- #1
ZTV
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I really don't get this group theory stuff at all. These should be simple questions, but alas not...
Assume that * is an associative operation on S and that a is an element of S.
Let C(a) = {x: x is an element of S and a*x = x*a}
Prove that C(a) is closed with respect to *
Unsure
To be perfectly honest, I don't understand the notation or anything. I have no clue where to start. I know that to prove closure I have to show that when g, h are elements of G that g*h is also an element of G. Does this mean I have to show that a*x is an element of S?
~~~~
Prove that Aut(Z3) is isomorphic to Z2 [Z3 and Z2 the group of modulo classes, eg. Z3: { [0][1][2]}
Once I've found Aut(Z3) which I think I've done, I need to show that Aut(Z3) is one-to-one, onto and operation preserving (homorphic)
I found Aut(Z3) to be {[0][1][2] , [0][2][1]}
With Aut(Z3) = {[0][1][2] , [0][2][1]}
and
Z2 = {[0][1]}
I can see how they are supposed to be isomorphic already.
I consider f: {Aut(Z3)) -> Z2}
I'm thinking that showing that they are one-to-one and onto may be trivial, but I'm not sure.
I also looked at f ([0][1][2]) = f ([0][2][1]) and proving [0]=[1] for one-to-one but this doesn't make sense because [0]=/=[1] ... hmmm... perhaps I'm being silly.
Homework Statement
Assume that * is an associative operation on S and that a is an element of S.
Let C(a) = {x: x is an element of S and a*x = x*a}
Prove that C(a) is closed with respect to *
Homework Equations
Unsure
The Attempt at a Solution
To be perfectly honest, I don't understand the notation or anything. I have no clue where to start. I know that to prove closure I have to show that when g, h are elements of G that g*h is also an element of G. Does this mean I have to show that a*x is an element of S?
~~~~
Homework Statement
Prove that Aut(Z3) is isomorphic to Z2 [Z3 and Z2 the group of modulo classes, eg. Z3: { [0][1][2]}
Homework Equations
Once I've found Aut(Z3) which I think I've done, I need to show that Aut(Z3) is one-to-one, onto and operation preserving (homorphic)
I found Aut(Z3) to be {[0][1][2] , [0][2][1]}
The Attempt at a Solution
With Aut(Z3) = {[0][1][2] , [0][2][1]}
and
Z2 = {[0][1]}
I can see how they are supposed to be isomorphic already.
I consider f: {Aut(Z3)) -> Z2}
I'm thinking that showing that they are one-to-one and onto may be trivial, but I'm not sure.
I also looked at f ([0][1][2]) = f ([0][2][1]) and proving [0]=[1] for one-to-one but this doesn't make sense because [0]=/=[1] ... hmmm... perhaps I'm being silly.