# automorphism of a group

• July 18th 2011, 10:42 AM
Siron
automorphism of a group
Hi,

Can someone help me with proving this proposition:

Let $G$ a group. For $g \in G$ we define:
$\lambda_g: G \to G: x \mapsto gxg^{-1}$

Proof $\lambda_g$ is an automorphism of $G$ (We call it an inside automorphism of $G$ determined by $g$).

If I wan't to proof $\lambda_g$ is an automorphism I've to demonstrate that $\lambda_g$ is an isomorphism or an bijective homomorphism.

1) $\lambda_g$ is an homorphism
But to prove if $\lambda_g$ is an homorphism in my inion I need a composite law for the group $G$.

I guess I've to proof for two arbitrary chosen numbers $x,y$ in $G$ that:
$f(x\perp y)=f(x)\perp f(y)$

But how do I now: $\perp$? Or doesn't that matter? ...

2) $\lambda_g$ is bijective:
I've to prove that:
$f(x)=f(y) \Rightarrow x=y$
In this case:
$gxg^{-1}=gyg^{-1} \Rightarrow x=y$
This is true because in a group a $g^{-1}$ is the symmetric element of $g$ and so: $gg^{-1}=1$

So the only thing I haven't proved yet is the homorphism. Does someone have a hint?

(Sorry If my English wouldn't be correct, but English mathematical terminology is not easy for me as a Belgian).
• July 18th 2011, 11:44 AM
girdav
Re: automorphism of a group
You have to prove that $\lambda_g(x\star y)=\lambda_g(x)\star\lambda_g( y)$: it's a consequence of associativity and the fact that $g^{-1}\star g=e$.
• July 18th 2011, 11:54 AM
Siron
Re: automorphism of a group
So that means I've to prove that:
$gxyg^{-1}=(gxg^{-1})\cdot(gyg^{-1})$
$\Leftrightarrow xy=xy$

So I've proved that $\lambda_g$ is an automorphism and I'm done.
Am I right? ...
• July 18th 2011, 12:00 PM
girdav
Re: automorphism of a group
You are almost done, since you have not proved yet that $\lambda_g$ is surjective.
• July 18th 2011, 12:20 PM
Siron
Re: automorphism of a group
Oh, indeed because: surjection + injection = bijection. If I've to prove that $\lambda_g$ is surjective that means each argument $x \in G$ has an image $f(x) \in G$, I think that's right because: $x \mapsto gxg^{-1}$ so $x \mapsto x$ (because $g\cdot g^{-1}=1$).

Am I correct? ...
• July 18th 2011, 01:15 PM
ModusPonens
Re: automorphism of a group
No, that's not correct. you have to find an element of the group for wich f(x)=y, for every y. If the group was abelian you could commute g and g^-1 but it's not necessarily abelian (in fact the exercise is only fun when the group may not be abelian).
• July 19th 2011, 01:48 AM
Swlabr
Re: automorphism of a group
Quote:

Originally Posted by ModusPonens
No, that's not correct. you have to find an element of the group for wich f(x)=y, for every y. If the group was abelian you could commute g and g^-1 but it's not necessarily abelian (in fact the exercise is only fun when the group may not be abelian).

So what you need to do, Siron, is find some element, $g \in G$, such that $g^{-1}hg=h$ for all $h \in G$. Now, if $g^{-1}hg=h$ then $gh=hg$ so $g\in Z(G)$. Can you find such an element in any given group? What is this element?...
• July 19th 2011, 09:19 AM
Tinyboss
Re: automorphism of a group
Incidentally, it's good to get into the habit of thinking of an isomorphism as a homomorphism with homomorphic inverse, rather than a bijective homomorphism. In the category of groups, the latter is a consequence of the former, and either may be taken as the definition, but in many categories it's not true. For example, in the category of topological spaces, the equivalent notions to homomorphism and isomorphism are continuous map and homeomorphism--and a bijective continuous map need not be a homeomorphism. The correct definition is a continuous map with continuous inverse.
• July 20th 2011, 05:18 AM
obd2
Re: automorphism of a group
You have already shown that $\lambda_g$ is indeed a group homomorphism. Now we just have to show that it is both injective (one-to-one) and surjective (onto).

