A note on finitely generated/related/presented groups

Group presentation is a method to express groups as a quotient of a free group and a normal subgroup. In a presentation, there are generators and relators, and the presentation yields a ‘minimalist implementation’ of these generators and relations. Some cases of special interests are finitely generated/related/presented groups. From the definition, finite generation + finite relation is not linguistically equivalent to finite presentation. There has been some discussion on the relation of the three properties for several other algebraic structures. This entry proves that finite generation + finite relation implies finite presentation for groups.

For modules, finite generation and finite relation implies finite presentation. For abelian groups (i.e., $\mathbb{Z}$-modules), finite generation implies finite relation and finite presentation.

We’ll go through the necessary definitions and then prove the following proposition:

If $\lAng{S|R'}\rAng\cong\lAng{S'|R}\rAng$ for some finite $S,R$, then both are isomorphic to $\lAng{S''|R''}\rAng$ with finite $S'',R''$. That is, a group that is both finitely generated and finitely related must be finitely presented.

Definitions

Group, free group (constructive, $F\left(S\right)$), generated (normal) subgroup ($\lAng S\rAng$ and ${\lAng S\rAng}_N$) are omitted.

Presentation Given a pair of set $S,R$, where $R\subseteq F\left(S\right)$, the group ${F\left(S\right)}/{{\lAng R\rAng}_N}$ is denoted by $\lAng{S|R}\rAng$.

Finite -tion If a group $G$ is isomorphic to some $\lAng{S|R}\rAng$ where $S$ [resp. $R$, both $S$ and $R$] is [resp. is, are] finite, $G$ is said to be finitely generated [resp. related, presented].

Remarks Here the definition of finitely generated group is different from what is usually known in the textbooks. One can regard the textbook definition as the internal version and the one here as the external version. Indeed, they are equivalent.

If $G=\lAng S\rAng$ for some finite $S\subseteq G$, define $R=\left\{s_1s_2\cdots s_n:s_1s_2\cdots s_n=1,s_i\in S\vee s_i^{-1}\in S,n\in\mathbb{N}\right\},$ then $G\cong\lAng{S|R}\rAng$.

Conversely, if $\lAng{S|R}\rAng\cong G$ for some finite $S$ via isomorphism $\varphi$, then $G=\lAng\left\{{\varphi\left(s{\lAng R\rAng}_N\right):s\in S}\right\}\rAng.$

Proof

We now prove that if a group is both finitely generated and finitely related, it is finitely presented.

Suppose $\lAng{S|R'}\rAng\cong\lAng{S'|R}\rAng$ where $S,R$ are finite. Let $\varphi:\lAng{S|R'}\rAng\to\lAng{S'|R}\rAng$ be an isomorphism. Write $N={\lAng R\rAng}_N$ and $N'={\lAng R'\rAng}_N$.

We write $\varphi\left(sN\right)=s'N'$ for each $s\in S$, where $s'$ is a finite product of elements and their inverses in $S'$. I.e., they use finitely many symbols from $S'$. For each relator in $R$, it uses only finitely many symbols from $S'$. Collect these symbols and put them into a (finite) set $T$.

We claim $S'=T$. Suppose not, and $s'\in{S'\setminus T}$. Define $f:S'\to\mathbb{Z},x\mapsto{\mathbb{I}\left[{x=s'}\right]},$ where $\mathbb{I}$ is the truth indication function, i.e., $f$ maps $s'$ to $1$ and everything else in $S'$ to $0$. (Uniquely) extend $f$ to a homomorphism $\psi:F\left(S'\right)\to\mathbb{Z}$. Note that for all $t\in T$, by definition, $\psi\left(t\right)=f\left(t\right)=0$, and since $R$ only uses symbols in $T$, we conclude $N\leq\ker\psi$. Now, $s'N=\varphi\left({\varphi^{-1}\left(s'N\right)}\right)=tN,$ where $t$ is a finite product of elements and their inverses in $T$. To get this, first write $\varphi^{-1}\left(s'N\right)$ as a coset of $N'$ translated by a finite product of elements and their inverses in $S$. Then apply $\varphi$ on each term in the product, and write the outcome as a coset of $N$ translated by a product of finite products of elements and their inverses in $T$ (by our choice). Applying $\psi$ to both sides of the equation yields $1=0$, a contradiction.

Therefore, $S'=T$ and $\lAng{S'|R}\rAng$ is a finite presentation.