Let $F$ be a free group of rank at least $2$. Of course then, $F$ can not be direct product of two subgroups (except the trivial decomposition $1\times F$).

Q. Can $F$ be written as semi-direct product of two subgroups? If yes, can we choose the components in semi-direct product to be free (sub)groups of finite rank?

  • $\begingroup$ I am not sure but I don't think this is possible as semidirect products establish non-trivial relations on generators... $\endgroup$ – DonAntonio May 20 '17 at 11:34
  • $\begingroup$ If we have fixed-point-free action of one subgroup on other, I didn't see what could be problem. (I had partially thought in the direction you pointed before stating question, but, I didn't came to final answer myself.) $\endgroup$ – Beginner May 20 '17 at 11:37
  • $\begingroup$ @Be I'm not sure I understand your point (pun intended...): in $\;N\rtimes_\phi H\;$ we have a homomorphism $\;\phi:H\to\text{ Aut}\,N\;$, and in the semidirect product we're going to have $$(n,h)\cdot(n,h)^{-1}=1\;,\;\;\text{with}\;\;(n,h)^{-1}=(n^{-h^{-1}},\,h^{-1})$$I think the above establishes non-trivial relations on generators $\endgroup$ – DonAntonio May 20 '17 at 11:42
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    $\begingroup$ Any homomorphism of F to onto another free group Z has a splitting so it is semidirect product. $\endgroup$ – i. m. soloveichik May 20 '17 at 12:05
  • $\begingroup$ To complement i.m.soloveichik's comment, all semidirect decompositions arise in this way, since subgroups of free groups are free. $\endgroup$ – YCor May 20 '17 at 21:03

The answer to the first question is yes, $F$ can be expressed as a semidirect prodcut in many different ways. Following i.m.solovelchik's comment, let $f$ be any epimorphism from $F$ to a free group, and let $K:=\ker f$. For example, if $a$ is one of the free generators, then we could define $f$ by $f(x) = a$ for all members of a free generating set containing $a$.

Then $F/K$ is free, and so $K$ has a complement $H$ in $F$ and $F = K \rtimes H$.

But the answer to the second question is no, we cannot write $F = K \rtimes H$ with both $K$ and $H$ free of finite rank. It can be shown that any nontrivial normal subgroup of $F$ of infinite index in $F$ has infinite rank, and so if $H$ and $K$ are both nontrivial, then $K$ must have infinite rank.


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