If $G$ is non-Abelian, show that $Aut(G)$ is not cyclic.

Question

I can see that $G$ non-Abelian $\Rightarrow$ $G / Z(G)$ is not cyclic.

And $G / Z(G)$ is isomorphic to $Inn(G) \Rightarrow$ $Inn(G)$ is not cyclic.

But from here I can't see how I'd relate $Inn(G)$ and $Aut(G)$, anyone have any ideas?

$ \textrm{Inn}(G) $ is a subgroup of $ \textrm{Aut}(G) $. Can a cyclic group have a non-cyclic subgroup? — Ege Erdil, Aug 20 '16 at 13:13

score 2 · Answer 1 · answered Dec 20 '17 at 09:03

2

Suppose $Aut(G)$ is cyclic, then so is its subgroup $Inn(G)$. But then $G/Z(G)\cong Inn(G)$, so $G/Z(G)$ is cyclic, and so $G$ is abelian.

answered Dec 20 '17 at 09:03

score 1 · Answer 2 · answered Aug 20 '16 at 13:28

Suppose that $Aut(G)$ is cyclic and let $f$ its generator, let $x,y\in G$ such that $y$ does not commute with $yxy^{-1}$, (since $G$ is not commutative, you can find $z$ such that $z$ does not commute with $y$, write $x=y^{-1}zy, yxy^{-1}=z$) there exists $n$ such that $c_x(z)=xzx^{-1}=f^n, m: c_y(z)=yzy^{-1}=f^m$, this implies that $c_x$ commutes with $c_y$, we deduce that $c_y(c_x(x))=c_x(x))$, i.e

$yxy^{-1}=xyxy^{-1}x^{-1}$, i.e $(yxy^{-1})x=x(yxy^{-1})$ i.e $x$ commutes with $yxy^{-1}$. Contradiction.

I do not understand when you said that "This implies that $c_x$ commutes with $c_y$" do you have an explanation to this please? — , Oct 23 '21 at 16:55

If $G$ is non-Abelian, show that $Aut(G)$ is not cyclic.

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