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