I am willing to show that every group of order 15 is cyclic, using class equation.
Let $G$ be a group of order 15. If $G$ is abelian, then $G=Z(G)$ and so for each $a\in G=Z(G)$ we have $cl(a)=\{a\}$. Hence only possible class equation is
$$15=1+1+1+\cdots+1+1(15~\text{times})$$
In this case, $G$ is isomorphic to either $\mathbb{Z}_{15}$ or the external direct product $\mathbb{Z}_3\times \mathbb{Z}_5$. But since $\mathbb{Z}_{15}\simeq \mathbb{Z}_3\times \mathbb{Z}_5$, it follows that $G$ is cyclic.
We now show if $G$ is non-abelian then contradiction will appear.
When $G$ is non-abelian, then $G\neq Z(G)$. Here $|Z(G)|\in \{1,3,5\}$. But if $|Z(G)|=3$ then $|G/Z(G)|=5$ a prime, so $G/Z(G)$ is cyclic and hence $G$ becomes Abelian, contradiction.
Similalry $|Z(G)|\neq 5$ as well and so only possibility is $|Z(G)|=1$. Then the class equation reads
$$15=|G|=|Z(G)|+\sum |cl(a)|=1+\sum |cl(a)|$$
where the sum is taken over the orders of all non-singleton conjugacy classes $cl(a)$ in $G$.
Let there be $x_3$ and $x_5$ number of conjugacy classes of order 3 and 5 respectively in $G$. Then we must have
$$15=1+3x_3+5x_5\Rightarrow 14=3x_3+5x_5$$
which is satisfied by $x_3=3, x_5=1$ so that ultimately the class equation becomes
$$15=1+(3+3+3)+5$$
I do not know how to bring contradiction here. Any help?
Thanks in well advance.
Best Answer
All the elements in a single conjugacy class have the same order. You have to have an even number of elements of order $3$ and the number of elements of order $5$ has to be divisible by $4$. We already have the identity element.
Look at the elements of order $5$ - there must be $4, 8, 12$ of these. This means $10, 6, 2$ elements of order $3$ (given none of order $15$)
The only possibility from what you have is $3+5=8$ elements of order $5$ and $3+3=6$ elements of order $3$.
There are various ways of working from here - but with $8$ elements of order $5$ there would have to be two subgroups of order $5$ and conjugation (consider the larger class) would have to take at least one generator of one to a generator of the other, and you should be able to see that this can't split the eight elements as $3+5$ for a contradiction.