Let $F\_n$ be the free group with $n$ generators $\{ x\_1 , \dots , x\_n \}$. From my understanding, we say that a subgroup $K \leq F\_n$ is a free factor of $F\_n$ if there exists a subgroup $H \leq F\_n$ such that $F\_n = H \* K$. Furthermore, if I understand correctly, if we write that $K = \langle f\_1 , \dots , f\_s \rangle$ and $H = \langle g\_1 , \dots , g\_m \rangle$, then we have that $H \* K = \langle f\_1 , \dots , f\_s , g\_1 , \dots , g\_m \rangle$.

Now my confusion is as follows - with this definition, isn't EVERY subgroup of a free group a free factor? For a silly example, let $n=2$, $F\_2 = \langle a, b \rangle$ and let $K = \langle aba , b \rangle$. Then can't we just take $C = \langle a \rangle$, which gives $K \* C = \langle aba, b , a \rangle = \langle a, b \rangle = F\_2$ ? Isn't it always possible to do this for any $K$, by just letting $C$ to be the subgroup of $F\_n$ generated by the $x\_i$'s that don't occur in $K$ (so possibly $C = F\_2$)?