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I am trying to figure out whether or not the following property is first-order expressible in the language of groups. $$\text{$G$ has a subgroup $H$ with which it forms a Frobenius pair $(H,G)$.}$$ My search online yielded no answers. Even if there is a positive or negative result, it seems to be not very well-known.

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    $\begingroup$ Can you define what is a Frobenius pair? $\endgroup$ Feb 6, 2023 at 8:21
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    $\begingroup$ Frobenius pair: $H\neq 1$, $H\neq G$, and every element of $G$ has at most one fixed point on $G/H$ (i.e., $g\notin H$ implies $g^{-1}Hg\cap H=\{1\}$). $\endgroup$
    – YCor
    Feb 6, 2023 at 8:39
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    $\begingroup$ One natural variant of the question is: does there exist a 1st order sentence $F$ such that for every finite group $G$, the group $G$ is Frobenius over some subgroup iff it satisfies $F$. $\endgroup$
    – YCor
    Feb 6, 2023 at 8:55
  • $\begingroup$ By the way for infinite groups I'm not sure there's a single clear-cut definition of Frobenius group. One could pick as a definition, the existence of a proper nontrivial malnormal subgroup. But one could also add the (very strong, but automatic for finite groups) requirement that, for some nontrivial malnormal proper subgroup, the complement of the union of its conjugates, along with $\{1\}$, forms a normal subgroup. Hence, if you are really interested in infinite groups (besides pseudofinite ones), it is important you fully write down a definition of Frobenius pair. $\endgroup$
    – YCor
    Feb 6, 2023 at 14:08

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I'll answer here positively the natural variant of the question:

does there exist a 1st order sentence such that for every finite group $G$, the group $G$ is Frobenius over some subgroup iff the given sentence holds in $G$.

In a sense, this sound to me like a more natural question, since Frobenius pairs are really motivated by finite group theory, and such a result allows to extend the notion to pseudofinite groups.

(Concerning infinite groups, my bet would be that the result is false although I can't prove it at the moment.)

In a group $G$ and $g\in G$, write $C_g$ the centralizer of $g$, and $L_g=\{h:[g,hgh^{-1}]=g\}$. Let $A_G$ be the set of elements of $G$ which commute with all their conjugates (i.e., are contained in some abelian normal subgroup). Let $B_G$ be the set of $g\in G$ such that $C_g$ is a normal subgroup, and $B'_G$ the set of $g\in B_G$ such that every element of $G\smallsetminus C_g$ acts by conjugation on $C_g$ with no nontrivial fixed point. Let $D_G$ be the set of $g\in G$ such that $L_g$ is a subgroup.

Proposition. A group is Frobenius iff there exists $z\in B'_G\smallsetminus\{1\}$ and $h\in D_G\smallsetminus\{1\}$ such that $G=C_z\rtimes L_h$.

Proof: For basic facts on Frobenius pairs see the Wikipedia page. Suppose $G=K\rtimes H$ is Frobenius. So $K$ is nilpotent (Thompson) and nontrivial, choose any central $z\in G$: then $K=C_z$ and $z\in B'_G$. Also, $H$ is a Frobenius complement, and hence has a nontrivial solvable radical (follows from results of Zassenhaus: it is indeed either solvable and nontrivial, or has a normal copy of $\mathrm{SL}_2(\mathbf{F}_5)$, which has a nontrivial center), hence has a nontrivial abelian normal subgroup; choose $h$ therein. Then $L_h=H$ follows easily: choose $g\in L_h$ since if $[h,ghg^{-1}]=1$ and the centralizer of $h$ is contained in $H$ it follows that $ghg^{-1}\in H$ and in turn $g\in H$ follows.

Conversely, if $G$ has these properties, then it is straightforward that $L_h$ is malnormal (i.e. $L_h\smallsetminus\{1\}$ is disjoint each from its conjugates by elements not in $L_h$) and this implies that $G$ is Frobenius.$\Box$

Now the property of the proposition is obviously expressible by a first-order sentence $F_0$ in $G$. Hence for every finite group $G$, $G$ satisfies $F_0$ iff $G$ is Frobenius.

In particular, we can define a pseudofinite group to be Frobenius iff it satisfies $F_0$. Note that this does not depend on $F_0$. This can be defined directly as: a group $G$ is pseudofinite Frobenius if for every 1st order sentence $F$ satisfied by $G$ there exists a finite Frobenius group satisfying $F$.

While the latter could be defined in any case, the existence of $F_0$ above ensures the following: if a group $G$ is pseudofinite Frobenius, then there exists a 1st order formula $F$ such that every finite group $G'$ satisfying $F$ is Frobenius. (More informally: it means that a pseudofinite group cannot be both pseudo-Frobenius and pseudo-(non-Frobenius).)

[For context, there also exists a first-order formula that characterizes, among finite groups, the solvable ones (Wilson). But there is none that characterizes, among finite groups, the nilpotent ones (YC-Wilson).]

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  • $\begingroup$ Note: this also shows that the Frobenius kernels and complements are definable (by formulas that don't depend on the Frobenius group). The Frobenius is actually parameter-free definable, namely as the set of those elements whose conjugacy class has a nontrivial centralizer. $\endgroup$
    – YCor
    Feb 6, 2023 at 14:02
  • $\begingroup$ (Typo in the above comment: "The Frobenius" should be "The Frobenius kernel". Of course the Frobenius complement cannot be parameter-free definable since it is not a normal subgroup.) $\endgroup$
    – YCor
    Feb 14, 2023 at 13:37

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