Does a group $G$ with an element of finite order $> 1$ and two conjugacy classes necessarily have $|G|=2$? 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.
 A: 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 $|G|=2$.
A: 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!
A: $G$ 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.