It suffices to show that $\lambda_g$ has an inverse. When you look further into this then you will notice that $\lambda_{g^{-1}}$ is indeed the inverse of $\lambda_g$. You can see this by noting that if $\lambda_{g^{-1}}(x) = g^{-1}x (g^{-1})^{-1} = g^{-1}xg$ then it follows that $\lambda_g(\lambda_{g^{-1}}(x)) = \lambda_{g^{-1}}(\lambda_g(x)) = x$ for any element $x\in G$ (because $g^{-1}g = gg^{-1} = e$ for any element $g\in G$).

Does this make sense?
• July 20th 2011, 06:37 AM
Swlabr
Re: automorphism of a group
Quote:

Originally Posted by obd2
You have already shown that $\lambda_g$ is indeed a group homomorphism. Now we just have to show that it is both injective (one-to-one) and surjective (onto).

It suffices to show that $\lambda_g$ has an inverse. When you look further into this then you will notice that $\lambda_{g^{-1}}$ is indeed the inverse of $\lambda_g$. You can see this by noting that if $\lambda_{g^{-1}}(x) = g^{-1}x (g^{-1})^{-1} = g^{-1}xg$ then it follows that $\lambda_g(\lambda_{g^{-1}}(x)) = \lambda_{g^{-1}}(\lambda_g(x)) = x$ for any element $x\in G$ (because $g^{-1}g = gg^{-1} = e$ for any element $g\in G$).

Does this make sense?

This isn't entirely true - you have shown that every element in the image has an inverse. That is to say, you have shown that the function is 1-1. You still need to show that it is surjective!

(For example, if we take the function $\phi: \mathbb{N}\rightarrow \mathbb{N}, a\mapsto a^2$, then if we define $\varphi: \mathbb{N}\rightarrow \mathbb{N}, a\mapsto \sqrt{a}$ then $\phi^{-1}=\varphi$, but clearly $\phi$ isn't a bijection...(what I have just written abuses notation horrendously, but my point is that my abuse is the same as your abuse...))
• July 20th 2011, 10:43 AM
obd2
Re: automorphism of a group
Quote:

Originally Posted by Swlabr
This isn't entirely true - you have shown that every element in the image has an inverse. That is to say, you have shown that the function is 1-1. You still need to show that it is surjective!

(For example, if we take the function $\phi: \mathbb{N}\rightarrow \mathbb{N}, a\mapsto a^2$, then if we define $\varphi: \mathbb{N}\rightarrow \mathbb{N}, a\mapsto \sqrt{a}$ then $\phi^{-1}=\varphi$, but clearly $\phi$ isn't a bijection...(what I have just written abuses notation horrendously, but my point is that my abuse is the same as your abuse...))

Well... I think I might have shown that it is surjective. Given an element $x\in G$ then I have shown that the element $\lambda_{g^{-1}}(x)$ maps to $x$ with $\lambda_g$. That is to say set $y = \lambda_{g^{-1}}(x)$ then $\lambda_g(y) = x$.
• July 21st 2011, 12:49 AM
Swlabr
Re: automorphism of a group
Quote:

Originally Posted by obd2
Well... I think I might have shown that it is surjective. Given an element $x\in G$ then I have shown that the element $\lambda_{g^{-1}}(x)$ maps to $x$ with $\lambda_g$. That is to say set $y = \lambda_{g^{-1}}(x)$ then $\lambda_g(y) = x$.

Ah yes, sorry, you're right.
• July 21st 2011, 05:30 AM
Swlabr
Re: automorphism of a group
Quote:

Originally Posted by Swlabr
Ah yes, sorry, you're right.

Actually...doesn't this only work in a finite group? You are saying that both $\lambda_g$ and $\lambda_g^{-1}$ are injections, but that doesn't really mean that they are bijections, unless your group is infinite...
• August 1st 2011, 08:28 AM
pece
Re: automorphism of a group
$fog = id$ say that $g$ is an injection and $f$ a surjection.

So showing $fog = id = gof$ show that $f$ and $g$ are both bijections (and that $f = g^{-1}$)