1. Nov 12, 2012

### Zondrina

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.

Last edited: Nov 12, 2012
2. Nov 12, 2012

### jgens

No. It is easy to check that $\mathrm{Aut}(\mathbb{Z}) \cong \mathbb{Z}/(2)$ and that $\mathrm{Inn}(\mathbb{Z}) = 0$.

3. Nov 12, 2012

### Erland

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.

4. Nov 12, 2012

### Zondrina

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 } ?

5. Nov 12, 2012

### jgens

They mean the same thing. Both of them are shorthand for the statement that $\mathrm{Inn}(\mathbb{Z})$ is the trivial group.

6. Nov 12, 2012

### micromass

You switched the two around. It is injective but not surjective, which is what you proved next.

7. Nov 12, 2012

### Zondrina

Oh whoops, mistype. Thanks for noticing.

Thanks for clarifying jgens I think I understand this now.