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I'm reading G.Graetzer's Lattice Theory: First Concepts and Distributive Lattices and working on its exercises. One of them is to prove $(A, \subset)$, where $A$ is the set of finitely generated subgroups of a group $G$, is a join-semilattice, but not necessarily a lattice. I proved the former half, but I can't think of a counterexample for the latter half, partly because I'm not familiar to group theory. What is a simple example of a group whose finitely generated subgroups' intersection is not finitely generated?

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  • $\begingroup$ As t.b. pointed out when I gave an inaccurate answer, the question should be "give an example of a family $\mathcal{H}$ of finitely generated subgroups of a group $G$ such that $\cap_{H\in \mathcal{H}}H$ is not finitely generated." The intersection of all finitely-generated subgroups is trivial or of prime order (take a finite $p$-group). $\endgroup$ – user1729 Dec 9 '11 at 13:02
  • $\begingroup$ Yes, that's what I meant. $\endgroup$ – Pteromys Dec 9 '11 at 13:43
  • $\begingroup$ I've realized it doesn't suffice for the original question on textbook to show two finitely generated subgroups whose intersection is not finitely generated, because we can choose smaller subgroup as their meet. $\endgroup$ – Pteromys Dec 10 '11 at 0:22
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$F_2\times \mathbb{Z}\cong \langle a, b, c; [a, c], [b, c]\rangle$ works.

Let $P:=\langle a, bc\rangle$ and $Q:=\langle a, b\rangle$. Then $P\cap Q=\langle b^iab^{-i}:i\in\mathbb{Z}\rangle$, which is a free group as subgroups of free groups are free. It is therefore clearly infinitely generated. Write $R:=\langle b^iab^{-i}:i\in\mathbb{Z}\rangle$.

To see that $P\cap Q$ is this, note that $b^iab^{-i}=b^ic^iac^{-i}b^{-i}=(bc)^ia(bc)^{-i}$ so these generators generate a subgroup of the intersection ($R\leq P\cap Q$), while taking a word, $W(a, bc)$ say, then $W(a, bc)=W(a, b)c^x$ where $x$ is the exponent sum of $b$ in $W(a, b)$. Thus, if $W(a, b)\in P\cap Q$ we must have that $W$ has exponent sum zero in $b$, and so $W(a, b)\in R$. Thus, $P\cap Q\leq R$. Therefore, $P\cap Q=R$ as required.

This proof is from a paper of D. I. Moldavanskii, entitled "Intersection of finitely generated subgroups", 1968.

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  • $\begingroup$ I'm not familiar with those notations, but I understand $\langle a, b\rangle$ means the subgroup generated by the set $\{a, b\}$ and so on. Am I right? And what's the group in question $\langle a,b,c;[a,c],[b,c]\rangle$ exactly? $\endgroup$ – Pteromys Dec 9 '11 at 14:03
  • $\begingroup$ It is a cross product of the free group on two generators with the integers. $[a, b]=aba^{-1}b^{-1}$. An alternative presentation would be $\langle a, b, c; ac=ca, bc=cb\rangle$. Look up group presentations for more details. $\endgroup$ – user1729 Dec 9 '11 at 14:05
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The question was also asking to show that the poset $A$ of finitely generated subgroups of a group $G$, ordered by inclusion, is not a meet-semilattice in general.

In the example from user1729 with $G=F_2\times \mathbb{Z}$, the poset $A$ contains the finitely generated subgroups $P$ and $Q$ of $G$. $P$ and $Q$ do not have a meet (= greatest lower bound) in $A$.

Indeed, the set of lower bounds of $P$ and $Q$ in $A$ is the set of finitely generated subgroups of $R=P\cap Q$. But $R$ is a free group on a countable set of generators, say $R=\langle z_1,z_2,...\rangle$. Every word in the generators only makes use of a finite number of generators, so is contained in $\langle z_1,...,z_m\rangle$ for some $m$. So every finitely generated subgroup of $R$ is contained in some $\langle z_1,...,z_m\rangle$, and we can always find a strictly larger subgroup $\langle z_1,...,z_{m+1}\rangle$ in $A$. So the set of lower bounds of $P$ and $Q$ does not have a greatest element.

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