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Suppose $k \in \mathbb{Z}$. Can one get the classification of the torsion groups in which every element is conjugate to its $k$-th power?

I was able to solve this problem $\forall k \neq 1$

Case 1:

If $k = 0$, then the only such group is the trivial one.

Proof:

The group identity is conjugate only with itself.

Case 2:

If $k = 1$, then all groups satisfy this property.

Proof:

Every group element is conjugate with itself.

Case 3:

If $k > 1$, then the only such group is the trivial one.

Proof:

Suppose a non-trivial torsion group $G$ satisfies this condition. We can choose a non-trivial element $a$ with minimal possible order. This elements will be some prime $p$. If $p|k$, then $a^k = e$ is not conjugate with $a$. And if $p$ does not divide $k$, then $\exists g \in \mathbb{G}$, such that $gag^{-1} = a^k$, and conjugation by that $g$ induces an automorphism on $\langle a \rangle_p$. But $|Aut(\langle a \rangle_p)| = p - 1 < p$, thus $\langle g \rangle$ contains a non-trivial element with order less than the one of $a$. Contradiction.

Case 4:

If $k < -1$, then the only such group is the trivial one.

Proof:

The proof is exactly the same as for case 3.

My question however is:

What to do with $k = -1$?

For $k = -1$ I know that the following groups satisfy those conditions:

1) the trivial group

2) groups of exponent $2$

3) dihedral groups

4) symmetric groups

However, I do not know, whether there is anything else, and if there is, then how to prove it.

Chain Markov
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    Your argument in the torsion-free case is wrong. In fact, there are examples of (necessarily torsion-free) 2-generator groups with exactly two conjugacy classes. https://math.stackexchange.com/questions/2794639/does-a-group-g-with-an-element-of-finite-order-1-and-two-conjugacy-classes/2794952 I cannot imagine any meaningful classification of such groups. There are also infinitely generated groups with the same property, a proof uses the HNN construction. https://math.stackexchange.com/questions/88980/infinite-group-with-only-two-conjugacy-classes – Moishe Kohan Dec 08 '19 at 17:25
  • The condition for $k=-1$ is equivalent to all complex characters of the group being real-valued, see https://projecteuclid.org/download/pdf_1/euclid.pjm/1102983447. Examples of these groups are symmetric groups. – AnalysisStudent0414 Dec 08 '19 at 17:26
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    To pinpoint the error in your argument, it is not true that $gag^{-1} = a^k$ implies that conjugation by $g$ induces an automorphism of $\langle a \rangle$. This does not hold in the Baumslag-Solitar groups that are defined by the presentations $\langle g,a \mid gag^{-1} = a^k \rangle$. – Derek Holt Dec 08 '19 at 18:03
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    @AnalysisStudent0414 I think what you write applies only to finite groups. – Derek Holt Dec 08 '19 at 18:04
  • @Derek Holt you are right! – AnalysisStudent0414 Dec 08 '19 at 18:07
  • @MoisheKohan, as my argument still seems to work for torsion groups, I edited the question accordingly, thus making it more narrow. – Chain Markov Dec 08 '19 at 18:22
  • The original property (without the torsion assumption) for fixed $k$ with $|k|\neq1$ is a Markov property, and so cannot be detected. Therefore, meaningful classification of groups with this property would be hard. To see that it's a Markov property, note that Osin's groups satisfy this property for any such $k$, while the cyclic group $C_{|k|}$ does not embed into any group with this property (as $a^k$ is trivial for all $a\in C_{|k|}$). – user1729 Dec 09 '19 at 11:55
  • @user1729 but, well, the OP writes "classification" but means "characterization", not necessarily in an algorithmic way when inputting finite group presentations. – YCor Dec 09 '19 at 21:54
  • For $k=-1$: all Q-groups satisfy this property. This includes all finite Weyl groups, not only symmetric groups, see https://mathoverflow.net/questions/10635/why-are-the-characters-of-the-symmetric-group-integer-valued/10652#10652 and https://mathoverflow.net/questions/134581/ for instance. – YCor Dec 09 '19 at 21:58
  • @YCor I know. I tried to cover myself with the caveats "meaningful" and "hard", but clearly I needed more! ("Being trivial" is a Markov property, but it would be foolish to suggest that we haven't classified or characterised the trivial group!). [Also, my use of Osin's groups above is not allowed as they are not necessarily finitely presented. Take $C_{k+1}$ instead.] – user1729 Dec 10 '19 at 08:32

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