# Group homomorphisms!

• February 2nd 2008, 08:16 AM
Sooz
Group homomorphisms!
Hi, I've got this question which I'm sure should be easy but I just don't seem to be able to get started...
I have a centre of a group G defined as
Z(G) = {x is an element of G : for all y contained in G xy=yx}.
Also k: G is mapped to G is an automorphism of G and automorphisms form a group Aut(G) under composition.
I have to show that the function f: G is mapped to Aut(G) defined by
[f(g)](h) = ghg^-1 is a group homomorphism.
I know that I have to show that f(g1,g2) = f(g1)f(g2) but I don't know how to go about it when the function is defined the way it is.
Any hints or advice would be very much apprecaited.
THANKS!
• February 2nd 2008, 02:59 PM
ThePerfectHacker
Quote:

Originally Posted by Sooz
I have to show that the function f: G is mapped to Aut(G) defined by

Define $\phi: G\mapsto \mbox{Aut}(G)$ as $\phi (g) = gxg^{-1}$. Note this is a good mapping because it turns an element of G into an automorphism of G. Now $\phi (g_1g_2) = g_1g_2 x g_2^{-1} g_1^{-1}$. While $\phi (g_1) = g_1xg_1^{-1} \mbox{ and }\phi (g_2) = g_2 x g_2^{-1}$. Thus $\phi (g_1) \circ \phi (g_2) = g_1 (g_2 x g_2^{-1}) g_1^{-1} = \phi(g_1g_2)$.
• February 4th 2008, 02:24 AM
Sooz
Group homomorphisms
Thanks, that makes sense, can I just check something though? Are you saying to let phi(g) mean exactly the same thing as [f(g)](h) and in doing so just ingoring the wierd way in which the map is defined in the question?
Thanks!
• February 4th 2008, 07:11 AM
ThePerfectHacker
Quote:

Originally Posted by Sooz
Are you saying to let phi(g) mean exactly the same thing as [f(g)](h) and in doing so just ingoring the wierd way in which the map is defined in the question?

Yes. I understood what you meant by it, but I did not like how it was written, so I wrote it in a way I found nicer.

If $g\in G$ define $i_g: G\mapsto G$ as $i_g(x) = gxg^{-1}$ then $i_g$ is an automorphism by a straightforward computation. Now define $\phi: G\mapsto \mbox{Aut}(G)$ as $\phi(g) = i_g$ (which was just defined above), this is a valid mapping because it takes elements in $G$ an turns them into automorphisms.
• February 6th 2008, 04:14 AM
Sooz
Normal subgroups
Quote:

Originally Posted by Sooz
I have a centre of a group G defined as
Z(G) = {x is an element of G : for all y contained in G xy=yx}.
Also k: G is mapped to G is an automorphism of G and automorphisms form a group Aut(G) under composition.
We also have f: G is mapped to Aut(G) defined by [f(g)](h) = ghg^-1

Thanks again for your help. I've managed to get a lot further now. I have just got to a point where I have shown that Ker(f) = Z(G) but now I need to prove that Im(f) is a normal subgroup of Aut(G). I know the following:
• By a proposition in my notes im(f) is a subgroup of Aut(G)
• G/Z(G) = Inn(G) by the 1st isomorphism theorem (I'm not sure if I can use this as we haven't defined the inner automorphism group in class)
• N is normal in G implies that there is some homomorphism of G for which N is the Kernel.

I don't know if I'm on track with any of those but it does seem like the question has led me straight to the First Isomorphism Theorem by getting me to find the Kernel. I'm not sure how to use it though! Its so frustrating because I feel like I'm close but I don't quite understand all this stuff well enough yet to make any connections or have a brainwave!
Thanks,
Sooz
• February 6th 2008, 07:10 AM
ThePerfectHacker
Quote:

Originally Posted by Sooz
Thanks again for your help. I've managed to get a lot further now. I have just got to a point where I have shown that Ker(f) = Z(G) but now I need to prove that Im(f) is a normal subgroup of Aut(G).

