What is the co-rank of a group $$G=\langle a\_1,a\_2,\dots,a\_h\mid a\_1^2a\_2^2\dots a\_h^2=1\rangle,$$ that is, finitely generated group with $h$ generators and one relation? 


By co-rank, I mean the maximum rank $n$ of a free homomorphic image $F\_n$ of the group.


I suspect that the answer is $corank\ G=\left[\frac h2\right]$. At least it is obvious that $corank\ G\ge\left[\frac h2\right]$, by mapping all odd $a\_i\in G$ to generators of $F\_n$ and all even $a\_i$ to their inverse.


Motivation: This is the fundamental group of a non-oriented surface $N\_h$. For orientable surface $\Sigma g$ this value is $g$, but I need to know it for non-orientable surface $N\_h$. 


$corank\ G$ is the cut-number of a non-orientable surface $N\_h$: the number of non-intersecting two-sided circles such that cutting along them leaves the surface connected.


In case it helps, I know that abelianization ($G$ made abelian) of this group is $Z^{h-1}\oplus(Z/2Z)$. I know that this group contains a characteristic subgroup of index 2 isomorphic to the fundamental group of an orientable surface $\Sigma\_{g-1}$; the latter has co-rank $g-1$.



Co-rank of a group with $a^2b^2c^2=1$ (fundamental group of non-orientable surface)
If you know that corank is the cut number, then it is $[h/2]$. Each nonseparating two-sided circle is a handle. You can remove them one by one (remove the circle and patch the two holes with disks), each time reducing $h$ by $2$.


Or, even better, remove all $r$ circles at once and patch the resulting $2r$ holes with disks. You get Euler characteristic $1-h+2r\le2$.


