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Prove that $|\operatorname{Inn}(G)| = 1 \Rightarrow G$ is Abelian.

Since $|\operatorname{Inn}(G)| = |\{\phi_e, \phi_{a_1}, \phi_{a_2},\ldots:$ such that $a_i \in G$, and $\phi_i(x)$ is an inner automorphism$\}| = 1$ $\Rightarrow \operatorname{Inn}(G) = \{\phi_e\}$

$\Rightarrow \phi_e = \phi_a$ for all $a_i$ in $G$

$\Rightarrow \phi_y(x) = yxy^{-1} = \phi_e(x) = x$

$\Rightarrow yxy^{-1} = x$

$\Rightarrow yx = xy$

Does this make sense, or am I miss-interpreting the definition of $\operatorname{Inn}(G)$?

Michael Hardy
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Oliver G
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    It does make sense and it looks correct to me. Perhaps a little more order is needed in the exposition, but it is just fine. – DonAntonio Aug 11 '16 at 17:25
  • More generally, if the inner-automorphism group of a group $G$ is cyclic, then $G$ is abelian. – Batominovski Aug 11 '16 at 17:38
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    Alternative: take a look at the map $g\mapsto\phi_g$ to find out that $G/z(G)$ and $Inn(G)$ are isomorphic. The kernel of the map is $z(G)$ and it is surjective. – drhab Aug 11 '16 at 18:04

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