I`m supposed to solve a question for my geometric group theory course that states:
Suppose $H$ subgroup of the group $G$, with a retraction from $G$ to $H$ and $G$ is finitely presentable. Show that $H$ is finitely presentable.
I found a way to go about this described in the book Invitations to Geometry and Topology edited by Simon Salamon and Martin Bridson. See here.
They give the following steps:
Show $H$ is finitely generated(Done).
Take a finite generating set for $H$ and expand it to a finite generating set for $G$ by adding elements from the kernel of the retraction(Done).
Argue that one can take a finite presentation with this generating set for $G$(Not done, here is my problem).
Add the generators from the kernel to the relations to arrive at a finite presentation for $H$(Seems straightforward).
So my question is:
For point 3, do I need to construct something or is it true that if you have a finitely presentable group and a finite generating set that you can find finitely many accompanying relations to make it into a finite presentation? If so, why, if not, why also :)?
Thanks in advance.
P.s., my first post here, hope it`s all up to the standard
To answer point 3, if you already have a finite presentation $G=\langle g_1,...,g_m \mid r_1,...,r_n\rangle$, and you then add some new elements $h_1,...,h_p \in G$ to the generating set, then to get a presentation of $G$ all you need to do is add some new relators to the original presentation for $G$, one new relator for each new generator $h_i$, of the form $$h_i^{-1} w_i(g_1,..,g_m) $$ where $w_i(g_1,..,g_m)$ expresses $h_i$ as a word in the original generators of $G$.