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 $&gt;1$.

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.
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 $|G|=2$.
$G$ 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.