# projections of finitely presented groups

let's call an object $x$ of a cocomplete category (categorical) finitely generated if $\hom(x,-)$ commutes with filtered colimits of monomorphisms, and finitely presented if $\hom(x,-)$ even commutes with arbitrary filtered colimits. I know that in the literature there are some other definitions for these notions. at least in categories of algebraic structures, it would be nice that these notations coincide with the usual ones. let's concentrate on the category of groups, for these make problems ;).

it's easy to see that categorical finitely generated means the usual finitely generated. but I've figured out that a group $G$ is categorical finitely presented if and only if there is a finitely presented group $H$, over which $id_G : G \to G$ factors through. is this the same as beeing finitely presented?

the question is equivalent to: let $H$ be a finitely presented group and $p : H \to H$ a projection, i.e. a homomorphism satisfying $p^2 = p$. is then $im(p)$ also finitely presented?

• Sorry for maybe derailing your question slightly, but it seems that finitely presented groups in the traditional sense are exactly those which are obtained by localizing objectwise finite presheaves of sets in the usual sketch for the category of groups. Is there some general fact about presentable categories which guarantees that these are the finitely presentable objects in the localized category, i.e., groups? (I don't have a copy of Adamek and Rosicky handy...) – Reid Barton Dec 28 '09 at 2:10

Here is a direct proof that "categorically finitely presented" (which I think it is more usual to call "finitely presentable," since no actual presentation is necessarily given) is the same as "finitely presented" in the usual sense. This is a special case of Theorem 3.12 in Adamek and Rosicky's book "Locally presentable and accessible categories," which applies to all finitary varieties of algebras.

First suppose $G$ is finitely presented in the usual sense, with $X$ a finite set of generators and $R \subset F[X]$ a finite set of relations, where $F[-]$ is the free group functor. Then there is a coequalizer diagram $F[R] \rightrightarrows F[X] \twoheadrightarrow G$. Since finite colimits of c.f.p. objects are c.f.p. (since finite limits commute with filtered colimits in Set), it suffices to show that finitely generated free groups are c.f.p. But $Hom_{Grp}(F[X],H) \cong Hom_{Set}(X,U(H))$ where $U$ is the underlying-set functor, which preserves filtered colimits, and finite sets are c.f.p. in Set (this is easy to see).

Conversely, suppose $G$ is c.f.p. Then in particular it is c.f.g. and hence (as you say) finitely generated in the usual sense. One way to see that is to write $G$ as the directed union (= filtered colimit of monics) of its finitely generated subgroups; then since $G$ is c.f.g. its identity must factor through one of these subgroups, and hence that subgroup must be all of $G$. Now let $X$ be a finite set of generators and consider the diagram of groups whose vertices are all finitely presented quotients $F[X]\twoheadrightarrow H_i$ through which $F[X]\twoheadrightarrow G$ factors, and whose transition maps $k_{i,j}\colon H_i \to H_j$ are maps under $F[X]$ (and hence over $G$). Then the canonical maps $h_i\colon H_i \to G$ make $G$ into the colimit of this diagram, since any relation holding in $G$ must hold in some finitely presented quotient of $F[X]$. Moreover, this diagram is filtered, so $id_G\colon G\to G$ factors through some $H_i$ via a map $u\colon G \to H_i$ with $h_i u = id_G$.

I'm guessing this is the point to which you got. The clever argument from A&R then goes as follows. By the first part of the argument, $H_i$ is c.f.p. But we have two maps $id_{H_i}\colon H_i\to H_i$ and $u h_i\colon H_i\to H_i$ which become identified in the colimit, i.e. $h_i id_{H_i} = h_i u h_i$; thus there must be some $k_{i,j}\colon H_i \to H_j$ such that $k_{i,j} = k_{i,j} u h_i$. It then follows that $h_j\colon H_j \to G$ is an isomorphism with inverse $k_{i,j} u$, for on the one hand, $h_j k_{i,j} u = h_i u = id_G$, but on the other hand, $k_{i,j} u h_j k_{i,j} = k_{i,j} u h_i = k_{i,j}$ and $k_{i,j}$ is epic. Thus, $G$ is finitely presented since $H_j$ is so.

