Igor Protasov, Serhii Slobodianiuk
September 14, 2016

### Abstract

Let G , H be groups and let κ${{\kappa}}$ be a cardinal. A bijection f:G→H${f:G\to H}$ is called an asymorphism if, for any X∈[G]<κ${X\in[G]^{<{\kappa}}}$, Y∈[H]<κ${Y\in[H]^{<{\kappa}}}$, there exist X′∈[G]<κ${X^{\prime}\in[G]^{<{\kappa}}}$, Y′∈[H]<κ${Y^{\prime}\in[H]^{<{\kappa}}}$ such that for all x∈G${x\in G}$ and y∈H${y\in H}$, we have f(Xx)⊆Y′f(x)${f(Xx)\subseteq Y^{\prime}f(x)}$, f-1(Yy)⊆X′f-1(y)${f^{-1}(Yy)\subseteq X^{\prime}f^{-1}(y)}$. For a set S , [S]<κ${[S]^{<{\kappa}}}$ denotes the set {S′⊆S:|S′|<κ}${\{S^{\prime}\subseteq S:|S^{\prime}|<{\kappa}\}}$. Let κ${{\kappa}}$ and γ be cardinals such that ℵ0<κ≤γ${\aleph_{0}<{\kappa}\leq\gamma}$. We prove that any two Abelian groups of cardinality γ are κ${{\kappa}}$-asymorphic, but the free group of rank γ is not κ${{\kappa}}$-asymorphic to an Abelian group provided that either κ<γ${{\kappa}<\gamma}$ or κ=γ${{\kappa}=\gamma}$ and κ${{\kappa}}$ is a singular cardinal. It is known [9] that if γ=κ${\gamma={\kappa}}$ and κ${{\kappa}}$ is regular, then any two groups of cardinality κ${{\kappa}}$ are κ${{\kappa}}$-asymorphic.