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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 $F_\infty$?

Thanks

-Kevin

flag	asked Sep 6 at 1:39 Kevin Schreve 43●2

Todd Trimble · Accepted Answer · 2012-09-06 04:25:07Z UTC

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.

Kevin Schreve · Answer 2 · 2012-09-06 14:50:36Z UTC

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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.

answered Sep 6 at 14:50

Kevin Schreve
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	Sure, no problem! Yeah, you need 50 points to leave comments (and I just added another 10 for you). (Also, you get a few more points when you accept an answer as satisfactorily addressing your question -- there's a place where you can click for that.) Finally, I had earlier left a comment which I later thought was wrong and deleted; hope you didn't mind too much. – Todd Trimble Sep 6 at 16:07
	Thanks very much for the help Todd! I actually asked my question as an unregistered user and then registered, so there are now two Kevin Schreve accounts... I say this because I wanted to accept your answer but couldn't...your answer leads to a new question- which isn't very clear unfortunately- given some non-trivial homomorphism $\phi:\mathbb{Z} \rightarrow$ Aut($F_\infty$), is there any way to detect whether $F_\infty \rtimes_\phi \mathbb{Z}$is free? – Kevin Schreve Sep 6 at 17:50
	If you want to combine your accounts, you can go over to meta and make that request (near the top of the page there are 4 "sticky" threads -- you'll see the thread that's relevant here). As for the math, you can either edit this question, or you can start a new question (the former might be a better option; you can refer back to this question obviously). Without having thought about it much, I think a necessary condition is that $\phi$ be injective. Other people may have more to say. – Todd Trimble Sep 7 at 1:04

Could $F_\infty \rtimes Z$ be isomorphic to $F_\infty$?

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