Are the orders of Aut(G) and Inn(G) infinite for an infinite group G?

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SUMMARY

The discussion centers on the properties of automorphisms and inner automorphisms in the context of infinite groups, specifically questioning whether the orders of Aut(G) and Inn(G) are infinite when |G| is infinite. It is established that for the infinite group of integers, Z, the automorphism group Aut(Z) is isomorphic to Z/2, containing only two automorphisms: the identity map and the negation map. Additionally, Inn(Z) is confirmed to be the trivial group, represented as 0 or {0}, indicating that not all infinite groups have infinite automorphism or inner automorphism orders.

PREREQUISITES
  • Understanding of group theory concepts, specifically automorphisms and inner automorphisms.
  • Familiarity with the notation Aut(G) and Inn(G) in group theory.
  • Knowledge of the properties of the group of integers, Z, under addition.
  • Basic comprehension of bijective functions and their implications in group homomorphisms.
NEXT STEPS
  • Explore the structure of automorphism groups for other infinite groups, such as the additive group of rational numbers.
  • Study the implications of inner automorphisms in non-abelian groups.
  • Investigate the relationship between group properties and the cardinality of their automorphism groups.
  • Learn about the classification of groups based on their automorphism structures, particularly in the context of infinite groups.
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This discussion is beneficial for mathematicians, particularly those specializing in abstract algebra, group theorists, and anyone interested in the properties of automorphisms in infinite groups.

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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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Zondrina said:
If |G| is infinite, are the orders of Aut(G) and Inn(G) also infinite?

No. It is easy to check that \mathrm{Aut}(\mathbb{Z}) \cong \mathbb{Z}/(2) and that \mathrm{Inn}(\mathbb{Z}) = 0.
 
Zondrina said:
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.
 
Erland said:
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 } ?
 
Zondrina said:
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 \mathrm{Inn}(\mathbb{Z}) is the trivial group.
 
Zondrina said:
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.
 
micromass said:
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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