# Detail in bijection between Inn(G) and G/Z(G)?

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 $$z \in Z(G)$$

• Do you know what the quotient group is? – Randall Oct 20 '18 at 0:42

## 2 Answers

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.

• 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? – Vicky Oct 20 '18 at 1:01
• 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). – Guido A. Oct 20 '18 at 1:58
• 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 – Vicky Oct 20 '18 at 2:54
• 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, $G/H = \{gH : g \in G\}$. – Guido A. Oct 20 '18 at 3:00
• (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 $[g] = gZ \in G/Z$. – Guido A. Oct 20 '18 at 3:01

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, $$\ker\varphi=\{g\in G\mid \forall x\in G, \,gxg^{-1}=x\}=Z(G).$$