5
$\begingroup$

The Nielsen-Schreier theorem states that subgroups of a free subgroup are free. Is this hold also for groups with operations?

Explicitly, let $G$ be a fixed group. Let $F$ be a group with $G$-action which is free (as a group with $G$-action). Let $F'\subset F$ be a subgroup closed by the $G$-action. Then, must $F$ be free as a group with $G$-action?

Improve this question
$\endgroup$
15
  • 1
    $\begingroup$ @MarkGrant the definition of being a free $G$-group is not what you say, but a certain universal property in the category of $G$-groups. Yet, a characterization is: a $G$-group is free (as $G$-group) if and only if it admits a subset $X$ such that $X$ is a free $G$-subset, and $G$ is freely generated by $X$ as group. $\endgroup$
    – YCor
    Commented Oct 27, 2018 at 8:10
  • 1
    $\begingroup$ A counterexample is the free $(Z/2Z)$-group on a single generator, as replied below by მამუკა ჯიბლაძე. This is the first test-case for such a question... $\endgroup$
    – YCor
    Commented Oct 27, 2018 at 8:12
  • 1
    $\begingroup$ The question would be maybe less obvious if one restricts to $G$ free (on a set $I$), so that a $G$-group is a group endowed with a family (indexed by $I$) of automorphisms. $\endgroup$
    – YCor
    Commented Oct 27, 2018 at 8:18
  • 1
    $\begingroup$ (...) If $Y$ is a $G$-set, then the free group $F_Y$ on $Y$ is naturally a $G$-group. Now endow $X\times G$ with the structure of $G$-set $g\cdot (x,h)=(x,gh)$; this is a free $G$-set, with orbits naturally indexed by $X$. Then $F_{X\times G}$ is a free $G$-group, on the set $X\times\{1\}$. Indeed, consider any map $X\to W$, $W$ a $G$-group. It extends uniquely to a $G$-map $X\times G\to W$, and the universal property of $F_{X\times G}$ as a group yields a unique extension $F_{X\times G}\to W$, which is a group homomorphism. The uniqueness implies that the latter is also $G$-equivariant. $\endgroup$
    – YCor
    Commented Oct 27, 2018 at 13:02
  • 1
    $\begingroup$ @YCor: Ah, right. As I understand it, what you describe is left adjoint to the forgetful functor from $G$-groups to sets, while what I'm describing is left adjoint to the forgetful functor from $G$-groups to $G$-sets. Both seem equally valid and useful definitions to me. $\endgroup$
    – Mark Grant
    Commented Oct 27, 2018 at 17:08

1 Answer 1

Reset to default
9
$\begingroup$

Let $G$ be a two element group; the free group on two generators $x,y$ with the action of $G$ interchanging them is a free $G$-group (on one generator). Its subgroup generated by $xy^{-1}$ is closed under the $G$-action but is not a free $G$-group.

Improve this answer
$\endgroup$
1
  • $\begingroup$ of course... didn't thought of that $\endgroup$
    – user49822
    Commented Oct 27, 2018 at 14:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .