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Very close to this question but instead of specifying finite $G$, we are given a group element with finite order $>1$.

First, $e$ is its own conjugacy class. Let the element $a \neq e$ have order $n$. All elements of $G$ other than $e$ belong to $C(a)$.

Since $o(a)=n$, we can prove that all the other elements in $C(a)$ also have order $n$. Moreover, if $n$ was composite $n=mk$ then $o(a^k)$ would divide $m$ which is a contradiction. Therefore $n$ is prime.

I am stuck at this point. If $a^2\neq e$, $ a^2= xax^{-1}$. This should lead to a contradiction but I am unable to arrive at it. Any hints would be helpful.

user1729
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RSS
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  • I think there are some Tarski monsters with just two conjugacy classes. In particular, it is not true that $G$ is finite. – verret May 24 '18 at 18:09
  • @verret interesting. Would you have any references for this? – RSS May 24 '18 at 18:19
  • @verret I don't think Tarski monsters are known to have a two conjugacy classes. They are simple though. (Tarski monster's were constructed by Ol'shanskii in the early 80s, while in a 2010 Annals paper Osin produced the first examples of inifinite groups with two conjugacy classes. If Tarski monsters were known to have two-conjugacy classes then Osin's paper would not have been such a big deal. Osin's example are often torsion-free, and, glancing at his paper, probably always contain an element of infinite order (he uses HNN-extensions). Paper is here: https://arxiv.org/abs/math/0411039) – user1729 May 25 '18 at 11:29
  • @user1729 In fact, if a group has order $> 2$ and precisely two conjugacy classes, then it is torsion free, by an argument fairly similar to the one in the answer. – Tobias Kildetoft May 25 '18 at 12:57
  • @TobiasKildetoft Does RSS's answer not show precisely this? As in, should "by an argument fairly similar to" not be "by the argument of"? – user1729 May 25 '18 at 13:33
  • @user1729 Ahh, yes. Not sure why I wrote it like that. I may have been thinking of the slightly stronger statement that if all non-identity elements have the same finite order $p$, then the group has at least $p$ conjugacy classes, which does follow by a very similar argument. – Tobias Kildetoft May 25 '18 at 18:58

3 Answers3

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Here is a way to complete the proof:

Define $I_g(x) = gxg^{-1}$ (conjugation map).

Since $a^2$ and $a$ are in the same conjugation class, $\exists g\in G$ such that $$ I_g(a) = gag^{-1}=a^2$$ $$ I_{g^2}(a)= g(gag^{-1})g^{-1} = ga^2g^{-1} =gaag^{-1}=gag^{-1}gag^{-1} = a^4$$ and so on:

$$ I_{g^n}(a) = g^nag^{-n} = a^{2^n}$$

Since $g^n = g^{-n} = e,$ $$ a^{2^n}=a$$ $$ \implies 2^n = 1 \mod{n}$$ But using Fermat's little theorem, for prime n: $$ 2^n = 2 \mod{n} $$ Which leads to a contradiction. Thus $a^2=e$. This means $n=2$ and all elements are self-inverses. Thus for any two $a,b \in G$: $$(ab)^{-1}=b^{-1}a^{-1}=ab $$ $$ab=ba$$ G is abelian. This means $C(a) = \{a\} $ and hence $|G|=2$.

RSS
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As RSS has shown, (two conjugacy classes)+(some element of finite order) implies the group is finite. The following theorem is then the "best possible" result:

Theorem (D. Osin, 2010). Any countable group $G$ can be embedded into a $2$–generated group $C$ such that any two elements of the same order are conjugate in $C$. Moreover, the number of elements of finite order in $G$ is equal to the number of elements of finite in $C$.

For example:

  • if $G$ is torsion-free then $C$ has two conjugacy classes,
  • if $G$ is cyclic of prime order then $C$ has three conjugacy classes.

The proof of this theorem is highly non-trivial. It appeared in the Annals of Mathematics, so was big news at the time!

user1729
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$G$ is the union of $C(a) $ and the neutral element. The fact the order if $a$ is odd implies that the order of the the group is even. By Lagrange you have an element of order 2. Contradiction.

Tsemo Aristide
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  • How can I prove that $G = $? – RSS May 24 '18 at 17:50
  • Since every element $x$ distinct of the identity is in $C(a) $ and the order of $a$ is $2$, $C(a) ={a} $, $a=x$. – Tsemo Aristide May 24 '18 at 17:56
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    How do we know the order is 2? To use Lagrange theorem (as in your answer) we need to prove G is finite first. – RSS May 24 '18 at 17:58
  • My argument shows thatthe order is even and you have shown that it is a prime. – Tsemo Aristide May 24 '18 at 18:12
  • I have shown the order of the non-identity elements to be prime. I don't know how to show the group is finite. Your answer has the hidden assumption that $ G= $ as far as I can tell. – RSS May 24 '18 at 18:18
  • But every element distinct of the identity is in $C(a) $ and & C(a) $ is finite. – Tsemo Aristide May 24 '18 at 18:21
  • How do we prove that $C(a)$ is finite? Maybe you could add it to your answer. – RSS May 24 '18 at 18:23
  • @Derek do those only have two conjugacy classes? I can't see how elements can be conjugate to powers of themselves, as that seems to imply that the conjugating element normalizes the subgroup, which leads to a larger proper subgroup. – Tobias Kildetoft May 24 '18 at 20:23
  • Yes I got confused. I will delete my comment. There are Tarski Monsters in which all subgroups of order $p$ are conjugate- they have $p$ conjugacy classes of elements. – Derek Holt May 24 '18 at 20:43