Here's a geometric proof that a retract of a finitely presented group is finitely presented.

This uses the fact that a f.g. group is coarsely simply connected iff it is finitely presented (Gromov calls this a triviality; it takes a little time to realize he's right).

Coarsely simply connected passes to retracts: it's clear because if you have a loop in the retract, coarse fill it in the all space and apply the retraction to the homotopy to get a homotopy inside the retract.

[The same argument shows that the Dehn function decreases when passing to retracts.]

• The same argument applies even if the map $p$ is not a homomorphism and the retract equation $p^2=p$ only holds coarsely, as long as you can independently verify that $p$ is coarse Lipschitz. You can use the Dehn function decrease to get lower bounds on the Dehn function of the ambient group in this manner, e.g. my paper with Handel front.math.ucdavis.edu/1009.5018 – Lee Mosher Jul 12 '12 at 16:23
• @Lee. Indeed, they are many extensions; in this generality the image of $p$ need not be a subgroup. This argument is used by Gromov in his 1993 book: if you apply the convex projection into a boundary horoball to a nonuniform lattice in a semisimple group, the image is sometimes a "large" part of the horosphere and in many explicit cases for which the horosphere is known to have at least exponential Dehn function, the projection contains loops of at least exponential area. This way it can be obtained that $SL_3(\mathbf{Z})$ has at least exponential Dehn function and $SL_3(F_p[t])$ isn't f.p. – YCor Jul 12 '12 at 16:53

I think this is true. Let $G= \langle g_1, \ldots, g_n; R_1, \ldots, R_m \rangle$ be a finite presentation for $G$. Let $p:G\to G$ be a projection (or retract). Then $p(G)< G$ is generated by $p(g_1), \ldots, p(g_n)$. Let $F_n$ be the free group on $n$ generators, and $R\in F_n$ a relator of $p(G)$, so $R(p(g_1), \ldots, p(g_n))=1$. I claim that $p(R_1),\ldots, p(R_m)$ give finitely many relations for $p(G)$. Give a new presentation for $G$, $$\langle g_1, \ldots, g_n , y_1, \ldots, y_n ; y_1^{-1} p(g_1) , \ldots, y_n^{-1} p(g_n), R_1, \ldots, R_m \rangle.$$ With respect to this presentation, the homomorphism $p$ is given by $p(g_i)=p(g_i), p(y_i)=p(g_i)$. Then we must have the relation $R(y_1, \ldots, y_n)=1 \in G$, and therefore $R(y_1, \ldots, y_n)$ may be written as a product of conjugates of the relators. Then $p(R(y_1,\ldots,y_n)) = R(p(g_1),\ldots,p(g_n))$ is a product of conjugates of projections of the relators of the second presentation of $G$. This shows that $p(G)$ is finitely presented, and since the first $n$ relators project to the trivial relator, we see that the relators are given by $p(R_1),\ldots, p(R_m)$.

Here's a reformulation:

Let $G$ be a finitely presented group, with a projection $p:G\to G$, so $p^2=p$.

Then there is a free group $F = \langle g_1, \ldots, g_n\rangle$ and an epimorphism $\phi:F\to G$ such that $ker(\phi)$ is finitely normally generated. We may write the presentation as $$\langle g_1,\ldots, g_n; R_1,\ldots, R_m\rangle.$$ The kernel of $p$ is normally generated by $\phi(g_i) p(\phi(g_i))^{-1}$ (exercise)*. Let $p': F\to F$ be a lift of $p$, which exists by the universal property of $F$, so that $p\circ \phi = \phi \circ p'$. Then $ker(p\circ \phi) = \phi^{-1}( ker(p))$ is normally generated by elements projecting to the normal generators of $ker(p)$ togther with $ker(\phi)$. We may take $g_i p'(g_i)^{-1}$ together with the finitely many normal generators of $ker(\phi)$ to get normal generators of $ker(p\circ \phi)$. This gives a finite presentation of $p(G)$, given by $$\langle g_1,\ldots,g_n ; R_1,\ldots, R_m, g_1 p'(g_1)^{-1},\ldots, g_n p'(g_n)^{-1} \rangle.$$

In my previous "argument", I attempted to project this presentation by $p'$ to $p'(F)$, which is a free subgroup of $F$, but this is unnecessary, and I was incorrect that the new relators project to trivial relators in $p'(F)$.

