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Let $G$ and $\overline G$ be groups, and $f\colon G\to\overline G$ a bijection. Then: $$\forall g,h\in G: f(gh) = f(g)f(h) \tag1$$ if and only if: $$\mathfrak c=\psi_f\bar{\mathfrak c}f \tag2$$ where $\mathfrak c$ and $\bar{\mathfrak c}$ are Cayley's embedding of $G$ and $\overline G$ into $\operatorname{Sym}(G)$ and $\operatorname{Sym}(\overline G)$, respectively, and $\psi_f\colon\operatorname{Sym}(\overline G)\to\operatorname{Sym}(G)$ is defined by $\sigma\mapsto f^{-1}\sigma f$.

Question. Is there any weaker (than $(1)$) condition on the bijection $f$, which ensures the weaker (than $(2)$) result: $$\operatorname{im}(\mathfrak c)=\operatorname{im}(\psi_f\bar{\mathfrak c}f) \tag3$$ (here "$\operatorname{im}(\_)$" stands for "the image of")?

Motivation. Intuitively, I'd expect $(3)$ being sufficient to get $G$ and $\overline G$ "somehow isomorphic".

citadel
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1 Answers1

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Let us prove that $(3)$ is equivalent to $$\exists k\in\overline G,\forall g,h\in G:f(gh)=f(g)kf(h)\tag4$$

To begin with, $$(3)\iff A_f=B_f$$ where $A_f$ and $B_f$ are the two following sets of maps from $G$ to $\overline G:$ $$A_f:=\{h\mapsto f(g)f(h)\mid g\in G\}=\{h\mapsto vf(h)\mid v\in\overline G\},$$$$B_f:=\{h\mapsto f(gh)\mid g\in G\}.$$

  • If $A_f=B_f$ then $B_f\subset A_f,$ i.e. $$\forall g\in G,\exists v\in\overline G,\forall h\in G:f(gh)=vf(h),$$ which is equivalent to $$\forall g\in G,\forall h\in G:f(gh)f(h)^{-1}=f(ge)f(e)^{-1}$$ where $e$ denotes the identity element of $G.$

    Letting $k:=\left(f(e)\right)^{-1},$ this implies $(4).$

  • Conversely, assume $(4).$ Then, $$\begin{align}B_f&=\{h\mapsto f(g)kf(h)\mid g\in G\}\\ &=\{h\mapsto ukf(h)\mid u\in\overline G\}\\&=A_f, \end{align}$$ which ends the proof.

Edit, to answer today's comment below:

Let us prove that if $f:G\to\overline G$ is a bijection satisfying $(4)$ then, there exists some $c\in G$ such that for every $\bar g,\bar h\in\overline G:$ $$f^{-1}(\bar g\bar h)=f^{-1}(\bar g)cf^{-1}(\bar h)$$ i.e. $$\bar g\bar h=f\left(f^{-1}(\bar g)cf^{-1}(\bar h)\right),$$ which, according to $(4),$ is equivalent to $$\bar g\bar h=\bar gkf(c)k\bar h.$$ The solution is $c=f^{-1}(k^{-2}).$ Note that it satisfies the dual identity $k=f(c^{-2}),$ since $f(c^{-2})=f(c^{-1})kf(c^{-1})=k,$ because (similarly to $k=\left(f(e)\right)^{-1}$ above) the property of $c$ makes it necessarily equal to $\left(f^{-1}(\bar e)\right)^{-1}$ where $\bar e$ is the identity element of $\overline G.$

Anne Bauval
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  • As opposedly to true isomorphisms, the inverse of a bijection $f$ with the property $(4)$ does not fulfill the same property. – citadel Jul 12 '23 at 02:14
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    Ah ok, I had missed that, thanks. So, seemingly we can think of a "quasi-isomorphism" relation, say $G\approx \overline G$, between possibly non-isomorphic groups ($G\not\cong\overline G$)? I think I proved already that, e.g., still $C_4\not\approx K_4$, but perhaps other cases work. – citadel Jul 12 '23 at 09:12