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On the K-theory of certain extensions of free groups

Vassilis Metaftsis and Stratos Prassidis
From the journal Forum Mathematicum


Since Hol(Fn) embeds into Aut(Fn+1), one can construct inductively the subgroups (n) of Aut(Fn+1) by setting (1)=Hol(F2) and (n)=Fn+1(n-1). We show that the FJCw holds for (n). Moreover, we calculate the lower K-theory for the groups (n).

MSC 2010: 19D35; 20F65

Communicated by Andrew Ranicki

Funding source: European Social Fund

Award Identifier / Grant number: THALIS

Funding statement: Research supported by the European Union (European Social Fund – ESF) and Greek national funds through the Operational Program “Education and Lifelong Learning” of the National Strategic Reference Framework (NSRF) – Research Funding Program: THALIS.

A Appendix

We will show the result stated in Remark 6, that certain groups of type Fn(×/2)<(n-1) are CAT(0).

Our investigation, similar to the one in [14, Proposition 3.2], shows that subgroups of Aut(F2) isomorphic to F2(×/2) occur when the action of /2 on F2 is of the form

t(x1)=x1-1,t(x2)=x2 or t(x1)=x1,t(x2)=x2-1.

We show that in all the above cases these subgroups are CAT(0).

Lemma 1

Let Z/2Zt1<Aut(F2) be such that t1(x1)=x1-1 and t1(x2)=x2, as above. Let Zt2<Hol(F2) so that Z×Z/2Z<Hol(F2). Then, we have the following cases.

  1. If t2Aut(F2), then t2=x2kF2, k, k0.

  2. If t2Aut(F2), then t2(x1)=x2kx1±1x2-k, t2(x2)=x2±1, k0 (four cases).


First, we assume that t2Aut(F2). Because t1 and t2 commute, t1 acts trivially on t2. Let w(x1,x2) be the word in F2 representing t2. Then, w(x1-1,x2)=w(x1,x2). This means that x1 does not appear in w(x1,x2). Thus, w(x1,x2)=x2k.

Now, let t2Aut(F2). Then, t2(x1)=w1(x1,x2) and t2(x2)=w2(x1,x2). Since t1t2=t2t1, we have that


As before, the second relation implies that the word w2=x2k, k. Looking at the first relation, we get that w1=cx1c-1, . But w1 and w2 must be a generating set for F2. This means that k,{±1}. Also,


Since t1t2=t2t1, we have c(x1,x2)=c(x1-1,x2) and thus c=x2k, k. ∎

Lemma 2

Let G=F2(Z×Z/2Z), where the generator t1 of Z/2Z acts as


and the generator t2 of Z acts as


Then, G is a CAT(0)-group.


The group G has the presentation


We set ξ=x2-kt2. Then, the presentation becomes


Now, we consider four cases.

Case 1. In this case,


Now, let L1=x2,t1,ξ:t12=[t1,x2]=[t1,ξ]=[x2,ξ]=1<G, which is isomorphic to 2×/2 and, thus, it is CAT(0). Also, let L2=x1,t1,ξ:t1x1t1-1=x1-1,ξx1ξ-1=x1,t12=[t1,ξ]=1. Then,


The infinite dihedral group is a CAT(0)-group because it is a Coxeter group (see [15]). Thus, L2 is a CAT(0)-group, as a direct product of CAT(0)-groups. Also, L=t1,ξ:t12=[t1,ξ]=1×/2. But then, G=L1*LL2 is CAT(0) (see [6, Part II, Corollary 11.19]).

Case 2. We assume that


By setting ξ2=t1ξ and rewriting the presentation, we are back in Case 1.

Case 3. We assume that


We repeat the same method as before. Let


Therefore, L1/2×(). Now, the second group can be written as an HNN-extension *r, where r is the non-trivial automorphism of . Then, is a CAT(0)-group (see [6, Part II, Corollary 11.22]) and, thus, L1 is CAT(0). Now, L and L2 are as in Case 1 and, thus, G=L1*LL2 is CAT(0).

Case 4. We assume that


Again, set ξ2=t1ξ and rewrite the presentation to arrive at Case 3. ∎

Proposition 3

The group Fn(Z×Z/2Z)<H(n-1) is a CAT(0)-group.


Let t1 be the generator of /2 and let t2 be the generator of . We will consider two cases.

Case 1. Let t2Hol(F2) and t2Aut(F2). From Case 1 of Lemma 1, t2 is an element x2k, k, k0. Then, G=Fn(×/2) has the presentation


We change the generators by setting ξ=x2-kt2. First, notice that [t1,ξ]=1 because t1 commutes with t2 and x2. Then, the presentation becomes


Set K1=t1,ξ,x3,,xn:t12=[t1,ξ]=[t1,xi]=[ξ,xi]1,i=3,,n<Fn(×/2). Then, K1 is isomorphic to n-1×/2, which is a CAT(0)-group. Let K=×/2=t1,t2, which is a virtually infinite cyclic group. Also, set K2<G with presentation


Notice that G=K1*KK2. In order to show that G is CAT(0), it suffices to show that K2 is a CAT(0) group.

To that end, we change generators in K2 by setting ζ=x2kξ. Then, the presentation of K2 becomes


For Case 1 of Lemma 2, K2 is CAT(0) and we are done.

Case 2. We assume that t2Aut(F2). Using Case 2 of Lemma 2, the group G has the presentation




and K=t1,t2×/2, which are two subgroups of G. Finally, set K2 to be


It is obvious that G=K1*KK2. To show that G is CAT(0), it suffices to show that K2 is a CAT(0) group.

Set ζ=x2-kt2 and the presentation becomes


which is CAT(0) from Lemma 2. ∎

The reader should notice that there are more possibilities for the /2 action on the F2 subgroups of Aut(F2).

We would like to thank the anonymous referee for helping us improve our exposition.


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Received: 2014-12-11
Revised: 2015-7-4
Published Online: 2015-9-24
Published in Print: 2016-9-1

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