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

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.

• cool question. definitely seems true! Commented Jun 20, 2023 at 23:01
• We agree, but we’ve tried for a few days and can’t get the proof! Any thoughts? Commented Jun 20, 2023 at 23:19
• What does H generate G mean ? If H is a subgroup, it should mean H=G. Commented Jun 22, 2023 at 6:55
• @AndresMejia It's indeed true, but proof is not entirely trivial; for example, inclusion of free subgroup generated by $b, c$ in $\langle a, b, c \vert a^{-1}[c, a][c, b]\rangle$ not only induced isomorphism on abelianization, but also on all lower central quotients. Reason why it's nevertheless true is that we have full classification of subgroups; they are either finite index surface subgroups, or infinite index free. Maybe I'll write up a proof in a day if I have time.
– xsnl
Commented Jun 23, 2023 at 13:10
• @xsnl I would appreciate if you did write up your proof Commented Jun 23, 2023 at 16:23

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.

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.