I have the following Proposition in Fischer's Algebra
Proposition: Let $G$ be a group and $\alpha: G \times M \to M$ a $G$-Action on the finite set $M$. Let $a_1, \dots , a_r$ be a representatives system of the $G$-Orbits then we have $$|G|=\sum_{i=1}^r[G:G_{a_i}] $$
Corollarly he introduces the Class equations of groups
Prop2: Let $G$ be a finite group. Let $Z(G)$ denote the center of $G$ and let $x_1, \dots , x_k$ be the representatives of the $G$-Orbits with several elements of the conjugation, then $$|G|=|Z(G)| + \sum_{i=1}^k[G:Z_G(a_i)] $$
My problems: Here is what I understand:
I know that that Prop2 should follow immediately from the proposition if we consider a group action $G$ on itself by conjugation, that is $\alpha:G \times G \to G$ by $\alpha(g,a_i)=ga_ig^{-1}$
Then we have by definition $Z_G(a_i)=G_{a_i}$
Furthermore if I force the orbit to consist only of one element I obtain that $$ Ga=\lbrace a \rbrace \iff gag^{-1}=a \iff ga=ag \iff a \in Z(G)$$ Which shows that the center of $G$ consists precisely of the $G$-Orbits with one element.
I don't understand how to proceed from here and why Fischer immediately manages to deduce Prop2.
Maybe it will be more clear to write down the generalized class equation in place of your first proposition (though what I'm about to write is equivalent).
I will use the notation $$(g,x)\mapsto g.x$$ for the action of $g$ on $X$. For $x\in X$, I'll write $G.x$ for the orbit of $x$, $G_x$ for the stabilizer of $x$, and $X^G=\{x\in X\mid $g.x=x$\mbox{ for all }g\in G\}$ for the set of invariants under this action. Note that we have $$|G.x|=1\Leftrightarrow x\in X^G.$$
Proposition: Let $G$ act on a finite set $X$, and $S$ a set of representatives of the nontrivial orbits (i.e. orbits of size >1). Then, $$|X|=|X^G|+\sum_{x\in S}[G:G_x]$$
proof: Since $X^G$ consists of all elements $x\in X$ such that $|G.x|=1$ and the nontrivial orbits are precisely the $G.x$, $x\in S$, we have $X=X^G\sqcup\left(\bigsqcup_{x\in S} G.x\right)$ where $\sqcup$ denotes disjoint union. Therefore, \begin{align*} |X|&=|X^G|+\sum_{x\in S}|G.x|\\&=|X^G|+\sum_{x\in S}[G:G_x], \end{align*} where the last equality uses the equality $|G.x|=[G:G_x]$ (which can be seen immediately by identifying the fibers of the action map $G\to G.x$ as the cosets of $G_x$).
Now apply this to the action of $G$ on itself by conjugation. We have $G^G=Z(G)$ and, for $x\in G$, $G_x=Z_G(x)$.