If a subgroup of a surface group surjects onto first homology, does it generate the surface group?

Question

Let $\Sigma_g$ denote a closed genus-$g$ surface, write $G = \pi_1(\Sigma_g)$, and consider a subgroup $H \le G$. Suppose we know that $H$ generates the abelianzation $G^\text{ab} = H_1(\Sigma_g)$. Does it follow that $H$ generates $G$?

For an arbitrary group this is false (e.g. when $G^\text{ab}$ is trivial), but I wonder if the hypothesis that $G$ is a surface group saves us.

Answer 1

In fact the question has a negative answer (credit to Trent Lucas). It's interesting to ask what hypotheses on $H$ give the question a positive answer:

We first describe how to construct such a subgroup. Recall that $A_5 = \langle (1,2,3), (3,4,5) \rangle$, and every $3$-cycle is a commutator: $(1,2,3) = [x_1, y_1]$ and $(3, 4, 5) = [x_2, y_2]$ where $x_i, y_i \in A_5$. Let $G = \pi_1(\Sigma_4)$ have presentation $\langle a_1, b_1, \dots, a_4, b_4 : [a_1, b_1]\dots[a_4, b_4] = 1 \rangle$. Consider the homomorphism $\phi : G \to A_5$ that sends: $a_1 \mapsto x_1, b_1 \mapsto y_1, a_2 \mapsto x_2, b_2 \mapsto y_2$, and the other $a_i$ and $b_i$ to elements of $A_5$ such that $\phi([a_1, b_1]\dots[a_4, b_4]) = 1$. (I.e., make sure $\phi([a_3, b_3]) = \phi([a_2, b_2])^{-1}$ and so on.)

Write $\pi: G \to G^{\text{ab}}$ for the abelianization.

Claim: $\pi: \ker(\phi) \to G^{\text{ab}}$ is a surjection.

Since $\ker(\phi)$ is a proper subgroup of $G$ ($a_1$ is not killed), we conclude that $\ker(\phi)$ is a proper subgroup of a surface group $G$ that surjects onto its abelianization $G^{\text{ab}}$.

Proof of claim: We show that the class $[a_1] \in G^{\text{ab}}$ has a $\pi$-preimage in $\ker(\phi)$; the other cases follow from a similar argument. Observe that $\phi: [G, G] \to A_5$ is a surjection, as both generators $(1, 2, 3)$ and $(3, 4, 5)$ have commutators $[a_1, b_1]$ and $[a_2, b_2]$ as preimages. Recalling that $\phi(a_1) = x_1$, choose a word $w \in [G, G]$ that maps to $x_1$. Now observe that

1. $\phi(a_1 w^{-1}) = x_1 x_1^{-1} = 1$, i.e., $a_1 w^{-1} \in \ker(\phi)$, and
2. $[a_1 w^{-1}] = [a_1]$, since $w^{-1} \in [G, G]$.

We conclude that $[a_1]$ has a $\pi$-preimage in $\ker(\phi)$ as desired.

Answer 2

Here's a simple construction of a counterexample. I'll describe it in genus $2$ but it easily translates to higher genus.

Consider the standard presentation $$\pi_1(\Sigma_2) \approx \langle a, b, c, d \mid aba^{-1}b^{-1}cdc^{-1}d^{-1} \rangle $$ Now consider the words \begin{align*} w_a &= aba^{-2}b^{-1}a \\ w_b &= b a b^{-2} a^{-1} b \\ w_c &= c d c^{-2} d^{-1} c \\ w_d &= d c d^{-2} c^{-1} d \end{align*} Notice that for each of $x=a,b,c,d$, the elements $x$ and $xw_x$ have equal images in the abelianization. So the subgroup $H$ that is generated by $A = aw_a^{10}$, $B = bw_b^{10}$, $C = cw_c^{10}$, $D = dw_d^{10}$ has the same image in the abelianization as the whole group, that is to say that $H$ maps surjectively to the abelianization.

However the four elements $A,B,C,D$ form a free basis for a rank $4$ free subgroup of $\pi_1(\Sigma_2)$, and this subgroup has infinite index --- in general, every free subgroup of $\pi_1(\Sigma_g)$ has infinite index.

By the way, the reason for taking the $10^{\text{th}}$ power in the formulas for $A,B,C,D$ is to make it obvious that $A,B,C,D$ form a free basis for a rank $4$ free subgroup: when you take any nontrivial reduced word in the letters $A,B,C,D$, and then substitute the defining expressions $A=a w_a^{10}$ and so on, you can use Dehn's Algorithm to visually check that the word represents a nontrivial element of $\pi_1(\Sigma_2)$. I could have put any positive integer in place of that $10$ and the same would hold.