Group of inner automorphisms is isomorphic to a quotient

[Mr Davis 97]

Homework Statement

Let $G$ be any group. Recall that the center of $G$, or $Z(G)$ is $\{ x \in G ~ | ~ xg = gx, ~ \forall g \in G\}$. Show that $G / Z(G)$ is isomorphic to $Inn(G)$, the group of inner automorphisms of $G$ by $g$.

Homework Equations

The Attempt at a Solution

I am not sure where to get started. I know that I am trying to find a particular isomorphism, but not sure how to find what that isomorphism must be, or whether that map will go from $G / Z(G)$ to $Inn(G)$ or the other way around.

 

[fresh_42]

Can you establish a surjective homomorphism $G \twoheadrightarrow Inn(G)\;$?

 

[Mr Davis 97]

Can you establish a surjective homomorphism $G \twoheadrightarrow Inn(G)\;$?

Define a general element of $Inn(G)$ to be $\varphi_g(x) = gxg^{-1}$

Would $\mu : G \rightarrow Inn(G)$ where $\mu (g) = \varphi_g$ be a surjection?

 

[fresh_42]

Define a general element of $Inn(G)$ to be $\varphi_g(x) = gxg^{-1}$

Would $\mu : G \rightarrow Inn(G)$ where $\mu (g) = \varphi_g$ be a surjection?

Yes, because every inner automorphism (=conjugation) looks like a $\mu (g)$, so $g$ is the pre-imange. Now what is the kernel of $\mu$?

 

[Mr Davis 97]

Yes, because every inner automorphism (=conjugation) looks like a $\mu (g)$, so $g$ is the pre-imange. Now what is the kernel of $\mu$?

Scratch that

$Ker( \mu ) = Z(G)$

 

[fresh_42]

No. $\{e\} \subseteq \ker \mu$ but not necessarily the entire kernel. The kernel is defined as the set of all elements that maps to $e'$ in the codomain. Now what is this $e'$ then and what does it mean, that $\mu$ maps an element $g$ on it?

 

[Mr Davis 97]

No. $\{e\} \subseteq \ker \mu$ but not necessarily the entire kernel. The kernel is defined as the set of all elements that maps to $e'$ in the codomain. Now what is this $e'$ then and what does it mean, that $\mu$ maps an element $g$ on it?

Sorry, I made a mistake and was too slow to correct. I think that $Ker (\mu ) = Z(G)$

Will the fundamental homomorphism theorem be used?

 

[fresh_42]

Not sure what this theorem is, but sounds right. How do you now, that $\ker \mu = Z(G)\,$? This is the essential part of the proof, so you should drop a line on it.

 

[Mr Davis 97]

Not sure what this theorem is, but sounds right. How do you now, that $\ker \mu = Z(G)\,$? This is the essential part of the proof, so you should drop a line on it.

$\ker \mu = \{x \in G ~ | ~ \mu (x) = \mu (e) \} = \{x \in G ~ | ~ \varphi_x = \varphi_e \} = \{x \in G ~ | ~ \varphi_x (g) = \varphi_e (g), ~ \forall g \in G \} = \{x \in G ~ | ~ x g x^{-1} = g, ~ \forall g \in G \} = \{x \in G ~ | ~ xg = gx, ~ \forall g \in G\} = Z(G)$