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Let $F_n$ denote a free group of rank $n$. The set of its free factors is partially ordered by inclusion. Recall that a psoet is called a lattice if any two elements have a smallest upper bound and a greatest lower bound.

Is this true for this poset?

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  • $\begingroup$ The intersection of any finite family of free factors in a free group is again a free factor, though there are infinite families for which this does not hold (see On the intersections of free factors of a free group, by Burns, Chau, and Solitar, Proc. Amer. Math. Soc. 64 (1977) no 1, 43-44. Also, the intersection of two retracts of a free group is a retract (Bergman, G.M., Supports of derivations, free factorizations, and ranks of fixed subgroups in free groups, Trans. Amer. Math. Soc. 351 (1999) no 4, 1531-1550. As has been mentioned, join seems harder. $\endgroup$ Mar 25, 2013 at 21:51
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    $\begingroup$ The Burns-Chau-Solitar counterexample is in a free group of countably infinite rank. For finite rank, everything works; I've amended my answer. $\endgroup$
    – Lee Mosher
    Mar 25, 2013 at 22:17
  • $\begingroup$ One difference to the free Abelian case is the following: In the free abelian case the rank is additive, i.e. the sum of the ranks of the greatest lower bound and the least upper bound of two elements equals the sum of the ranks of those elements. This does not hold in this case. The elements $c$ and $[a,b]c$ generate free factors of the free group generated by $a,b,c$. Their intersection is trivial. Their least upper bound $L$ contains $[a,b]$. Thus $L\cap F(a,b)$ is a free factor that contains $[a,b]$. But this commutator is not contained in rank $1$ free factor. $\endgroup$ Mar 26, 2013 at 12:27
  • $\begingroup$ Thus it is $F(a,b)$ and hence $L=F(a,b,c)$. $\endgroup$ Mar 26, 2013 at 12:27

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ORIGINAL ANSWER, ADDRESSING A SLIGHTLY DIFFERENT QUESTION: There is a closely related poset for which greatest lower bounds and least upper bounds indeed exist. Instead of an individual free factor $A$, first consider its conjugacy class $[A]$. Then, instead of individual conjugacy classes of free factors $[A]$, consider a "free factor system" (in the language of Bestvina-Feighn-Handel): a finite set $\mathcal{F} = \{[A_1],\ldots,[A_k]\}$ such that there exists a free factorization of the form $F_n = A_1 * \cdots * A_k * B$ where $B$ may or may not be trivial but all the $A_i$'s are nontrivial. The partial ordering $\mathcal{F}\sqsubset \mathcal{F}'$ is defined by requiring that for each $[A] \in \mathcal{F}$ there exists $[A'] \in \mathcal{F}'$ such that $A$ is conjugate to a subgroup of $A'$.

The Kurosh Subgroup Theorem can be translated into the statement that this poset has greatest lower bounds. The greatest lower bound of $\mathcal{F}$ and $\mathcal{F}'$ is $$\mathcal{F} "meet" \mathcal{F}' = \{[A \cap A'] \, | \, [A] \in \mathcal{F}, [A'] \in \mathcal{F}', A \cap A' \ne 1\} $$ (I don't know how to get the "meet" operator in this version of TeX).

CORRECTION: You have to allow the "trivial" free factor system in order for this meet to be well-defined, because it is possible that the definition above produces the emptyset. In which case I should have left out the condition "$A \cap A' \ne 1$" in my definition of the meet.

The meet operator can then be extended to an operator on arbitrary sets of free factor systems, and using this one gets least upper bounds too: the least upper bound of $\mathcal{F}$ and $\mathcal{F}'$ is the meet of all free factor systems $\mathcal{F}''$ (including the improper free factor system $\{[F]\}$) such that $\mathcal{F} \sqsubset \mathcal{F}''$ and $\mathcal{F}' \sqsubset \mathcal{F}''$. This $\mathcal{F}''$ is called the ``free factor support'' of $\mathcal{F}$ and $\mathcal{F}'$.

ADDITION, ADDRESSING THE ORIGINAL QUESTION: Now that I've had a chance to think more about the Kurosh subgroup theorem myself, I realize that the answer to the original question is just "yes".

First, for greatest lower bound: the intersection of any collection of free factors of the finite rank free group $F_n$ is a free factor. For two free factors $A,B$ this is a consequence of the Kurosh subgroup theorem which says that the set of nontrivial intersections of $A$ with conjugates of $B$ is a finite set $\{C_1,\ldots,C_K\}$ with the property that there exists a free factorization $A = C_1 * \cdots * C_k * D$ where $D$ may be trivial; so $A \cap B$, if nonempty, is one of the $C$'s, and is therefore a free factor of $A$, and a free factor of a free factor of $F_n$ is a free factor of $F_n$. In general, for any collection of free factors, write them in a sequence, intersect each initial segment of the sequence, and use the fact that in a finite rank free group, a free factor strictly included in another has smaller rank (proved by abelianizing).

Then, for least upper bound: given any collection $\mathcal A$ of free factors, the intersection of all free factors (including the improper free factor $F_n$) containing each element of $\mathcal A$ is the least a free factor containing each element of $\mathcal A$.

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  • $\begingroup$ Let me emphasize that my assumption that the group in consideration is free (almost) not needed at all. The argument that the intersection of two free factors is again a free factor does not need it; thus we always have greatest lower bounds of finite sets of free factors. The least upper bound really just uses that this poset does not have infinite descending chains; which is true for any finitely generated group: The statement about intersection can be used to pruduce from such a chain a free product decomposition of the whole group with infinitely many, nontrivial factors. $\endgroup$ Mar 26, 2013 at 12:17
  • $\begingroup$ The meet symbol $\land$ is \land or \wedge. If you prefer $\sqcap$ to match $\sqsubset$, it’s \sqcap. $\endgroup$ Mar 26, 2013 at 12:34

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