# Extension of surface group by cyclic is residually finite

Let $$G$$ be the fundamental group of a surface, and consider an extension $$1 \to \mathbb{Z}/p\mathbb{Z} \to E \to G \to 1$$. Is $$E$$ residually finite?

I'm interested in proving the injectivity of the map $$\mathbb{Z}/p\mathbb{Z} \to \widehat{E_p}$$, where the latter group is the pro-$$p$$ completion of $$E$$. I know that this map is injective (at least for some surface groups $$G$$, for example when the surface is orientable with genus $$g \geq 1$$), but I'm not really sure how to proceed, and this seems the most promissing way.

• It is even linear. – Moishe Kohan Mar 8 at 22:17
• @MoisheCohen $G$ is linear, but why would $E$ be linear? In general, extensions of linear groups doesn't seem to be linear. Is there a reference for this? – Henrique Augusto Souza Mar 9 at 23:40

The hint for your group is to notice that the presentation of this group (after passing to a finite index subgroup if needed) has the form $$\langle t, a_1,b_1,...,a_g,b_g| [a_1,b_1]\cdots[a_g,b_g]t, [a_i,t], [b_j,t], t^q, 1\le i, j\le g\rangle.$$ Take the quotient by the normal closure of the subgroup generated by $$a_2, b_2,...,a_g,b_g$$. The result is a 2-generated nilpotent group to which the center of your group maps injectively. Now, use the fact that finitely generated nilpotent groups are linear. Hence, you obtain an (unfaithful) linear representation of your group such that the center maps injectively. The rest you should be able to figure out.
• Isn't this presentation only for orientable surfaces? If the surface isn't orientable and has genus $g \geq 2$, isn't it's presentation $\langle x_1,\cdots,x_g \mid x_1^2\cdots x_g^2 \rangle$? – Henrique Augusto Souza Mar 11 at 14:40