Re: automorphism of a group

You have to prove that : it's a consequence of associativity and the fact that .

Re: automorphism of a group

So that means I've to prove that:

So I've proved that is an automorphism and I'm done.

Am I right? ...

Re: automorphism of a group

You are almost done, since you have not proved yet that is surjective.

Re: automorphism of a group

Oh, indeed because: surjection + injection = bijection. If I've to prove that is surjective that means each argument has an image , I think that's right because: so (because ).

Am I correct? ...

I appreciate your help.

Re: automorphism of a group

No, that's not correct. you have to find an element of the group for wich f(x)=y, for every y. If the group was abelian you could commute g and g^-1 but it's not necessarily abelian (in fact the exercise is only fun when the group may not be abelian).

Re: automorphism of a group

Re: automorphism of a group

Incidentally, it's good to get into the habit of thinking of an isomorphism as a homomorphism with homomorphic inverse, rather than a bijective homomorphism. In the category of groups, the latter is a consequence of the former, and either may be taken as the definition, but in many categories it's not true. For example, in the category of topological spaces, the equivalent notions to homomorphism and isomorphism are continuous map and homeomorphism--and a bijective continuous map need not be a homeomorphism. The correct definition is a continuous map with continuous inverse.

Re: automorphism of a group

You have already shown that is indeed a group homomorphism. Now we just have to show that it is both injective (one-to-one) and surjective (onto).

It suffices to show that has an inverse. When you look further into this then you will notice that is indeed the inverse of . You can see this by noting that if then it follows that for any element (because for any element ).

Does this make sense?

Re: automorphism of a group

Quote:

Originally Posted by

**obd2** You have already shown that

is indeed a group homomorphism. Now we just have to show that it is both injective (one-to-one) and surjective (onto).

It suffices to show that

has an inverse. When you look further into this then you will notice that

is indeed the inverse of

. You can see this by noting that if

then it follows that

for any element

(because

for any element

).

Does this make sense?

This isn't entirely true - you have shown that every element in the image has an inverse. That is to say, you have shown that the function is 1-1. You still need to show that it is surjective!

(For example, if we take the function , then if we define then , but clearly isn't a bijection...(what I have just written abuses notation horrendously, but my point is that my abuse is the same as your abuse...))

Re: automorphism of a group

Quote:

Originally Posted by

**Swlabr** This isn't entirely true - you have shown that every element in the image has an inverse. That is to say, you have shown that the function is 1-1. You still need to show that it is surjective!

(For example, if we take the function

, then if we define

then

, but clearly

isn't a bijection...(what I have just written abuses notation horrendously, but my point is that my abuse is the same as your abuse...))

Well... I think I might have shown that it is surjective. Given an element then I have shown that the element maps to with . That is to say set then .

Re: automorphism of a group

Quote:

Originally Posted by

**obd2** Well... I think I might have shown that it is surjective. Given an element

then I have shown that the element

maps to

with

. That is to say set

then

.

Ah yes, sorry, you're right.

Re: automorphism of a group

Quote:

Originally Posted by

**Swlabr** Ah yes, sorry, you're right.

Actually...doesn't this only work in a finite group? You are saying that both and are injections, but that doesn't really mean that they are bijections, unless your group is infinite...

Re: automorphism of a group

say that is an injection and a surjection.

So showing show that and are both bijections (and that )