Can this be the class-equation of a finite group $G$ of order 10? Can this be the class-equation of a finite group $G$ of order 10?

$10=1+1+2+2+2+2$

I know that if conjugacy class of an element has order one then it must belong to $Z(G)$ and vice versa. Here $o(Z(G))=2$ .Also order of the conjugacy class must divide order of the group .
Keeping these two facts in mind I feel the it must be true.however the answer is no .Why?

A quotient-free argument goes as follows. The centralizer of both singleton points is the whole $G$, while the centralizer of any other point has order $5$. Therefore, the intersection of all the centralizers, which is $Z(G)$, has order either $5$ or $1$. Contradiction.
More generally, if a group of order $pq$ (with $p,q$ primes, $p<q$) has center of order $p$, then the noncentral elements must have the centralizer of order $p$, and the Class Equation reads: $$pq=p+kq$$
for some integer $k>1$: contradiction, because $q\nmid p$. And since $p\nmid q$ either, by the same argument the center can't have order $q$. So, either $G$ is Abelian, or it is centerless.
