Could we get any information from about class equation from index of sylow subgroups I have the class equation of a group of order $84$ 
as,  class equation  $84 = 1+1+12+21+21+28$ to check it is right or wrong 
i understands definition of class equation,
but what i wondering is that is there any way that if we knew about the Sylow subgroups of the group could we get any information about class equation 
I can calculate index of Sylow subgroups 
 A: In general, the center $Z(G) \subseteq C_G(g)$, the centralizer of each $g \in G$. As is well-known, the size of the conjugacy class $Cl_G(g)$ of $g$ is $|G:C_G(g)|$. It follows that $\#Cl_G(g)$ divides $|G:Z(G)|$ for all $g \in G$. The order of the center can be read off the class formula (count the number of $1's$). Hence in your case $|Z(G)|=2$ and $|G:Z(G)|=42$. And clearly $12$ and $28$ do not divide $42$.
Still another approach is the following. Suppose you can find a non-trivial normal subgroup $N$ of $G$. Then observe that $N$ is the disjoint union of conjugacy classes of $G$, that is $N =\{1\} \cup Cl_G(n_1) \cup \cdots \cup Cl_G(n_t)$ for certain $n_i \in N-\{1\}$. So the conjugacy classes are building blocks of normal subgroups (but not every union of conjugacy classes gives rise to a normal subgroup).
And yes, now we can leverage Sylow Theory. Take $P \in Syl_7(G)$, it is easy to see that $P \lhd G$ (the only divisor of $(84:7)=12$ being $\equiv 1$ mod $7$ is $1$). So by the previous remarks $P=\{1\} \cup$ "some conjugacy classes". But $|P|=7$ and we cannot find distinct conjugacy classes to fill the remaining $6$ elements. Again, a contradiction.
