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 $z \in Z(G)$

  • 1
    $\begingroup$ Do you know what the quotient group is? $\endgroup$ – Randall Oct 20 '18 at 0:42

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ (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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.