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Let $G$ be a discrete group. Consider the Banach $\mathbb{Z}_p$-algebra: $$c_0(G, \mathbb{Z}_p) = \{ F : G \to \mathbb{Z}_p \mid \lim_{g \to \infty} |F(g)|_p = 0 \}$$ with the product given by the convolution: $$(F \ast G)(h) = \sum_{g \in G} F(g)G(hg^{-1})$$ When does an element of this algebra have a left inverse? Is this algebra well-studied? I am searching for references on these algebras.

P.S. $\lim_{g \to \infty} |F(g)|_p = 0 $ simply means that for every $\varepsilon > 0$, there is a finite subset of $G$ such that outside of this subset, $|F(g)|_p < \varepsilon$.

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    $\begingroup$ If I am not mistaken, this is the $p$-adic completion of the group algebra $\mathbf{Z}[G]$ (or $\mathbf{Z}_p[G]$). An element should have a left inverse if and only if its image in $\mathbf{F}_p[G]$ does: indeed, the assumption signifies that $yx=1+pz$ for some $y$ and $z$, but then the power series $(1+pz)^{-1} = \sum p^i z^i$ converges and we can multiply both sides of the equality by it to get a left inverse of $x$. $\endgroup$
    – Piotr Achinger
    Commented Oct 11, 2023 at 18:43
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    $\begingroup$ I took the liberty of adding some top-level tags which seem more likely to attract comments/answers from experts $\endgroup$
    – Yemon Choi
    Commented Oct 11, 2023 at 19:42
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    $\begingroup$ A small remark: for a general group the question of identifying units in $R[G]$ for any domain $R$ is old and notoriously hard; for $R = \Bbb Z$ it has a name of Kaplansky's conjecture (conjecture is that units in integral group ring of a torsion-free group are exactly group elements times +-1). It's not much easier for p-adics. $\endgroup$
    – Denis T
    Commented Oct 11, 2023 at 20:37
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    $\begingroup$ The formula for convolution is wrong, it should be $$(F \ast G)(h) = \sum_{g \in G} F(g)G(g^{-1}h)$$ $\endgroup$
    – Alcides Buss
    Commented Oct 26, 2023 at 9:35
  • $\begingroup$ You're right, edited. @AlcidesBuss $\endgroup$
    – Luiz Felipe Garcia
    Commented Oct 26, 2023 at 13:37

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