I'm trying to compute some homology for related groups and was wondering if the homology of these groups have been studied. Using the description of these groups as semidirect products of $\mathbb{Z}[1/k] \ltimes_k \mathbb{Z}$, I thought this would just pop out from the Lyndon-Hothschild-Serre spectral sequence, but I'm finding it a bit more difficult.
Has the homology of the Baumslag Solitar groups B(1,k) been computed
combinatorial-group-theorygroup-cohomologygroup-theory
Related Solutions
Here are my solutions from when I was studied under Ken Brown as an undergraduate:
V.3.5: From the semi-direct product representation $S_3=\mathbb{Z}_3\rtimes \mathbb{Z}_2$ where $\mathbb{Z}_2$ acts on $\mathbb{Z}_3$ by conjugation, $H^*(S_3)=H^*(S_3)_{(2)}\oplus H^*(S_3)_{(3)}\cong H^*(S_3)_{(2)}\oplus H^*(\mathbb{Z}_3)^{\mathbb{Z}_2}$ by Theorem III.10.3, with $H^*(S_3)_{(2)}$ isomorphic to the set of $S_3$-invariant elements of $H^*(\mathbb{Z}_2)$. Exercise III.10.1 showed that $H^*(S_3)_{(2)}\cong \mathbb{Z}_2$, so it suffices to compute $H^*(\mathbb{Z}_3)^{\mathbb{Z}_2}$ where we know that $\mathbb{Z}_2$ acts by conjugation on $\mathbb{Z}_3$ ($1\mapsto 1$, $x\mapsto x^2$, $x^2\mapsto x$). But this action can be considered as the endomorphism $\alpha(2)$ from Exercise V.3.4 since $(1)^2=1$ and $(x)^2=x^2$ and $(x^2)^2=x^4=x$, and that exercise implies that the induced map on the $(2i)^{th}$-cohomology is multiplication by $2^i$ [we know that the cohomology is trivial in odd dimensions]. Now $2^1\equiv 2\,\text{mod}3$ and $2^2\equiv 1\,\text{mod}3$, so by multiplying both of those statements by $2$ repeatedly we see that $2^i\equiv 1\,\text{mod}3$ for $i$ even and $2^i\equiv 2\,\text{mod}3$ for $i$ odd. Thus the largest $\mathbb{Z}_2$-submodule of $H^{2i}(\mathbb{Z}_3)\cong\mathbb{Z}_3$ on which $\mathbb{Z}_2$ acts trivially is $\mathbb{Z}_3$ (itself) for $i$ even, and is $0$ for $i$ odd. It now follows that the integral cohomology $H^*(S_3)$ is the same as that which was deduced in Exercise III.10.1, namely, it is $\mathbb{Z}_2$ in the $2\,\text{mod}4$ dimensions and is $\mathbb{Z}_6$ in the nonzero $0\,\text{mod}4$ dimensions and is $0$ otherwise (besides the $0^{th}$-dimension in which it is $\mathbb{Z}$).
III.10.1: From the semi-direct product representation $S_3=\mathbb{Z}_3\rtimes \mathbb{Z}_2$ where $\mathbb{Z}_2$ acts on $\mathbb{Z}_3$ by conjugation, $H^*(S_3)=H^*(S_3)_{(2)}\oplus H^*(S_3)_{(3)}\cong H^*(S_3)_{(2)}\oplus H^*(\mathbb{Z}_3)^{\mathbb{Z}_2}$ by Theorem III.10.3. Now $S_3$ is the unique nonabelian group of order 6, so $D_6\cong S_3$ and we can use a result (given below as "additional exercise") which implies that the $\mathbb{Z}_2$-action on $H_{2i-1}(\mathbb{Z}_3)\cong H^{2i}(\mathbb{Z}_3)$ is multiplication by $(-1)^i$. Thus $H^n(\mathbb{Z}_3)^{\mathbb{Z}_2}$ is isomorphic to $\mathbb{Z}_3$ for $n=2i$ where $i$ is even, and is trivial for $n$ odd and $n = 2i$ where $i$ is odd. Taking any Sylow $2$-subgroup $H\cong \mathbb{Z}_2$, Theorem III.10.3 states that $H^*(S_3)_{(2)}$ is isomorphic to the set of $S_3$-invariant elements of $H^*(H)$. In particular we have the monomorphism $H^{2i-1}(S_3)_{(2)}\hookrightarrow H^{2i-1}(H)=0$, so $H^{2i-1}(S_3)_{(2)}=0$. An $S_3$-invariant element $z\in H^{2i}(H)\cong\mathbb{Z}_2$ must satisfy the equation $\text{res}^H_Kz=\text{res}^{gHg^{-1}}_Kgz$, where $K$ denotes $H\cap gHg^{-1}$. If $g\in H$ then $gHg^{-1}=H$ and the above condition is trivially satisfied for all $z$ ($hz=z$ by Proposition III.8.1). If $g\notin H$ then $K=\lbrace 1\rbrace$ because $H$ is not normal in $S_3$ and only contains two elements, so the intersection must only contain the trivial element. But then the image of both restriction maps is zero, so the condition is satisfied for all $z$; thus $H^{2i}(S_3)_{(2)}=\mathbb{Z}_2$. Alternatively, a theorem of Richard Swan states that if $G$ is a finite group such that $\text{Syl}_p(G)$ is abelian and $M$ is a trivial $G$-module, then $\text{Im}(\text{res}^G_{\text{Syl}_p(G)})=H^*(\text{Syl}_p(G),M)^{N_G(\text{Syl}_p(G))}$. It is a fact that $N_{S_3}(\mathbb{Z}_2)=\mathbb{Z}_2$, so taking $G=S_3$ and $H=\text{Syl}_2(S_3)\cong\mathbb{Z}_2$ and $M=\mathbb{Z}$ we have $\text{Im}(\text{res}^{S_3}_H)=(\mathbb{Z}_2)^{\mathbb{Z}_2}=\mathbb{Z}_2$ in the even-dimensional case. Since any invariant is in the image of the above restriction map (by Theorem III.10.3), the result $H^{2i}(S_3)_{(2)}=\mathbb{Z}_2$ follows.
