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Automorphism

I came across this word today, when my lecturer explained it to us but I didn't understand him properly and the explanations on the internet are just as confusing. I was just wondering if anyone has a simple way of explaining this to me with examples. Thanks.
Reply 1
Avatar for nuodai
nuodai
17
In the context of what? Groups? Vector spaces? Modules? Fields?

In any case, if XX is some object (e.g. group, space, module, field) then θ:X→X\theta: X \to X is an automorphism if θ\theta is an isomorphism. That is, an automorphism is an isomorphism from an object to itself.

So for example if GG is a group then the identity map G→GG \to G which maps g↦gg \mapsto g is a group automorphism. And so is the inversion map g↦g−1g \mapsto g^{-1}. As another example, the complex conjugation map C→C\mathbb{C} \to \mathbb{C} given by z↦zˉz \mapsto \bar{z} is a field automorphism. A final example is if VV is a vector space of dimension nn over a field FF and A∈GLn(F)A \in \text{GL}_n(F) then the map v↦Avv \mapsto Av is a vector space automorphism.
(edited 12 years ago)
Reply 2
Avatar for Zhen Lin
Zhen Lin
12
For a more interesting example, consider field extensions. Let K be some field. A field extension is then a field L containing K as a subfield, and a morphism of field extensions is a homomorphism of fields leaving K invariant. An automorphism of the field extension L/K is then an isomorphism of fields L→LL \to L leaving K invariant. So, for example, let K=RK = \mathbb{R} and L=CL = \mathbb{C}; then complex conjugation is an automorphism of the field extension L/K. But there are other field isomorphisms C→C\mathbb{C} \to \mathbb{C}... they just don't happen to keep R\mathbb{R} invariant.
Reply 3
Avatar for SsEe
SsEe
13
Another example. If GG is a group and g∈Gg \in G then the map ϕg:G→G\phi_g : G \rightarrow G that conjugates by gg (ie, ϕg(x)=g−1xg\phi_g(x) = g^{-1} x g ) is an automorphism of GG. These conjugation maps are called inner automorphisms. In fact, the set of these, {ϕg:g∈G}\{ \phi_g : g \in G \} , is a group under composition and is written Inn(G)\text{Inn}(G). You can show using the first isomorphism theorem that Inn(G)≅G/Z(G)\text{Inn}(G) \cong G / Z(G) where Z(G)={g∈G:∀x∈G,gx=xg}Z(G) = \{ g \in G : \forall x \in G, gx = xg \} is the centre of GG
Reply 4
Avatar for Hasufel
Hasufel
16
a 1-2-1 correspondence mapping the elements of a Set onto itself, so the domain and range of the function are the same. e.g. f(x)=x+1 is an automorphism on R, but g(x)=sin (x) isn`t - range (- inifn,+infin), range (-1,1).
(edited 12 years ago)
Thanks. So if I wanted to show that θ \theta : G→G G \to G defined by θ(g)=g−1 \theta (g) = g^{-1} is an automorphism of G where G is an Abelian group, I just show that it is isomorphic?
Reply 6
Avatar for Slumpy
Slumpy
16
Original post by JBKProductions
Thanks. So if I wanted to show that θ \theta : G→G G \to G defined by θ(g)=g−1 \theta (g) = g^{-1} is an automorphism of G where G is an Abelian group, I just show that it is isomorphic?


Yep.
Original post by Slumpy
Yep.


To show isomorphism, I have to show that it is a bijective map but I can't seem to show that it's both injective and surjective.
Let x,y∈G x, y \in G . Since G is Abelian x∗y=y∗x x * y = y * x then θ(x∗y)=θ(y∗x) \theta (x*y) = \theta (y*x) and I'm stuck there. Am I doing it the wrong way to show injectivity?
Reply 8
Avatar for Zhen Lin
Zhen Lin
12
Original post by JBKProductions
To show isomorphism, I have to show that it is a bijective map but I can't seem to show that it's both injective and surjective.
Let x,y∈G x, y \in G . Since G is Abelian x∗y=y∗x x * y = y * x then θ(x∗y)=θ(y∗x) \theta (x*y) = \theta (y*x) and I'm stuck there. Am I doing it the wrong way to show injectivity?


What does that have to do with injectivity?

The easiest way to proceed in this case is to show that θ\theta has an inverse. And it's obvious what the inverse is: θ\theta itself. So θ\theta is not only an automorphism but also an involution.
Reply 9
Avatar for Slumpy
Slumpy
16
Original post by JBKProductions
To show isomorphism, I have to show that it is a bijective map but I can't seem to show that it's both injective and surjective.
Let x,y∈G x, y \in G . Since G is Abelian x∗y=y∗x x * y = y * x then θ(x∗y)=θ(y∗x) \theta (x*y) = \theta (y*x) and I'm stuck there. Am I doing it the wrong way to show injectivity?


To show injectivity, suppose θ(y)=θ(x) \theta (y) = \theta (x) . Now show x=y.
Original post by Zhen Lin
What does that have to do with injectivity?

The easiest way to proceed in this case is to show that θ\theta has an inverse. And it's obvious what the inverse is: θ\theta itself. So θ\theta is not only an automorphism but also an involution.


Thanks I had forgot about that. I just thought that since they mentioned Abelian, I had to use that as part of it. So it's enough to just say θ(g)=g−1 \theta (g) = g^{-1} then g=θ−1(g−1)⇒θ−1=θ g = \theta^{-1} (g^{-1}) \Rightarrow \theta^{-1} = \theta therefore bijective and so on?
Reply 11
Avatar for Zhen Lin
Zhen Lin
12
Original post by JBKProductions
Thanks I had forgot about that. I just thought that since they mentioned Abelian, I had to use that as part of it.


You do.

So it's enough to just say θ(g)=g−1 \theta (g) = g^{-1} then g=θ−1(g−1)⇒θ−1=θ g = \theta^{-1} (g^{-1}) \Rightarrow \theta^{-1} = \theta therefore bijective and so on?


That establishes that θ\theta is a bijection. But how do you know it's a homomorphism in the first place?
Original post by Zhen Lin
You do.



That establishes that θ\theta is a bijection. But how do you know it's a homomorphism in the first place?


To show that it is a homomorphism I have to show θ(x∗y)=θ(x)∗θ(y) \theta (x*y) = \theta (x) * \theta (y) right?
Here's what I tried (Which could be and probably is very wrong!):
θ(x∗y)=(x∗y)−1=y−1∗x−1=θ(y)∗θ(x)=θ(x)∗θ(y) \theta (x*y) = (x*y)^{-1} = y^{-1} * x^{-1} = \theta (y) * \theta (x) = \theta (x) * \theta (y) since Abelian so its a homomorphism?
Reply 13
Avatar for Zhen Lin
Zhen Lin
12
Original post by JBKProductions
To show that it is a homomorphism I have to show θ(x∗y)=θ(x)∗θ(y) \theta (x*y) = \theta (x) * \theta (y) right?
Here's what I tried (Which could be and probably is very wrong!):
θ(x∗y)=(x∗y)−1=y−1∗x−1=θ(y)∗θ(x)=θ(x)∗θ(y) \theta (x*y) = (x*y)^{-1} = y^{-1} * x^{-1} = \theta (y) * \theta (x) = \theta (x) * \theta (y) since Abelian so its a homomorphism?


That works.

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