*(exercise) Here's the details to show the kernel is normally generated by $\phi(g_i) p(\phi(g_i))^{-1}$. For simplicity of notation, I'll denote the generators as $g_i$ instead of $\phi(g_i)$, and assume that they are semigroup-generators of $G$. Then $ker(p)$ is normally generated by $g_i p(g_i)^{-1}$. Call this normal subgroup $K$. Notice that it suffices to show that $g p(g)^{-1} \in K$, for all $g\in G$, since if $g\in ker(p)$, then $g p(g)^{-1} = g \in K$. We may show this by induction on the length of the word representing $g$. Clearly it's true if $g= g_i$. Now, suppose it is true for products of $k$ generators, $k\in \mathbb{N}$. Then if $g= g_{i_1} \cdots g_{i_{k+1}}$, we have $$g p(g)^{-1}= g_{i_1} \cdots g_{i_{k+1}} p(g_{i_{k+1}})^{-1} \cdots p(g_{1})^{-1} = g_{i_1} p(g_{i_1})^{-1} ( p(g_{i_1}) ( g_{i_2} \cdots g_{i_{k+1}} p(g_{i_{k+1}})^{-1} \cdots p(g_{i_2})^{-1} ) p(g_{i_1})^{-1}) \in K$$ by induction and the fact that $K$ is closed under products and conjugations.

• I think that we have to distinguish between elements of $G$ and words in free groups more carefully. the word $R(y_1,....,y_n)$ is trivial in $G$, thus in the free group on the $g_i,y_j$ it can be written as a product of conjugates of the relators in the second presentation. but how can we apply $p$ now; what is meant by that? finally, we want to express $R$ in $F_n$ via the $p(R_i)$. – Martin Brandenburg Dec 28 '09 at 11:36
• you're right, I've attempted to clarify and correct the argument I had in mind. – Ian Agol Dec 28 '09 at 19:09
• Historical note: this is Lemma 1.3 of Wall's "Finiteness conditions for CW-complexes". If I remember correctly, Wall says that Stallings told him about it. – HJRW Dec 28 '09 at 19:40
• "The kernel of $p$ is normally generated by $\phi(g_i) p(\phi(g_i))^{-1}$ (exercise)." when you know that the kernel of $p$ is finitely normally generated, then $im(p) = G/ker(p)$ is finitely presentable. so this is actually the whole point. I don't believe that this is a generating set. – Martin Brandenburg Dec 29 '09 at 10:57
• I believe Ian's argument, but this all feels like a bit of a mess. You can simplify things a lot by modifying your presentation. G is the semidirect product of kerp by p(G), so you can take a generating set for G that consists of generators for p(G) together with a finite set whose images under the action of p(G) generate ker p. Now you have a generating set with the property that p(g)=g or p(g)=1 for every generator. It's now easy to prove that the desired presentation for p(G) is obtained by killing every generator that dies under p. – HJRW Dec 29 '09 at 18:03

the Category of groups $Gr$ is locally finitely presentable ([LP] Cor.3.7, p.141) and a object $X\in Gr$ is finitely presentable in categorical sense (i.e. $h^X$ commute by finite filtrant colimits) iff is finitely presentable in the usual algebraic sense ($X$ can be expressed as finite generators and finite equations) ([LP] Cor.3.12, p.143).
Then if you have a idempotent $p: G\to G$ with $G$ finitely presentable, $Im(p)=Cok(1_G, p)$ (classical splitting os a idempotent), and any finite colimits of finitely presentable objects is a presentable objects ([LP] Cor.1.3, p.12)