Prove or contradict: A finite group with exactly $2$ conjugacy classes always isomorphic to $\mathbb{Z}_2$.
At first I was trying to work with familiar groups to contradict it(permutations, cyclic, dihedral and general linear) but I could not find any counter example.
So I was trying to prove it.Let $G$ be a group that has exactly $2$ conjugacy classes, obviously one of those classes is $e$, and the other class is the rest of the elements.
So I'll define $f:G \rightarrow \mathbb{Z}_2$ as follows - $f(e) = 0$, and $f(g) = 1$ for every other $g \in G$. It is easy to see that $f$ is homomorphism, and also that $\ker (f) = e$, and by the first isomorphism theorem, we get that $G \cong \mathbb{Z}_2$.
Is it correct?
Hint: use the Orbit-Stabiliser Theorem, with $G$ acting on itself by conjugation. (So the orbits are the conjugacy classes of $G$).