# Given a conjugation by $g$ when and how can one determine $g$?

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 (?!).
• $M$ is the group ${\rm Inn}(G)$ of inner automorphisms of $G$, and the map $f$ has kernel the centre $Z(G)$ of $G$, so $G/Z(G) \cong {\rm Inn}(G)$. – Derek Holt Mar 9 at 9:45

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 $$G/Z(G)$$