4
$\begingroup$

An element $s$ of a group $G$ is a logical generator of $G$ iff every element of $G$ can be defined in the first order language of groups with $s$ as a parameter. In this case we may call $G$ a logically cyclic group. This is equivalent to say that for any elementary extension $G^{\ast}$ of $G$ and any automorphism $\alpha:G^{\ast}\to G^{\ast}$, the condition $\alpha(s)=s$ implies $\alpha_{|_G}=\mathrm{id}$.

As a result, the set $\{ s\}$ is an $\mathrm{Aut}$-basis of $G$ in terms of [1] (i.e., every automorphism of $G$ is uniquely determined by its value on $s$).

I started to study logically cyclic groups in [2]. Recently, I realized that in fact I don't have any example of a group which is not logically cyclic but it has a singleton $\mathrm{Aut}$-basis. So, here is my question:

Is it true that if $\{ s\}$ is an $\mathrm{Aut}$-basis of $G$, then $s$ is a logical generator of $G$?

[1] G. Cutolo, C. Nicotera: Subgroups defining automorphisms in locally nilpotent groups, Forum Mathematicum, 15 (2003).

[2] M. Shahryari: On logically cyclic groups, J. Group Theory, 18 (2015).

$\endgroup$
1
  • 3
    $\begingroup$ An observation: if $\{s\}$ is an Aut-basis onf $G$, then $G$ is abelian. Indeed, the inner automorphism $g\mapsto g^s$ fixes $s$, hence it is the identity, i.e., $s\in Z(G)$; but then all inner automorphisms fix $s$, hence they are the identity, i.e., $G$ is abelian. $\endgroup$ Commented Aug 13, 2022 at 7:07

1 Answer 1

6
$\begingroup$

$\{1\}$ is an Aut-basis of the group of profinite integers $\hat{\mathbb Z}$, but since this group has cardinality $2^\omega$, it has no logical generator. However, every element of $\hat{\mathbb Z}$ is type-definable from $1$ (i.e., any pair of distict elements are distinguishable by a first-order formula with parameter $1$); using Ali Enayat’s terminology, the structure $\langle\hat{\mathbb Z},+,1\rangle$ is Leibnizian. (Conversely, if $G$ is the group reduct of a $\mathbb Z$-group such that $\{1\}$ is an Aut-basis of $G$, then all elements of $G$ are type-definable from $1$, and $G$ embeds in $\hat{\mathbb Z}$; see my paper Rigid models of Presburger arithmetic.)

Similarly, if $p$ is any prime, the group of $p$-adic integers $\mathbb Z_p$ has no logical generator (being uncountable), but all its elements are type-definable from $1$, thus $\{1\}$ is an Aut-basis.

Note that “every element of $G$ is type-definable from $s$” gives you a general property to consider that is intermediate between “$\{s\}$ is an Aut-basis of $G$”, and “$s$ is a logical generator of $G$”.

$\endgroup$
1
  • $\begingroup$ The best natural example. Thank you so much. The intermediate property is also very nice generalization. $\endgroup$
    – Sh.M1972
    Commented Aug 13, 2022 at 12:10

You must log in to answer this question.

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