Let $F$ be a free group of countably infinite rank, and let $L \leq F$ be a finite index subgroup. For some prime $p$ and $k \in \mathbb{N}$, let $\sigma \colon L \to \mathrm{GL}_k(\mathbb{Z}/p\mathbb{Z})$ be an irreducible representation.

The representation $\sigma$ induces a left action of $L$ on $A = (\mathbb{Z}/p\mathbb{Z})^k$, making $A$ a simple elementary $p$-abelian $L$-module. Therefore the set of crossed homomorphisms $$C^1(L,A) = \{f \colon L \to A \ | \ \forall x,y \in L. \ f(xy) = f(x) + xf(y) \}$$ is defined in the usual way (as in cohomology of groups). For $f \in C^1(L,A)$ define: $$\mathrm{Ker}(f) = \{x \in L \ | \ f(x) = 0 \}$$ and note that this is a subgroup of $L$. Set: $$M = \bigcap_{f \in C^1(L,A)} \mathrm{Ker}(f)$$ and observe that this is a subgroup of $F$. My question is:

Is it possible that there exists some $a \in F$ such that $\langle M \cup \{a\} \rangle = F$ ?

I hope that the answer is negative, so let me sketch a proof in case that $\sigma = 1$ (so $k=1$). Since the action of $L$ on $A$ induced by $\sigma$ is trivial, $C^1(L,A) = \mathrm{Hom}(L,\mathbb{Z}/p\mathbb{Z})$. Thus: $$M = \bigcap_{\phi \in \mathrm{Hom}(L,\mathbb{Z}/p\mathbb{Z})} \mathrm{Ker}(\phi) \leq \bigcap_{\psi \in \mathrm{Hom}(F,\mathbb{Z}/p\mathbb{Z})} \mathrm{Ker}(\psi) \cap L \leq \bigcap_{\psi \in \mathrm{Hom}(F,\mathbb{Z}/p\mathbb{Z})} \mathrm{Ker}(\psi) = N$$ so if $\langle M \cup \{a\} \rangle = F$ then $\langle N \cup \{a\} \rangle = F$, and then $F/N$ is cyclic which is absurd.

Also, note that the case $L = F$ is not difficult.

  • $\begingroup$ When you say "I hope that the answer is negative", do you mean that you hope that "For all $L$, $p$, $k$, and $\sigma$ the answer is negative"? $\endgroup$ – Lee Mosher May 18 '15 at 23:58
  • $\begingroup$ @LeeMosher yes. I hope that for every choice of $F,L,p,k,\sigma$ with the conditions stated above, there is no $a \in F$ s.t $<M,a> = F$. $\endgroup$ – Pablo May 19 '15 at 5:41

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