Could $F_\infty \rtimes Z$ be isomorphic to $F_\infty$? Let $F_\infty$ be the free group on infinitely many generators, and $\phi: \mathbb{Z} \rightarrow$ Aut$(F_\infty)$ be any group homomorphism. 
My question is: If we form the semi-direct product  $F_\infty \rtimes Z$ could this group be free- in particular could it be isomorphic to $F_\infty$? 
Thanks
-Kevin
 A: It's certainly not free for every group homomorphism $\phi: \mathbb{Z} \to Aut(F_\infty)$; for example, if $\phi$ is trivial, then the semidirect product would be the cartesian product $F_\infty \times \mathbb{Z}$. This cannot be a free group, because every subgroup of a free group is free, but $F_\infty \times \mathbb{Z}$ contains a subgroup of the form $\mathbb{Z} \times \mathbb{Z}$ which is not free. 
On the other hand, there are plenty of examples of $\phi$ where it is free. Take for example any surjective group homomorphism $F_\infty \to \mathbb{Z}$, and let $K$ be the kernel. Then $K$, being a subgroup of a free group, is free. Moreover, the exact sequence 
$$1 \to K \to F_\infty \to \mathbb{Z} \to 1$$ 
splits, and this implies $F_\infty$ is a semidirect product of $\mathbb{Z}$ with $K$ in some way (see for example Wikipedia). Notice also $K$ cannot be finitely generated (if it were, then so would be the semidirect product), so $K$ must be a countably generated free group, and we are done. 
