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?*



Does the poset of free factors of a free group form a lattice?
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 $\mathcal A$.


