
#1
Nov1212, 03:42 PM

P: 1,320

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. 



#2
Nov1212, 03:55 PM

P: 1,623





#3
Nov1212, 04:06 PM

P: 302





#4
Nov1212, 04:30 PM

P: 1,320

Question about automorphismsSo 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(xy) = 0. Since 20, 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
Nov1212, 04:57 PM

P: 1,623





#6
Nov1212, 05:02 PM

Mentor
P: 16,565





#7
Nov1212, 08:12 PM

P: 1,320

Thanks for clarifying jgens I think I understand this now. 


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