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Question about automorphisms

by Zondrina
Tags: automorphisms
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Zondrina
#1
Nov12-12, 03:42 PM
P: 1,538
So I've been reading a bit about automorphisms today and I was wondering about something. I'm particularly talking about groups in this case, say a group G.

So an automorphism is a bijective homomorphism ( endomorphism if you prefer ) of a group G.

We use Aut(G) to denote the set of all automorphisms of G.

There is also an inner automorphism induced by a in G. The map f := G → G such that f(x) = axa-1 for all x in G is called the inner automorphism of G induced by a.

We use Inn(G) to denote the set of all inner automorphisms of G.

My question now :

If |G| is infinite, are the orders of Aut(G) and Inn(G) also infinite? It makes sense to me that they would be.
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jgens
#2
Nov12-12, 03:55 PM
P: 1,622
Quote Quote by Zondrina View Post
If |G| is infinite, are the orders of Aut(G) and Inn(G) also infinite?
No. It is easy to check that [itex]\mathrm{Aut}(\mathbb{Z}) \cong \mathbb{Z}/(2)[/itex] and that [itex]\mathrm{Inn}(\mathbb{Z}) = 0[/itex].
Erland
#3
Nov12-12, 04:06 PM
P: 345
Quote Quote by Zondrina View Post
If |G| is infinite, are the orders of Aut(G) and Inn(G) also infinite? It makes sense to me that they would be.
No, not always. The integers, Z, is a group under addition. |Z| is infinite, but there are only two automorphisms on Z, given by f(x)=x (identity map) and g(x)=-x, respectively.

Zondrina
#4
Nov12-12, 04:30 PM
P: 1,538
Question about automorphisms

Quote Quote by Erland View Post
No, not always. The integers, Z, is a group under addition. |Z| is infinite, but there are only two automorphisms on Z, given by f(x)=x (identity map) and g(x)=-x, respectively.
Ahh yes, that makes sense now because every other map doesn't generate all of Z.

So f(x) = 2x wouldn't be an automorphism for example. While it is a homomorphism, it is not injective, only surjective I believe.

Short proof to make sure : So f : Z → Z is already well defined for us. So we assume f(x) = f(y) → 2x = 2y → 2(x-y) = 0. Since 2|0, clearly x = y and f is injective. Now notice that the n elements of Z get mapped to fewer elements, so f is not surjective. Easy to show it's a homomorphism.

So f(x) = x and g(x) = -x are the only things that could be in Aut(Z) since they are the only maps which turn out to be automorphisms.

As for Inn(Z) = 0. That confused me a bit.

EDIT : Wouldn't it be Inn(Z) = { 0 } ?
jgens
#5
Nov12-12, 04:57 PM
P: 1,622
Quote Quote by Zondrina View Post
As for Inn(Z) = 0. That confused me a bit.

EDIT : Wouldn't it be Inn(Z) = { 0 } ?
They mean the same thing. Both of them are shorthand for the statement that [itex]\mathrm{Inn}(\mathbb{Z})[/itex] is the trivial group.
micromass
#6
Nov12-12, 05:02 PM
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Quote Quote by Zondrina View Post
So f(x) = 2x wouldn't be an automorphism for example. While it is a homomorphism, it is not injective, only surjective I believe.
You switched the two around. It is injective but not surjective, which is what you proved next.
Zondrina
#7
Nov12-12, 08:12 PM
P: 1,538
Quote Quote by micromass View Post
You switched the two around. It is injective but not surjective, which is what you proved next.
Oh whoops, mistype. Thanks for noticing.

Thanks for clarifying jgens I think I understand this now.


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