Additional Group Theory Issues

[ZTV]

I really don't get this group theory stuff at all. These should be simple questions, but alas not...

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.

 

[Focus]

For the first question take two elements of the C(a), say x and y, and show that x*y is in C(a), i.e. (x*y)*a=a*(x*y)

Second question, you need to find the Automorphisms on Z3 which you have. Aut is a group under composition. Label the two automorphisms you have and make an isomorphism (which is pretty obvious) to Z2. The problem you are having is that you got the wrong impression what Aut is. It is a set of functions, so Aut(Z3) = {[0][1][2] , [0][2][1]} is not true. It consists of two functions, one of which is the identity, and the other one swaps [1] and [2].

 

[HallsofIvy]

I really don't get this group theory stuff at all. These should be simple questions, but alas not...

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?

~~~~
You are asked to prove that C(S) is closed with respect to *. That means "if a and b are both in C(S) then a*b is in C(S)". And, of course, you need to use the definition of C(S). If a and b are such that a*x= x*a and b*x= x*b for all x in S, what can you say about (a*b)*x and x*(b*a)? (Note that a*x and x*b are also in S. Do you know why?)

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]}

Am I to assume that "[0][1][2]" is the automorphism that maps 0 to 0, 1 to 1 and 2 to 2 while "[0][2][1]" is the automorphis that maps 0 to 0, 1 to 2, and 2 to 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.

Yes, that last statement is not true. One of the things you should have learned about "isomorphisms" is that an isomorphism always maps "identity" to "identity". [0][1][2] is obviously the identity function in Aut(Z3) so it must be mapped to the identity of Z2. There is only one other element left in Aut(Z3) and one left in Z2! You should be able to write down the only possible isomorphism from Aut(Z3) to Z2.

 

[ZTV]

So what are the elements of Aut(Z3) then?

Functions f that map Z3->Z3 such that fZ3 = Z3 ?

If this is the case...

I now have aut(Z3) = {fa fb} and Z2 = {[0][1]}

I'm trying to show that they are homomorpic/operation preserving.

I have to define a function y that maps
y: Aut(Z3) -> Z2 is this correct? If so, how do I show it is operation preserving?

Once I have that it is homomorphc, the fact that it is one-to-one and onto is trivial because fa -> [0] if it is homomorphic, so fb must go to [1] hence one-to-one and onto. Is this correct also?

Thanks

 

[Focus]

So what are the elements of Aut(Z3) then?

Functions f that map Z3->Z3 such that fZ3 = Z3 ?

If this is the case...

I now have aut(Z3) = {fa fb} and Z2 = {[0][1]}

I'm trying to show that they are homomorpic/operation preserving.

I have to define a function y that maps
y: Aut(Z3) -> Z2 is this correct? If so, how do I show it is operation preserving?

Once I have that it is homomorphc, the fact that it is one-to-one and onto is trivial because fa -> [0] if it is homomorphic, so fb must go to [1] hence one-to-one and onto. Is this correct also?

Thanks

Yes, mind you fa is the identity map. If you have shown it to be a homomorphism then it preserves the operation (which is what homomorphisms are). The bijection is clear, homomorphism shouldn't be too hard to prove.

As a general rule of thumb you want to specify how y maps, its the map as you said that takes fa to [0] and fb to [1].

Think of homomorphism as structure preserving, if f is a homomorphism then f(ab)=f(a)f(b), so they practically have the same operation. Isomorphism tells you that you just essentially relabeled your elements, because it is bijective and operation preserving.