1

Is a quotient of a finitely presentable group $G$ by a subgroup $N$ necessarily finitely presentable? What about if the subgroup $N$ is also finitely presentable?

Shaun
  • 47,747
user101010
  • 5,491

2 Answers2

5

Since there exist finitely generated groups without finite presentations, the answer to your first question is clearly no, since free groups of finite rank are finitely presentable.

Mariano Suárez-Álvarez
  • 139,300
4

If $N$ is finitely generated, then you can just add the generators to the relation set of $G$ to get $G/N$, which will be a finite presentation. (after all a presentation is just a list of generators and things which normally generate some normal subgroup).

And Mariano's answer is right, although it isn't completely trivial to construct/show a group is infinitely presented, but finitely generated. Some well known groups are $B \wr A$ (restricted wreath product) where $B$ is nontrivial and $A$ is infinite. Another way to construct infinitely presented groups is through small cancellation theory.

  • 2
    Examples were discussed at MSE at least twice: http://math.stackexchange.com/questions/547087/finitely-generated-group-which-is-not-finitely-presented?rq=1, http://math.stackexchange.com/questions/1323510/is-every-normal-subgroup-of-a-finitely-generated-free-group-a-normal-closure-of?noredirect=1&lq=1. Personally, the simplest example and a proof that I know is the amalgam $G=F_2 \star_{F_\infty} F_2$. Then $b_2(G)=\infty$ by the Mayer-Vietoris sequence, hence, $G$ is not finitely presentable. – Moishe Kohan Dec 03 '16 at 05:08