Take the 2-minute tour ×  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
$F_\infty$?    share|improve this question add comment  1 Answer
up vote 16 down vote accepted  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