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



Could $F_\infty \rtimes Z$ be isomorphic to $F_\infty$?
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](http://en.wikipedia.org/wiki/Splitting_lemma#Non-abelian_groups)). 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. 


