这是一份kcl伦敦大学学院 5CCM232A作业代写的成功案
Thus
$$
\sum_{g \in S}\left|G_{g}\right|=\sum_{g \in G}|F(g)| .
$$
Let $s_{1}, s_{2}, s_{3}, \cdots g^{s_{t}}$ be representatives of the $t$ orbits $(1 .$ e. we take just one $s_{i}$ from each orbit). Then
$$
\sum_{i=1}^{t} \sum_{s \in G\left(s_{i}\right)}\left|G_{s}\right|=\sum_{g \in G}|F(g)| .
$$
By theorem, if $s \in G\left(s_{i}\right)$, then
$$
G_{s_{i}}=g G_{g} g^{-1},
$$
where $g(s)=s_{i}$. Thus
$$
\left|G_{b_{i}}\right|=\left|G_{g}\right|,
$$
when $\varepsilon \in G\left(s_{i}\right)$. Hence
$$
\sum_{i=l}^{t}\left|G\left(s_{i}\right)\right|\left|G_{s_{i}}\right|=\sum_{g \in G}|F(g)| .
$$
By theorein $4.6 .3$ (4),
$$
\left|G\left(s_{i}\right)\right|\left|a_{g_{i}}\right|=|G| .
$$
Hence finally
$$
t|G|=\sum_{g \in G} F(g) .
$$
5CCM232A COURSE NOTES ：
Note that
$$
\bar{g}^{r}=\overline{\left(g^{r}\right)}=g^{r} Z(G) .
$$
Th1s means that $G$ consists of the cosets
$$
g Z(G), \quad g^{2} Z(G), \quad g^{3} Z(G), \ldots, g^{p-1} Z(G), \quad g^{P} Z(G)=Z(G) .
$$
Thus
$$
G=g Z(G) \cup g^{2} Z(G) \cup \ldots U g^{p-1} 2(G) \cup 2(G)
$$
Now let $x, y$ be any two elements of $G$. Then
$$
\text { and } \quad x \in g^{i} Z(G)
$$