a question on rank of fundamental group

Assume $G$ be the fundamental group of a closed orientable hyperbolic 3-manifold. Let $G_{1} = \langle a_{1},...,a_{k} \rangle$ be a free subgroup of $G$, and let $G_{2}=\langle a_{k+1} \rangle$ be a free cyclic subgroup of $G$.

My question is: If the set $\{a_{1},...,a_{k}, a_{k+1}\}$ generates $G$, does the rank of $G$ then satisfy $r(G)\geq k$?

• This doesn't answer your question, but you might be interested in the Schreier index formula, which roughly says, that finite index subgroups of a free group on $n$ generators are free of rank $> n$. The smaller the subgroup is (as long as it remains finite index), the larger its rank. Of course infinite index subgroups can have smaller rank. May 10 '14 at 18:54

Suppose that $M$ fibers over circle; $G=\pi_1(M)\cong P_g \rtimes <a>$, where $P_g$ is the geneus $g$ surface group. Then for each sufficiently large $k$, you can find a free rank $k$ subgroup $F_k<P_g<G$ such that $F_k$, and $a$ generate $G$.

Edit: Let $D$ be a generating set of $P_g$. Partition $D$ in two disjoint nonempty subsets $D=B\sqcup C$. Then for all sufficiently large $n$, the group generated by $$a^nBa^{-n}, a^{-n}Ca^n$$ is free (this follows from Klein combination and the fact that conjugation by $a$ corresponds to a pseudo-Anosov homeomorphism of the surface). This is your subgroup $F_k$.

Thus, the answer to your question is negative. In view of the virtual fibration conjecture, the same will be probably true for all compact hyperbolic 3-manifolds.

• Would you mind explain these words: you can find a free rank k subgroup Fk<Pg<G such that Fk, and a generate G May 10 '14 at 3:52

Complementing Misha's answer, the following is true for any non-(virtually cyclic) hyperbolic group $G$:

Claim: for any sufficiently large $k \in \mathbb{N}$ there is a free quasiconvex subgroup $H_k \le G$ of rank $k$ and an infinite order element $g \in G$ such that $G=\langle H_k,g \rangle$.

By sufficently large, I mean that $k \ge K-1$, where $K$ is the smallest integer such that $G$ can be generated by $K$ elements of infinite order. Thus $K=rank(G)$ if $G$ is torsion-free. It is not difficult to show that $K \le rank(G)+1$ in general.

The above claim is especially easy to prove if $G$ is torsion-free and is generated by two elements $h,g \in G$.

Indeed, since the subgroup $H_1:=\langle h \rangle$ is quasiconvex and has infinite index in $G$, by a theorem of Goulnara Arzhantseva ["On quasiconvex subgroups of word hyperbolic groups", Geometriae Dedicata, 87 (2001), 191-208.], there exists $h_2\in G\setminus\{1\}$ such that the subgroup $H_2:=\langle H,h_2\rangle\le G$ is quasiconvex and is isomorphic to the free product $H*\langle h_2\rangle$, which is evidently a free group of rank $2$. Clearly, $G=\langle H_2,g \rangle$. Now, without loss of generality, we can assume that $|G:H_2|=\infty$ (one can replace $h_2$ with its square to ensure this). Applying Arzhantseva's result again one finds $h_3 \in G\setminus\{1\}$ such that $H_3:=\langle H_2,h_3 \rangle\le G$ is free of rank $3$, quasiconvex in $G$ and has infinite index. And so on.

Of course, if $r:=rank(G) >2$ then the proof becomes more technical, as the first $r-1$ generators may not generate a free quasiconvex subgroup of infinite index. However, one can construct an argument by induction, using generalizations of Arzhantseva's theorem.

• Hi, thanks for the case of non hyperbolic group. Since I can only choose one of yours and Misha's as an answer and Misha's is more closer to my question, I choose Misha's as an answer. May 14 '14 at 5:59