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Let $G$ be a group and $f_g : G \to G, f_g(h) = ghg^{-1}$ be conjugation by $g$. Let $M = \{f_g|g\in G\}$ be the set of all conjugations.

Given $t \in M$, what can be said about "extracting" $g$? (That is finding the element $g$ such that $t(g)(h) = ghg^{-1}$.)

So far I have convinced myself of the following:

  • Let $f : G \to M, f(g) = f_g$. Then $f$ is a homomorphism and I am looking for $f^{-1}$ (the inverse of $f$).
  • I suspect the inverse does not exist in general but might exist for some groups. I wonder if something more could be said though.
  • In the case of $\tilde{f}_g(h)=gh$, I could "extract" $g$ with the neutral element $e$, since $\tilde{f}_g(e) = g$, similarly for $g^{-1}$ one has $f_g(g^{-1}) = g^{-1}$, that is if I have a way of "testing" all elements of $G$ I could identify $g$ by searching for a fixpoint (?!).
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    $\begingroup$ $M$ is the group ${\rm Inn}(G)$ of inner automorphisms of $G$, and the map $f$ has kernel the centre $Z(G)$ of $G$, so $G/Z(G) \cong {\rm Inn}(G)$. $\endgroup$ – Derek Holt Mar 9 at 9:45
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You cannot find $g$ from $f_g$ in general because if $g$ is in the center of $G$, so $gh=hg$ for all $h\in G$, then you have that $f_g(h)=ghg^{-1}=hgg^{-1}=h$ so $f_g=\textrm{Id}_G$ and the identity of $G$ has not information about the element $g$ that induces $f_g$.

In general you cannot find $g$ also if $g$ is not in the center of $G$ because, for example, if you consider $r\in Z(G)$ then $f_g=f_{gr}$ so $g$ can be identified up to elements of $Z(G)$.

If $g,r\in G$ are such that $f_g=f_r$ then you can prove that $g^{-1}r\in Z(G)$ so you have that $M$ can be identify as the quotient $G/Z(G)$

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