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 September 2nd, 2011, 01:39 AM #1 Member   Joined: Sep 2010 Posts: 60 Thanks: 0 Category-theory (finite group theory) prove Let G be a group, $H \triangleleft G$, and for $g \in G$ define $g\pi: H \rightarrow H$ by $x(g\pi)=x^g, x\in H$. Let $\mathcal{C}$ be the category of groups and homomorphisms. Prove $\pi$ is a $\mathcal{C}$-representation of G with kernel $C_G(H)$. $\pi$ is the representation by conjugation of G on H. If H=G, the image of G under $\pi$ is the inner automorphism group of G and is denoted by Inn(G). Prove $Inn(G)\triangleleft Aut(G)$. (Define Out(G)=Aut(G)/Inn(g) to be the outer automorphism group of G. I would be really appreciate if you could help me.
 September 2nd, 2011, 05:04 AM #2 Senior Member   Joined: Sep 2008 Posts: 150 Thanks: 5 Re: Category-theory (finite group theory) prove Well, where did you get stuck? This is an exercise that mainly consists of understanding the concepts. After that you simply check the axioms. But that are one-line proofs without any fancy tricks.
 September 2nd, 2011, 10:17 AM #3 Member   Joined: Sep 2010 Posts: 60 Thanks: 0 Re: Category-theory (finite group theory) prove To tell the truth, I didn't get stuck, unfortunatelly I can't start with. I would appreciate your help!
 September 3rd, 2011, 02:33 AM #4 Senior Member   Joined: Sep 2008 Posts: 150 Thanks: 5 Re: Category-theory (finite group theory) prove Ok, let's start with the representation. I would say a representation of a group G on an Object O of a category C is a Homomorphism $G\rightarrow Aut_C(O)$ i.e. a Homomorphism. In your case the object is H in the category of groups and you have to show that $g\mapsto \pi_g$ is a homomorphism. For that you have to check two things: 1)\pi_g is in fact a Group_automorphism of H (i.e., the described map actually goes into $Aut_{groups}(H)$) 2)if h is an other element of G then $\pi_{gh}=\pi_g \circ \pi_h$ (i.e. the described map is a homomorphism.) Once you've done that you should compute the kernel as the centralizer. But that should basically consist of writing down the definition of the map and the centralizer. Do you understand that and can you do 1 and 2? There is really not much to do!
 September 3rd, 2011, 06:58 AM #5 Member   Joined: Sep 2010 Posts: 60 Thanks: 0 Re: Category-theory (finite group theory) prove Thank you for your help. I am not sure, if i am right, but I try... 1) is true because $\pi$ is the conjugation of g on H and H is a normal subgroup of G (which is invariant under conjugation by members of the group). 2) is true because$x^gh=(x^g)^h)$ <== $(gh)^{-1}xgh=h^{-1}g^{-1}xgh=(x^g)^h$ The elements of the centralizer are $\{g: g^{-1}xg=x}$, and why is it the same as the kernel? Why is it true that $Inn(G) \triangleleft Aut(G)$? I am really grateful for your help!
September 3rd, 2011, 07:47 AM   #6
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Re: Category-theory (finite group theory) prove

Quote:
 Originally Posted by butabi Thank you for your help. I am not sure, if i am right, but I try... 1) is true because $\pi$ is the conjugation of g on H and H is a normal subgroup of G (which is invariant under conjugation by members of the group). 2) is true because$x^{gh}=(x^g)^h)$ <== $(gh)^{-1}xgh=h^{-1}g^{-1}xgh=(x^g)^h$
Good. There where some brackets missing to put gh both in the exponent. But apart from the typo it is correct.

Quote:
 The elements of the centralizer are $\{g: g^{-1}xg=x}$, and why is it the same as the kernel?
Well what is the kernel? The elements that go to 1 in the "Aut" group. What is the one in the Automorphisms? It is the identity. So what does it mean for some g that the conjugation with g on H is the identity map? Well, we need for all x in H that $g^{-1}xg=x$, thus we are back at the definition of the centralizer.

Quote:
 Why is it true that $Inn(G) \triangleleft Aut(G)$?
You have to show for all Automorphisms $\phi$ that $x\mapsto \phi^{-1}(\phi(x)^g)$ is again an inner automorphism, i.e., the conjugation by some element g' in G. Write down, what the conjugation is, and use that $\phi$ and $\phi^{-1}$ are homomorphisms and you will see, what element that is.

 September 3rd, 2011, 09:02 AM #7 Member   Joined: Sep 2010 Posts: 60 Thanks: 0 Re: Category-theory (finite group theory) prove Thank you again. I am very grateful indeed. So I don't know if I am right: $\phi^{-1}(\phi(x)^g)=\phi^{-1}(\phi(g^{-1}xg)=\phi^{-1}(\phi(g^{-1})\phi(x)\phi(g))=\phi^{-1}(\phi(g^{-1}))\phi^{-1}(\phi(x))\phi^{-1}(\phi(g))=g^{-1}xg$
September 3rd, 2011, 01:03 PM   #8
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Re: Category-theory (finite group theory) prove

Quote:
 Originally Posted by butabi So I don't know if I am right: $\phi^{-1}(\phi(x)^g)=\phi^{-1}(\phi(g^{-1}xg)=\phi^{-1}(\phi(g^{-1})\phi(x)\phi(g))=\phi^{-1}(\phi(g^{-1}))\phi^{-1}(\phi(x))\phi^{-1}(\phi(g))=g^{-1}xg$
Not quite, my notation was meant to indicate, that $\phi(x)$ is conjugated by $g$:

$\phi^{-1}(\phi(x)^g)=\phi^{-1}(g^{-1}\phi(x)g)=...$

But you certainly have got the point now.

 September 3rd, 2011, 02:52 PM #9 Member   Joined: Sep 2010 Posts: 60 Thanks: 0 Re: Category-theory (finite group theory) prove Yeah, sorry, I missunderstood your notation. Thank you very much for your help once again, you helped me a lot!

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