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Suppose we have a discrete group $G$, finite or infinite, on which we form the group algebra $\mathbb{F}_2[G]$. Suppose also that we have a map $S$:

$$S : \mathbb{F}_2[G] \to \mathbb{F}_2[G]: \sum{a_g \cdot g} \to \sum{a_g \cdot g^{-1}}$$

The problem I'm trying to solve is the following : given an element $x \in \mathbb{F}_2[G]$, find all $y \in \mathbb{F}_2[G]$ such that $y \cdot S(y) = x$.

For small groups $G$, we can write down a system of equations on the $a_g$ but I cannot find a general way... any help or lead appreciated !

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This might be useless but I think the problem is equivalent to a combinatorial one because any element $\mathbb{F}_2[G]$ is entirely determined by its support. Given a finite subset $A \subset G$, then I think your problem amounts to finding a set $B$ such that for all $g \in G$, we have $$|B \cap g^{-1}B| \equiv 1_A(g) \mod 2$$ Setting $x = 1_A$ and $y = 1_B$, then $y$ is a solution to $y S(y) = x$. Also, because $S(ab) = S(b) S(a)$, solutions are unique up to right multiplication by solutions of $y S(y) = 1$ (which includes all $g \in G$). – Joel Cohen Jan 5 '12 at 20:52
When the group algebra, $\mathbb{F}[G]$, is given its standard Hopf algebra structure, $S$ is the antipode map. Maybe a text on Hopf algebras might be helpful. Wish I knew more about them :) – Bill Cook Jan 5 '12 at 21:54
@Joel Cohen: doesn't this look like a convolution of some sort ? I'm asking because convolutions also arise in the Hopf algebras Bill is talking about... however I'm surely not proficient enough in those to tell more about it. – AlexPof Jan 6 '12 at 18:52
@AlexPof : Yes, the product in the group algebra amount to a convolution product : if $x = \sum_{g \in G} a_g g$ and $y = \sum_{g \in G} b_g g$ then $xy = \sum_{g \in G} c_g g$ where $(c_g)_{g \in G}$ is the convolution product of $(a_g)_{g \in G}$ and $(b_g)_{g \in G}$ (which means we have $c_g = \sum_{h \in G} a_h b_{g^{-1}h}$). But I don't know much more either, I was merely sharing some random thoughts about the problem. – Joel Cohen Jan 6 '12 at 19:01
The map induced by $g \mapsto g^{-1}$ is called the classical involution in group ring theory. An element of a group ring is called symmetric if it is fixed under the classical involution, and a unit of a group ring is called unitary if the classical involution inverts it. Related questions were studied intensively by Victor Bovdi and others, you may find further references in MathSciNet and a collection of preprints in arXiv. Hope that may give further leads. – Alexander Konovalov Jul 1 '13 at 19:57

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