Problem 1. Let F be the free group of finite rank n ≥ 2. Suppose an endomorphism φ of F takes every primitive element to a primitive one. Is it true that φ is actually an automorphism of F?

For n = 2, the answer was proved to be “yes” by S. Ivanov.

Vladimir Shpilrain

Problem 2. (Closure operators and finite solvable groups) Let G be a finite group, let L(G) be its lattice of subgroups and let σ : L → L(G) be a closure operator on L(G). This means that:

1. (i) if H, K ≤ G, then σ(H) ≤ σ(K);

2. (ii) if H ≤ G, then H ≤ σ(H); and

3. (iii) if H ≤ G, then σ(H) = σ(σ(H)).

Assume now that σ is a non–trivial closure operator, i.e. σ can be defined for every L(G), where G is a finite group and there exists a finite group X such that σ : L(X) → (X) is not trivial. The following problem arises: if σ : L(G) → L(G) is as above and if every subgroup of G is closed, i.e. σ(H) = H, ∀H ≤ G, is it true that G is solvable?

Arguments for the “yes” answer:

1. If σ(H) := HG = 〈Hg|g ∈ G〉 and if σ(H) = H, ∀H ≤ G, then G is Dedekind, so nilpotent.

2. If σ(H) := CG(CG(H)) and if σ(H) = H, ∀H, ≤ G, then G is supersolvable (Gaschütz).

3. If σ(H) := ∩{M|M is a maximal subgroup of G and H ≤ M} and if σ(H) = H, ∀H ≤ G, then G is supersolvable (Menegazzo, di Martino et al.)