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Let $G$ be a group. Is $G$ abelian if $f : G \to \mathrm{Aut}(G)$, defined by $f(g) = f_g:x \mapsto g^{-1}xg$, is a homomorphism ?

I am aware of this question, but notice that I have $g^{-1}xg$ instead of $gxg^{-1}$. My condition $f_{ab} = f_a \circ f_b$ is equivalent to : $$\forall a,b \in G, \;\forall x \in G \quad b^{-1}a^{-1}xab=a^{-1}b^{-1}xba$$ Notice that $\forall x \in G, \; c^{-1}xc = d^{-1}xd \quad\implies\quad c=d$ iff $Z(G) = \{1\}$.

I tried to work with the possible counter-example $G=D_8$. I'm not sure if there is a counter-example. At least, the converse is trivial, i.e. if $G$ is abelian then $f$ is a homomorphism.

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  • $\begingroup$ This map is always a homomorphism (in groups): $f_g(xy) = g^{-1} xy g = g^{-1}x(gg^{-1})yg=f_g(x) f_g(y)$ $\endgroup$ – Ranc Oct 13 '16 at 12:16
  • $\begingroup$ @Ranc: No, $f$ is only an antihomomorphism in general. $\endgroup$ – Alphonse Oct 13 '16 at 12:17
  • $\begingroup$ @Alphonse: What's an anti-homomorphism? Are my calculations wrong? $\endgroup$ – Ranc Oct 13 '16 at 12:19
  • $\begingroup$ @Alphonse oh i get it now. You are talking about the map: $f\colon G \rightarrow \mathrm{Aut} (G) \colon (g\mapsto \left(x\mapsto g^{-1}xg\right) ) $ $\endgroup$ – Ranc Oct 13 '16 at 12:21
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    $\begingroup$ I think your condition is equivalent to $G/Z(G)$ being abelian. So it would hold in some nonabelian grouops, such the dihedral group of order $8$. $\endgroup$ – Derek Holt Oct 13 '16 at 12:35
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Let $\varphi_g\colon x\mapsto gxg^{-1}$. Then we know that $\varphi$ defines a homomorphism $G\to\operatorname{Aut}(G)$, with image the inner automorphisms. Also the map $f$ has the same set as image. Since $f_g=\varphi_{g^{-1}}$, the statement $f_{gh}=f_g\circ f_h$ becomes $$ \varphi_{(gh)^{-1}}=\varphi_{g^{-1}}\circ\varphi_{h^{-1}} $$ or as well, since $g$ and $h$ are arbitrary, $$ \varphi_{hg}=\varphi_g\circ\varphi_h $$ This amounts to saying that $\operatorname{Inn}(G)$ is abelian. Since $\operatorname{Inn}(G)\cong G/Z(G)$, we have that the condition is equivalent to $G/Z(G)$ being abelian, that is, $G$ nilpotent with nilpotency class $\le2$.

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