Okay, thus we established that $\phi: G\mapsto \mbox{Aut}(G)$ defined as $\phi(g) = i_g$ is a homomorphism. Now by the fundamental isomorphism theorem is means $G/\ker \phi \simeq \phi (G)$. Now $\phi(G) = \mbox{Inn}(G)$ where $\mbox{Inn}(G)$ is the group of all inner automomorphism*. The kernel $\ker \phi$ is the set of all $y\in G$ such that $\phi (y)$ gets mapped into the identity of $\mbox{Aut}(G)$, i.e. the identity automorphism. Thus, we require that $\phi (y) = yxy^{-1}$ to leave all element unpremuteted meaning $yxy^{-1} = x \mbox{ for all }x\in G$ if and only if $yx = xy$ if and only if $x\in \mbox{Z}(G)$ (the center of the group) by definition.
Hence, we have shown that, $G/\mbox{Z}(G) \simeq \mbox{Inn}(G)$.

*)The automorphism $i_g$ is called an inner automorphism. It can be shown that the set of all inner automorphisms forms a normal subgroup of the automorphism group $\mbox{Aut}(G)$.
• February 6th 2008, 10:53 AM
Sooz
Quote:

Originally Posted by ThePerfectHacker
It can be shown that the set of all inner automorphisms forms a normal subgroup of the automorphism group $\mbox{Aut}(G)$.

Thanks but I'm still a bit confused. The bit you put at the end is the thing I need to do (about showing its a normal subgroup) and hints on how to show this? Thanks again!
• February 6th 2008, 02:48 PM
ThePerfectHacker
Quote:

Originally Posted by Sooz
Thanks but I'm still a bit confused. The bit you put at the end is the thing I need to do (about showing its a normal subgroup) and hints on how to show this? Thanks again!

Let $\mbox{Inn}(G) = \{\theta \in \mbox{Aut}(G)| \theta (x) = gxg^{-1} \mbox{ for some }g\in G\}$. First we need to prove it is a subgroup. It is closed because if $g_1 xg_1^{-1}$ is one inner automorphism and $g_2 xg_2^{-1}$ is another inner automorphism then their product (which is function composition) is $(g_1g_2)x(g_1g_2)^{-1}$ which is an inner automorphism. Also, $exe^{-1}$ is the identity automorphism. It is associative because in general function compositions are associative. And finally it has an inverse. Thus, the set of inner automorphism form a subgroup. To show that $\mbox{Inn}(G) \triangleleft \mbox{Aut}(G)$ we need to show that for all $\phi \in \mbox{Aut}(G)$ and $\theta\in \mbox{Inn}(G)$ we have $\phi \theta \phi^{-1} \in \mbox{Inn}(G)$. Can you do this last part? If not post what you can do.
• February 8th 2008, 01:58 AM
Sooz
Quote:

Originally Posted by ThePerfectHacker
Can you do this last part? If not post what you can do.

I can't seem to do very much at all! I started off with
$\phi\theta\phi^{-1}$ but do I just look at the whole thing in one go or just look at $\phi\theta(x)$ first? By doing this I get that this is equal to $\theta\phi\theta^{-1}$ but I don't see how that helps or if it is even right! I understand the subgroup bit though!
Thanks again,
Sooz
• February 8th 2008, 08:33 AM
ThePerfectHacker
Quote:

Originally Posted by Sooz
I can't seem to do very much at all! I started off with
$\phi\theta\phi^{-1}$ but do I just look at the whole thing in one go or just look at $\phi\theta(x)$ first? By doing this I get that this is equal to $\theta\phi\theta^{-1}$ but I don't see how that helps or if it is even right! I understand the subgroup bit though!
Thanks again,
Sooz

Let $\theta$ be an inner automorphism, i.e. $\theta (x) = gxg^{-1}$, and let $\phi$ be any automorphism. Now $\phi \theta \phi^{-1} (x) = \phi (\theta (\phi^{-1} (x) ) ) = \phi ( g \phi^{-1} (x) g^{-1} ) = \phi (g) \phi (\phi^{-1} (x)) \phi (g^{-1}) = \phi(g)x\phi(g)^{-1}$.
• February 8th 2008, 10:46 AM
Sooz
Thanks
Thanks so much for your help, I really feel like I understand this all now. Thanks for your patience!
Sooz