Higman embedding theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T02:48:20Z http://mathoverflow.net/feeds/question/73076 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73076/higman-embedding-theorem Higman embedding theorem Mustafa Gokhan Benli 2011-08-17T17:11:56Z 2011-08-18T07:40:48Z <p><a href="http://en.wikipedia.org/wiki/Higman%2527s_embedding_theorem" rel="nofollow">The Higman Embedding theorem</a> says that any finitely generated and recursively presented group can be embedded in a finitely presented group.</p> <p>My question is if one can embed such a group as <strong>a normal subgroup</strong> into a finitely presented group?</p> http://mathoverflow.net/questions/73076/higman-embedding-theorem/73078#73078 Answer by Mark Sapir for Higman embedding theorem Mark Sapir 2011-08-17T17:23:35Z 2011-08-18T07:40:48Z <p>No. Take a f.g. non-finitely presented group $G$ with trivial $Out(G)$ and trivial center (such groups clearly exist; in fact one can even assume that $Out(G)$ is locally finite, say $G$ is the Grigorchuk group of intermediate growth, by <a href="http://www.math.tamu.edu/~grigorch/publications/sidkigrig3.pdf" rel="nofollow">the result</a> of Grigorchuk and Sidki). Suppose that $G$ is a normal subgroup of $H$. Then every $h\in H$ acts on $G$ by conjugation $x\to x^h$. Since $Aut(G)=Inn(G)$, there exists $g\in G$ such that $x^g=x^h$ for every $x\in G$. Hence $g^{-1}h$ centralizes $G$. Therefore $H=G Z_H(G)$ (if $Out(G)$ is locally finite, then instead of $G$ here you will get a finite extension of $G$, the image of $H$ in $Aut(G)$ under the natural homomorphism) where $Z_H(G)$, the centralizer of $G$ in $H$, intersects $G$ trivially (since the center of $G$ is trivial) and is a normal subgroup. Hence $H$ is a direct product of $G$ and $Z=Z_H(G)$. Since $H$ is finitely generated, $Z$ is finitely generated (being a homomorphic image of $H$). Hence if $H$ is finitely presented, then $G$ is presented by the presentation of $H$ plus finitely many relations killing the generators of $Z$, so $G$ must be finitely presented, a contradiction. </p>