Additional Exercise: The cyclic group $C_m$ is a normal subgroup of the dihedral group $D_m=C_m\rtimes C_2$ (of symmetries of the regular $m$-gon). There is a $C_2$-action on $C_m=\langle \sigma\rangle$ given by $\sigma\mapsto \sigma^{-1}$. Determine the action of $C_2$ on the homology $H_{2i-1}(C_m,\mathbb{Z})$, noting that there is an element $g\in D_m$ such that $g\sigma g^{-1}=\sigma^{-1}$.
Solution: Letting $c(g):C_m\rightarrow C_m$ be conjugation by $g$, we can apply Corollary III.8.2 to obtain the induced action of $D_m/C_m\cong C_2$ on $H_*(C_m,\mathbb{Z})$ given by $z\mapsto c(g)_*z$. It suffices to compute $c(g)_*$ on the chain level, using the periodic free resolution $P$ of $C_m$, and using the trivial action on $\mathbb{Z}$. Using the condition $\tau(hx)=[c(g)](h)\tau(x)=h^{-1}\tau(x)$ on the chain map $\tau:P\rightarrow P$ (for $h\in C_m$), we claim that $\tau_{2i-1}(x)=\tau_{2i}(x)=(-1)^i\sigma^{m-i}x$ for $i\in\mathbb{N}$ and $\tau_0(x)=x$. Assuming this claim holds, the chain map $P\otimes_{C_m}\mathbb{Z}\rightarrow P\otimes_{C_m}\mathbb{Z}$ [in odd dimensions] is given by $x\otimes y\mapsto (-1)^i\sigma^{m-i}x\otimes gy=(-1)^i\sigma^{m-i}x\otimes y=(-1)^ix\otimes \sigma^{i-m}y=(-1)^ix\otimes y$, and so $c(g)_*$ [hence the $C_2$-action] is multiplication by $(-1)^i$ on $H_{2i-1}(C_m,\mathbb{Z})$. It suffices to prove the claim. Seeing that $N\tau_{2i}(1)=\tau_{2i-1}(N)=N\tau_{2i-1}(1)$ where $N$ is the norm element, we can restrict our attention to $\tau_{2i-1}$ and use induction on $i$ since $(\sigma-1)\tau_1(1)=(\sigma-1)(-\sigma^{m-1})=\sigma^{m-1}-1=\sigma^{-1}-1=\tau_0(\sigma-1)$. This chain map must satisfy commutativity $(\sigma-1)\tau_{2i-1}(1)=\tau_{2(i-1)}(\sigma-1)$, and this is indeed the case because $(\sigma-1)\tau_{2i-1}(1)=(-1)^i(\sigma^{m-i+1}-\sigma^{m-i})$ and $\tau_{2(i-1)}(\sigma-1)=(\sigma^{-1}-1)(-1)^{i-1}\sigma^{m-i+1}=(-1)^i(\sigma^{m-i+1}-\sigma^{m-i})$.
Recap: $H^n(S_3)\cong$
$\mathbb{Z}$ for $n=0$,
$\mathbb{Z}_6$ for $n\equiv 0\;\text{mod}\,4$ with $n\ne 0$,
$\mathbb{Z}_2$ for $n\equiv 2\;\text{mod}\,4$,
$0$ otherwise
The only nilpotent Baumslag-Solitar group is $\mathbb{Z}^2=\langle a,b\mid [a,b]\rangle$. More generally:
Theorem. A one-relator group is nilpotent if and only if it is abelian, if and only if it is a cyclic group or $\mathbb{Z}^2=\langle a,b\mid [a,b]\rangle$.
