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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 $F_\infty$?



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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 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.

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Your «plenty of examples» are, in a way, all, no? – Mariano Suárez-Alvarez Sep 6 '12 at 4:54
Yes, that's true, Mariano. – Todd Trimble Sep 6 '12 at 5:29
Let me remark that this works for any infinite cardinal, despite the question does not make precise what infinite is taking to define $F_\infty$. – Fernando Muro Sep 6 '12 at 5:39
Oh, you're absolutely right, Fernando. I just assumed $\aleph_0$, but... good point. – Todd Trimble Sep 6 '12 at 6:12
This isn't an answer- but this was my first question posted so I can't comment yet? Anyway, thanks very much for your answer. – user26257 Sep 6 '12 at 14:50

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