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$G$ is a group. $Z(G) = \{g \in G: \forall x \in G (gx = xg) \}$ is called the center of $G$. It is easy to verify that $Z(G) \triangleleft G$. $\text{Inn}(G)$ is the inner automorphism group of $G$. As a consequence of the first isomorphism theorem, we have $$G / Z(G) \cong \text{Inn}(G).$$ The homomorphism involved here is defined as $a \in G \mapsto \sigma_{a} \in \text{Inn}(G)$ where $\sigma_{a}$ is a bijection from $G$ to $G$ with $\sigma_{a}(x) = axa^{-1}$. The details can be found here: Factor Group over Center Isomorphic to Inner Automorphism Group.

The isomorphism is too abstract. Could anyone offer some intuitive (or even visual) explanations?

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    $\begingroup$ Actually this isomorphism looks very concrete to me: every element $g\in G$ defines an automorphism of $G$ by conjugation. Two elements define the same automorphism exactly when they "differ" by a central element. $\endgroup$ – Andrea Mori Mar 19 '14 at 13:51
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    $\begingroup$ What is the central elements? $\endgroup$ – hengxin Mar 19 '14 at 13:54
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    $\begingroup$ A central element is an element in the center $\endgroup$ – Andrea Mori Mar 19 '14 at 13:55
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Well, "conjugation by $a$" shuffles the elements of $G$: each $g$ goes to $aga^{-1}$, right?

There are two possibilities: (i) if $a$ happens to commute with every element of $g$, then $aga^{-1}$ is the same as $gaa^{-1} = g$, so the "shuffling" is just the identity. (ii) If not, then some element $g$ is sent to a different element, and the shuffling is nontrivial.

So not every element $a$ of $G$ leads to an interesting shuffle: only the ones that fail to commute with at least one other element. So you might think that the inner automorphisms would correspond to $G - Z(G)$. But in fact, if $a$ produces an interesting automorphism, $\phi$, but $b$ produces the identity, then $ab$ will also produce $\phi$. So really the right correspondence is with $G/Z(G)$.

Does that help at all?

Let me add a concrete example: Suppose that $G$ is the group of all rotations and dilations of 3-space, where the rotations leave the origin fixed, and by "dilations" I mean maps like $\mathbf v \mapsto c\mathbf v$, where $c$ is a nonzero scalar. Then the subgroup $D$ of dilations is the center of $G$: if you scale up, rotate, and then scale down, it's the same as just rotating. And if you scale up by $a$, then by $b$, then by $1/a$, it's the same as scaling just by $b$.

And $G/D$ looks just like $SO(3)$ (i.e., the group of origin-fixing rotations of 3-space, known as the "special orthogonal" group -- "orthogonal" because the matrices are orthogonal; "special" because their determinant is $+1$). Why does this correspond to the inner automorphisms of $G$? Because each rotation $R$ corresponds to an automorphism of $G$, namely, the change of basis induced by $R$.

I don't know if that helps clarify things or not, but it was worth a shot...

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  • $\begingroup$ Yes, it does. Thanks. I will accept it if no other "competitive" answers appear (for some time). $\endgroup$ – hengxin Mar 19 '14 at 14:18
  • $\begingroup$ Nice example! Possible typos: the subgroup $D$ is dilations is the center of $G$? In addition, what does $SO$ stand for? $\endgroup$ – hengxin Mar 20 '14 at 8:13
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I would suggest looking at it in the following way.

If we define $Aut(G)$ to be the group of automorphisms of $G$, i.e.

$$ Aut(G) = \{f : G \to G \mid f \text{ is a bijective homomorphism}\} $$

then there is a map $\phi : G \to Aut(G)$ given by

$$ \phi(g) = \big(h \mapsto ghg^{-1}\big) $$

that is, $\phi(g)$ is the automorphism given by conjugation by $g$.

Now, it is (hopefully) clear that the image of $G$ in $Aut(G)$ is exactly the set of inner automorphisms, i.e. $Inn(G)$. What is the kernel? Well, the kernel is exactly those $g \in G$ such that $h = ghg^{-1}$ for all $h \in G$. i.e. the kernel is $Z(G)$!

Thus it follows that $im(\phi) = G / \ker\phi$ (this is a restatement of the first isomorphism theorem), i.e. $Inn(G) = G / Z(G)$.

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