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: From the given class equation, $|Z(G)| =2$. Then $G/Z(G)$ has order 5, and since 5 is prime we must have that $G/Z(G) \simeq \mathbb{Z}/5\mathbb{Z}$. In particular, the quotient is cyclic, which implies that $G$ is abelian. But this cannot be the case because the class equation shows us that there are conjugacy classes with size larger than 1, a contradiction.
A: Hint: Look at the quotient of such a group with the center. Can you infer that such a group must be abelian?
A: Using Z-test one can also prove this way :
Suppose |G|=N and $|z(G)|=m$ if the class equation for G, $N=1+1+...+1\text{(m times)}+n_{m+1}+n_{m+2}+...+n_{k}$ where each $n_{m+1}>1 \  \ 'm'$ must divides $\frac{N}{n_i}$ for each i.
Here $|z(G)|=2$ but $2$ does not divides $\frac{N}{n_3}=\frac{10}{2}=5$
Therefore given equation is not a class equation.
A: $G/Z(G)$ is cyclic group of order $5$ and hence it implies that $G$  is abelian see, which means $G=Z(G)$ but class equation tells us $|Z(G)|=2$. Hence not possible.
A: I don't see an easy way to eliminate the class equation, but using the Sylow theorems you can classify the groups of order 10 and see whether that class equation matches any of them 
By the third Sylow theorem, $G$ has one Sylow 5-subgroup because $n_5 \mid 2$ and $n_5 \equiv 1 \pmod{5}$.
Also by the third Sylow theorem, $G$ can have either 1 or 5 Sylow 2-subgroups; if it has one, then $G$ is cyclic (it is isomorphic to $C_2\times C_5$), so it is abelian and its equation is $10=1+1+\cdots +1$.
The only other option is that it has 5 Sylow 2-subgroups $H_1,H_2, \dots, H_5$. Then by the second Sylow theorem, these subgroups are conjugate of each other.
For $i\neq j$, we have $H_i=gH_jg^{-1}$ for some $g \in G$. Then the nonidentity element $h_i\in H_i$ can be written as $gh_jg^{-1}$, where $h_j$ is the nonidentity element of $H_j$. Thus, $h_j$ and $h_i$ are in the same conjugacy class.
This is true for any pair of distinct indices, so their conjugacy class has 5 elements.
Since there is no 5 in the equation $1+1+2+2+2+2$, it cannot be the equation of our group.
A: 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.
