On group theory

[Papapetros Vaggelis]

Let \(\displaystyle{\left(G,\cdot\right)}\) be a group such that the group \(\displaystyle{\left(\rm{Aut}(G),\circ\right)}\) is cyclic.

Prove that the group \(\displaystyle{\left(G,\cdot\right)}\) is an abelian group.

Is the converse true ?

[Demetres]

Let us consider the group $\mathrm{Inn}(G)$ of the inner automorphisms of $G$. This is a subgroup of $\mathrm{Aut}(G)$ and so it must by cyclic as well.

Suppose that $\mathrm{Inn}(G)$ is generated by conjugation by the element $x$. This conjugation fixes the element $x$ of $G$. Thus every inner automorphism must also fixed $x$. I.e. $y^{-1}xy = x$ for every $y \in G$ and thus $x$ commutes with $y$. Since $x$ commutes with every element of $G$, then conjugation by $x$ is the identity automorphism. Thus every inner automorphism is the identity automorphism. Therefore every two elements of $G$ commute, i.e. $G$ is abelian. [Given elements $g,h \in G$ we know that conjugation by $g$ is the identity automorphism, thus $g^{-1}hg = g$ and thus $g$ and $h$ commute.]

The converse is not true. For example the automorphism group of the abelian group $C_2 \times C_2$ is isomorphic to $S_3$ which is not cyclic. [Every automorphism of $C_2 \times C_2$ must fix the identity element and permute the other three elements of $C_2 \times C_2$. Furthermore, every such permutation gives rise to an automorphism.]

[Papapetros Vaggelis]

Thank you mr.Demetres.

Here is a more analytical solution about \(\displaystyle{\left(\rm{Aut}(V_{4}),\circ\right)\simeq \left(S_{3},\circ\right)}\)

where \(\displaystyle{V_{4}=C_{2}\times C_{2}=\left\{e,a,b,c\right\}}\) is \(\displaystyle{\rm{Klein's}}\) group.

Let \(\displaystyle{X=\left\{a,b,c\right\}\subseteq V_{4}}\) .

We define a left action

\(\displaystyle{\cdot: \rm{Aut}(V_{4})\times \left\{a,b,c\right\}\longrightarrow \left\{a,b,c\right\}\,,\left(f,x\right)\mapsto f\cdot x=f(x)}\)

Therefore, there exists a homomorphism \(\displaystyle{\rho:\rm{Aut}(V_{4})\longrightarrow S(X)}\) which is also

onto \(\displaystyle{S(X)}\) because, if \(\displaystyle{\sigma\in S(X)}\) , then we define \(\displaystyle{f:V_{4}\longrightarrow V_{4}}\)

by \(\displaystyle{f(e)=e\,,f(a)=\sigma(a)\,,f(b)=\sigma(b)\,\,,f(c)=\sigma(c)}\) and then \(\displaystyle{f\in\rm{Aut}(V_{4})}\)

with \(\displaystyle{\rho(f)=\sigma}\) .

Futhermore, since \(\displaystyle{f(e)=e\,,\forall\,f\in\rm{Aut}(V_{4})}\), we get :

\(\displaystyle{\begin{aligned} \rm{Ker}(\rho)&=\left\{f\in\rm{Aut}(V_{4}): \rho(f)=Id_{X}\right\}\\&=\left\{f\in\rm{Aut}(V_{4}): \rho(f)(x)=x\,,\forall\,x\in X\right\}\\&=\left\{f\in\rm{Aut}(V_{4}): f(x)=x\,,\forall\,x\in V_{4}\right\}\\&=\left\{Id_{V_{4}}\right\}\end{aligned}}\)

So,

\(\displaystyle{\left(\rm{Aut}(V_{4}),\circ\right)\simeq \left(S(X),\circ\right)\simeq \left(S_{3},\circ\right)}\)

and the group \(\displaystyle{\left(S_{3},\circ\right)}\) is not cyclic since it does not contain elements of order \(\displaystyle{6=|S_{3}|}\) .