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I've seen other posts (Inner automorphisms form a normal subgroup of $\operatorname{Aut}(G)$) about the topic of $\operatorname{Inn}(G) \simeq G/Z(G)$, but what I want to ask is a detail. When we ensure that $\operatorname{Inn}(G) \simeq G/Z(G)$, we want to say that there exists some isomorphism $F$ such that:

$$F: G/Z(G) \rightarrow \operatorname{Inn}(G), \quad F(gz) = \tau_g$$

where $g \in G, z \in Z(G)$ and $\tau_g(h) = ghg^{-1}$ for all $h \in G$

But is this $F$ really an isomorphism? I'm not pretty sure due to there is not ONE $\tau_g$ for ONE $gz$, what we have is a total of $|Z(G)| = \text{order of }Z(G)$ elements of the form $gz$ for just ONE $\tau_g$ due to $\tau_g = \tau_{gz}$ if $z \in Z(G)$

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    $\begingroup$ Do you know what the quotient group is? $\endgroup$ – Randall Oct 20 '18 at 0:42
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Well, to see that $F$ is a bijection, we can show that it is surjective and inyective. I will note $Z := Z(G)$.

  • Surjectivity: let $\tau_g(x) = gxg^{-1}$ an element of $Inn(G)$. If we now consider the class of $g$ in $G/Z$, it has image precisely $\tau_g$ via $F$, as you described.

  • Injectivity: suppose that $F(gZ) = F(hZ)$. Thus, $\tau_g \equiv \tau_h$ and thus

$$ gxg^{-1} = \tau_g(x) = \tau_h(x) = hxh^{-1} \ (\forall x \in G) $$

or equivalently, $h^{-1}gx = xh^{-1}g$. Therefore, $h^{-1}g \in Z$ which implies $gZ = hZ$.

Furthermore, $F(1_{Z/G}) = F(1 \cdot Z) = \tau_1 = id_G$ and $$F(gZ)F(hZ) = \tau_g\tau_h = \tau_{gh} = F(ghZ) = F(gZhZ)$$

thus proving that $F$ is not only a bijection but an isomorphism.

Intutively, what is happening is that conjugation by some $g \in G$ is 'not affected by the component of $g$ in $Z$'. By that I mean that if $g = sz$ with $z \in Z$, $\tau_{sz} = \tau_s$. So we can group elements which only differ in a traslation via an element of the center, since their conjugation will be the same: that is exactly what $G/Z(G)$ is.

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  • $\begingroup$ Then, the key is that $F$ acts not on each $gz$ but on the WHOLE $gZ$, so we have an isomorphism between each $gZ$ and each $\tau_g$. I was thinking about $F(gz)$ instead of $F(gZ)$. That is the point, right? $\endgroup$ – Vicky Oct 20 '18 at 1:01
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    $\begingroup$ Yes, more or less. We have a correspondence between $gZ$ and $\tau_g$, and an isomorphism between $G/Z$ and $Inn(G)$. Since the domain of $F$ is $G/Z = \{gZ : g \in G \}$, writing $F(gz)$ would not make sense. We are sending each class of the quotient to the conjugate that they all represent (since they differ by an element of the center). $\endgroup$ – Guido A. Oct 20 '18 at 1:58
  • $\begingroup$ What do you mean with class of the quotient? I know what a conjugacy class is, but how do you relate that concept with G/Z? Sorry for these basic questions, but I'm a rookie in Group Theory and this is the last question I have $\endgroup$ – Vicky Oct 20 '18 at 2:54
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    $\begingroup$ Okay so, in general if you have a set $X$ and an equivalence relation $\sim$ on $X$, the equivalence classes are the sets $[x] = \{y \in X : y \sim x\}$. These make a partition of the space: $X = \cup_{x \in X}[x]$ and either $[x] \cap [y] = \emptyset$ or $[x] = [y]$. Now, if $G$ is a group and $H \leq G$ a subgroup, there is an equivalence relation on $G$ given by $g \sim g' \iff g^{-1}g' \in H$. This partitions $G$ into equivalence classes. We call this set $G/H$, and one can actually see that the classes have the form $[g] = gH$. Thus, $G/H = \{gH : g \in G\}$. $\endgroup$ – Guido A. Oct 20 '18 at 3:00
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    $\begingroup$ (cont.) when $H$ is normal, $G/H$ has a natural group structure, which you have probably studied in class, via $gH \cdot g'H := (gg')H$. So when I say class of the quotient $G/Z$, I mean an element $[g] = gZ \in G/Z$. $\endgroup$ – Guido A. Oct 20 '18 at 3:01
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You have a map, defined by \begin{align} \varphi:G&\longrightarrow \operatorname{Inn}G\\ g&\longmapsto(x\mapsto gxg^{-1}) \end{align} which is surjective by definition, and which you can check to be a group homomorphism. By the first isomorphism theorem you obtain an isomorphism $$G/\ker\varphi \simeq \operatorname{Inn}G,$$ and clearly, $$\ker\varphi=\{g\in G\mid \forall x\in G, \,gxg^{-1}=x\}=Z(G).$$

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