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Let $F$ be a free group of rank $\aleph_0$ and let $H$ be a subgroup for which there exists some $a \in F$ such that $\langle H \cup \{a\} \rangle = F$. Must there be some free basis $B$ for $F$ and a subset $B_0 \subseteq B$ for which $\langle H \cup B_0 \rangle = F$ and $|B_0|$ is bounded by some absolute finite constant $K$?

Can we take $K = 1$?

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    $\begingroup$ I don't think you are asking what you mean to ask, because as written, you can just take $B$ to be an arbitrary free basis, and $B_0=B$, and $K=\aleph_0$. I suspect you want $B_0$ finite, for one thing... $\endgroup$
    – Arturo Magidin
    Commented May 21, 2015 at 18:11
  • $\begingroup$ @ArturoMagidin thanks. I meant that $K$ is finite. $\endgroup$
    – Pablo
    Commented May 21, 2015 at 18:23

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