concerns about how you view the above.You are asked to show that \(\displaystyle G'\) itself is a group. It is not a subgroup of anything, (btw, what the heck is S?)
I see what I did wrong there, and never mind what S is. I was confused.
The elements of \(\displaystyle G'\) are all the permutations on the group \(\displaystyle G\). The operation is function composition.Show closure, associativity, identity, & inverse. Inverse is given by the elements being onto-to-one onto functions,Again do you see that \(\displaystyle G~\&~G'\) are different groups with different operations? That is important for part (3).
Vaguely. Thanks for the help!
 
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