How Do You Find Automorphism in a Group G?

[semidevil]

So an automorphism is a group G to itself.

So how do you prove that in general? Is it the same to proving an isomorphism?

So suppose they ask to find Aut(Z). what do I do?

Z is all the ingegers, from 1 to n...so how do prove that?

how do i start?

 

[Galileo]

Your sentences sound a bit confused, making it unclear as to what you are asking.

So an automorphism is a group G to itself.

An automorphism of a group G is a bijective homomorphism from the group G to itself.

So how do you prove that in general? Is it the same to proving an isomorphism?

What is there to prove? That is simply the definition of an automorphism.
If you are given a function and you wish to prove it is an automorphism, just see if it has all the properties of an automorphism.

So suppose they ask to find Aut(Z). what do I do?

Ok, so Aut(Z) is the set of all automorphisms on Z (the integers).
I don't think there's a general way to approach these kind of problems, but in any case: write down the conditions these automorphisms must have.
The trivial automorphism f(x)=x is one. I have a gut feeling it's also the only one. So see if you can try to prove that.

 

[gravenewworld]

An automorphism of a group G is just an isomorphism from G to itself. Just do the same thing you do with isomorphisms, that is show that the automorphism is 1-1, onto, and preserves operation.

 

[jcsd]

Obviously there exists an isomorphism between a group and itself so you needen't prove that, secondly there is only one isomorphism between a group and itself (cleraly an automorphism maps I to I, therefore for any given a: a*I = a'*I therefore a = a' for all a in any given group)

 

[Dogtanian]

Ok, so Aut(Z) is the set of all automorphisms on Z (the integers).
I don't think there's a general way to approach these kind of problems, but in any case: write down the conditions these automorphisms must have.
The trivial automorphism f(x)=x is one. I have a gut feeling it's also the only one. So see if you can try to prove that.

I'd have thought f(x)= -x is a automorphism from Z to Z. But given what has been said here I have tried to figure out why it isn't one, but with no sucess...

 

[semidevil]

I have another question.

When people say to "find a function that maps from G to H(the first step to proving isomorphism)", does it mean to find a function that belongs to both G and H. like, if I want to find an isomorphism from integers to even integers, I can use 2x, because 2x belongs to even integers, and also the integers. Is that pretty much what I am doing here?

I'm asking, because I need to do a lot of proof, and can I use that statement to prove problems(if it is true). It's just that if I start out wiht this statement, everything seems to fall into place, so just want to make sure I'm not messing with the logic...

 

[matt grime]

Obviously there exists an isomorphism between a group and itself so you needen't prove that, secondly there is only one isomorphism between a group and itself (cleraly an automorphism maps I to I, therefore for any given a: a*I = a'*I therefore a = a' for all a in any given group)

NO! there are (potentially) many automorphisms for every group.

 

[matt grime]

I'm not sure if it's a language thing or an understanding thing, but a function from G to H doesn't belong to either G or H.

A function is a mapping from one set to the other.

 

[jcsd]

NO! there are (potentially) many automorphisms for every group.

What's wrong with my reasoning? Actually I think I've seen it, due to sloppy notation I haven't proved that a' = a, I've proved a' = a'.

 

[matt grime]

Well it produces an answer that is wrong for a start, and sicne in your reasoning you didn't at any point use an isomorphism or even a homomorphism then you should have cause to think.

You appear to be saying that an automorphism satisfies:

f(ab)=af(b) and since f(1)=1 it uniquely determines the map as the identity map. Well, that isn't the definition of an automorphism, or even homomorphism.


Let G be any non-abelian group. Let x be any element not in centre of G (such exist as it is not abelian). Then the map G to G, given by f(y)=xyx^{-1} is an automorphism that is not the identity automorphism (or x would commute with all elements #).

x maps to x^{-1} is also a non-trivial automorphism for almost any abelian group (which is trivial iff every element has order dividing 2, admittedly).

 

[jcsd]

What I did wrong was to say a*b = f(a)*f(b) when I should of said f(a*b) = f(a)*f(b), therefore if I correce twhat I did all i proved was that f(a) = f(a)