Just my curiosity... Are there proofs the following which does not involve Hall's matching theorem:

A group $\Gamma$ is amenable if and only if it does not admit a paradoxical decomposition.

Def: A group $\Gamma$ has a paradoxical decomposition if there are pairwise disjoint subsets $F_1,\ldots, F_n$, $E_1,\ldots, E_m$ of $\Gamma$, and elements $g_1,\ldots g_n$, $h_1,\ldots,h_m\in \Gamma$ such that $\Gamma$ can be expressed as $$\Gamma= \bigsqcup_{i} g_i F_i= \bigsqcup_{j} h_j E_j$$

Just my curiosity... Are there proofs the following fact, which does not involve Hall's matching theorem:

A group $\Gamma$ is amenable if and only if it does not admit a paradoxical decomposition.

Def: A group $\Gamma$ has a paradoxical decomposition if there are pairwise disjoint subsets $F_1,\ldots, F_n$, $E_1,\ldots, E_m$ of $\Gamma$, and elements $g_1,\ldots g_n$, $h_1,\ldots,h_m\in \Gamma$ such that $\Gamma$ can be expressed as $$\Gamma= \bigsqcup_{i} g_i F_i= \bigsqcup_{j} h_j E_j$$

just my curiosity... are there proofs the following which does not involve Hall's matching theorem:

A group $\Gamma$ is amenable if and only if it does not admit a paradoxical decomposition.

Def: A group $\Gamma$ has a paradoxical decomposition if there are pairwise disjoint subsets $F_1,\ldots, F_n$, $E_1,\ldots, E_m$ of $\Gamma$, and elements $g_1,\ldots g_n$, $h_1,\ldots,h_m\in \Gamma$ such that $\Gamma$ can be expressed as $$\Gamma= \bigsqcup_{i} g_i F_i= \bigsqcup_{j} h_j E_j$$

Just my curiosity... Are there proofs the following which does not involve Hall's matching theorem:

A group $\Gamma$ is amenable if and only if it does not admit a paradoxical decomposition.

Def: A group $\Gamma$ has a paradoxical decomposition if there are pairwise disjoint subsets $F_1,\ldots, F_n$, $E_1,\ldots, E_m$ of $\Gamma$, and elements $g_1,\ldots g_n$, $h_1,\ldots,h_m\in \Gamma$ such that $\Gamma$ can be expressed as $$\Gamma= \bigsqcup_{i} g_i F_i= \bigsqcup_{j} h_j E_j$$