Show that each non-abelian finite group $G$ has at least $2$ conjugacy classes which contain at least $2$ elements.
I have a solution which uses the class equation. However, as we have not dealt with this equation in class, we do not have it at our disposal.
Is it possible to prove this proposition without the class equation?
My attempt looks like this:
As $G$ is non-abelian, there are $a$ and $b$ with $a\neq b$ such that $aba^{-1}\neq b $ and also $bab^{-1}\neq a $. Hence, in the conjugacy class of $b$, there are $b$ itself and $aba^{-1}$, which is why this class contains at least two distinct elements. The same can be done for the conjugacy class of $a$ which contains at least $a$ and $bab^{-1}$. At first glance, both classes seem to be distinct.
However, could it be possible that both conjugacy classes are equal?
Then, this attempt of a proof would be invalid, of course.
Best Answer
To arrive at a contradiction assume for the moment that $G$ has only a single conjugacy class of cardinality greater than $1$. This implies that all non-central elements form a conjugacy class, say $G-Z(G)=Cl_G(x)$ for some $x \notin Z(G)$. Note, since $x\neq x^2$, $|C_G(x)| \geq 2$. This yields $|G:C_G(x)|=\#Cl_G(x) \leq |G|/2$, whence $|G|-|Z(G)| \leq |G|/2$, being equivalent to $|G|/2 \leq |Z(G)|$. But $G$ is not abelian, that is, $|Z(G)| \lt |G|$ and we conclude $|G|/2=|Z(G)|$ so $G/Z(G)$ is cyclic (of order $2$), implying $G$ is abelian, a contradiction!
Note (added January 19th 2022) Here is a somewhat different and streamlined proof.
Proposition There do not exist finite groups $G$ having a single conjugacy class of cardinality $\gt 1$.
Proof Let $n \gt 1$ be the cardinality of the conjugacy class. Then by the Class Formula we have $$|G|=|Z(G)|+n$$ Since $n$ is the index of some centralizer, by taking the equation mod $n$ we see that $|Z(G)| \equiv 0$ mod $n$, that is, $n \mid |Z(G)|$. Taking the equation mod $|Z(G)|$, we also see that $|Z(G)| \mid n$. We conclude that $|Z(G)|=n$ and $|G|=2n$. So, $|G/Z(G)| \cong C_2$ is cyclic and $G$ must be abelian, contradicting $n \gt 1$. $\square$
This line of reasoning also generalizes nicely to the non-existence of finite groups $G$ having exactly 2 conjugacy classes of the same cardinality.