Proof. Let $G$ be a one-relator group which is not cyclic nor isomorphic to $\mathbb{Z}^2$, and which has non-trivial centre. Then $G$ is two generated and its centre is infinite cyclic [1]. Moreover, Pietrowski [2] proved that such a group $G$ admits a presentation $$\mathcal{P}:=\langle x_1, \ldots, x_n\mid x_1^{p_1}=x_2^{q_1}, \ldots, x_{n-1}^{p_{n-1}}=x_n^{q_{n-1}}\rangle.$$ Indeed, in this form the centre is the subgroup $\cap\langle x_i\rangle$. (Given a presentation of this form it is not easy to verify if it defines a one-relator group, although an algorithm exists [3], but a simple example is $\langle a,b; a^m=b^n\rangle$.)
Now, abelianise this presentation $\mathcal{P}$. Every one-relator group surjects onto $\mathbb{Z}$, and so this abelianisation also maps onto $\mathbb{Z}$. Therefore, some generator $x_i$ has infinite order in the abelianisation. As the centre is wholly contained in $\langle x_i\rangle$, the centre intersects the derived subgroup trivially. This means that $G$ is not nilpotent, as required, because non-abelian nilpotent groups have a centre which intersects the derived subgroup non-trivially (see, for example, here). QED
In fact, we have a stronger result:
Corollary. A one-relator presentation $\langle \mathbf{x}\mid R\rangle$ defines a nilpotent group if and only if one of the following holds
- $|\mathbf{x}|=1$ (i.e. the group is obviously cyclic)
- $|\mathbf{x}|=2$, so $\mathbf{x}=\{a, b\}$ say, and $R$ is conjugate to $[a, b]$ or $[a, b]^{-1}$ in $F(a, b)$.
- $|\mathbf{x}|=2$ and $R$ is a "primitive element" of $F(\mathbf{x})$, i.e. there exists some word $S\in F(\mathbf{x})$ such that $\langle R, S\rangle=F(\mathbf{x})$.
I won't detail a proof, but the point is that in each case the group is "obviously" nilpotent (1. and 3. correspond to cyclic groups, 2. is $\mathbb{Z}^2$), while certain well-known results in combinatorial group theory give us that a one-relator presentation of a cyclic group or $\mathbb{Z}^2$ must have one of these forms (for example, the main result for (2) is Theorem 3.9 of the book Combinatorial Group Theory by Magnus, Karrass and Solitar).
[1] Murasugi, K. "The center of a group with a single defining relation." Mathematische Annalen 155.3 (1964): 246-251.
[2] Pietrowski, A. "The isomorphism problem for one-relator groups with non-trivial centre." Mathematische Zeitschrift 136.2 (1974): 95-106.
[3] Metaftsis, V. "An algorithm for stem products and one-relator groups." Proceedings of the Edinburgh Mathematical Society 42.1 (1999): 37-42.
Best Answer
By Magnus(-Karrass-Solitar) theorem, one-relator group only has torison in case its relator is a proper power. Then, by Cockroft's theorem (W. H. Cockroft, On two-dimesional alpherical complexes, 1954, rediscovered later by Lyndon and other people, see, for example, "Lyndon Identity theorem"), presentation complex of a torsion-free one-relator group is aspherical, so it is a classifying space for that group.
So... we're lucky, because we can use explicit cellular structure of presentation complex — which happens to be $K(G, 1)$ — to compute (co)homology. Cellular complex for $BS(r, s)$ will be $$\Bbb Z \overset{(0, r-s)}\longrightarrow \Bbb Z^2 \overset{0}\longrightarrow \Bbb Z$$
So, $H_1$ is $\Bbb Z \oplus \Bbb Z/(r-s)$ (which is already clear from presentation), and $H_2$ is zero. Turns out there's not much to study after all.
I want to remark that vanishing of $H_2$ can be obtained without (pretty hard) asphericity result. To make a classifying space out of presentation complex, we only add cells in dimensions 3 and higher, but that can only kill 2-dim homology classes, not add new ones. Now we notice that cellular differential should be nonzero on (the only) 2-cell of presentation complex (because it's a 2-generated group with rank 1 abelianization, and we need something to kill a 1-dim cycle). Same reasoning tells us that $k$-generated $(k-l)$-relator group with abelianisation having rank $l$ will have vanishing $H_2$.
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A different avenue for calculating homology would be using "Mayer-Vietoris" sequence from I. M. Chiswell. Exact sequences associated with a graph of groups — especially if you want to compute homology with nontrivial coefficients.
P. S. I vaguely remember a paper by D. J. S. Robinson, where he computers homology of more general graphs of groups with cyclic edge and vertex groups, but cannot remember the title. It would be interesting to know homology of "cyclic BS-type" groups, like Higman group - i. e. you have generators indexed by $\Bbb Z/n$, and $i$-th generator conjugates the $i+1$-st one into some power. Relation modules of presentations of this type occurred in potential (counter)examples to the relation gap conjecture.