# Co-rank of a group with $a^2b^2c^2=1$ (fundamental group of non-orientable surface)

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 $g-1$.

• The free homomorphic images of non-orientable surfaces were worked out in: R. I. Grigorchuk and P. F. Kurchanov. 'On quadratic equations in free groups' in Proceedings of the International Conference on Algebra, Part 1 (Novosibirsk, 1989). But, as Alex's answer makes clear, you can also work this out using elementary topology. (The case $h=3$ in the title is a famous theorem of Lyndon.)
– HJRW
Dec 29, 2014 at 14:57

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 $1-h+2r\le2$.