By A Baudisch, D Seese, HP Tuschik, M Weese

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G := g∈G Note that Gg = g G = G for all g ∈ G, hence G is central in R[G]. It is easy to see that if G is ﬁnite, then ann(ω) = RG; cf. 2]. The following observation is standard (cf. [Pa77, p. 4. Suppose G is a p-group, and R is a ﬁeld of characteristic p. Then every non-zero ideal I of R[G] contains G. Proof. Let I = 0 be an ideal of R[G]. The ideal ω is nilpotent, so we can choose an integer m ≥ 0 maximal with Iω m = 0. Then clearly 0 = Iω m ⊆ I ∩ ann(ω) = I ∩ RG and thus G ∈ I since R is a ﬁeld.

4. Let H be a normal subgroup of G such that G/H satisﬁes P. Then π1 (G/H) is residually P, where H = {H ∩ Gv }v∈V (Y ) . Proof. 3 (the proof of which didn’t need ﬁniteness of the Gv /Hv ). Digression: trees of elementary abelian p-groups. Suppose the underlying graph Y of G is a tree, and ﬁx a vertex v0 of Y . Assume that for each edge e of Y , the morphism fe ◦ fe−1 extends to an isomorphism ϕe : Go(e) → Gt(e) , such that ϕ−1 = ϕe for all e. For each vertex v of Y we then have an isomorphism e ϕv : Gv → Gv0 given as the composition ϕv := ϕen ◦ ϕen−1 ◦ · · · ◦ ϕe1 where (e1 , .

We t(e) = v2 . Let ni be the number of components of πv−1 i get the cover X → X by taking lcm(n1 , n2 )/n1 copies of Xv1 and lcm(n1 , n2 )/n2 copies of Xv2 and by connecting each copy of a component of πv−1 (fe (Xe )) with a 1 copy of a component of πv−1 (f (X )) via a copy of X . e e e 2 We now turn to the case of graphs with more than one topological edge. Since Y is connected, either there is an edge e such that the graph Y \ {e} obtained by deleting the edge e from Y is connected, or there is an edge e such that one of the components of Y \ {e} consists of a single vertex with no topological edge.

### Decidability and Generalized Quantifiers by A Baudisch, D Seese, HP Tuschik, M Weese

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