Cycle indicator symmetric function and Polya's theorem Let $G$ be a subgroup of the symmetric group $S_n$. Given a partition $\lambda$ of $n$, denote by $n_G(\lambda)$ the number of elements in $G$ of cycle-type $\lambda$. The cycle indicator function of $G$ is the symmetric function $c(G)\in\Lambda_{\mathbb{Q}} = \Lambda \otimes_{\mathbb{Z}} \mathbb{Q}$ defined by
$$
c(G) = \frac{1}{|G|} \sum_{\lambda \vdash n} n_g(\lambda)p_\lambda,
$$
where $p_\lambda$ denotes the power sum symmetric function defined by
$$
p_r = \sum_{i\geq 1} x_i^r, \quad r\geq 1
$$
and
$$
p_\lambda = p_{\lambda_1} p_{\lambda_2}\cdots, \quad \text{where} \quad \lambda = (\lambda_1,\lambda_2,\dots).
$$
Now, let $\Sigma = \{\alpha = (\alpha_1,\dots,\alpha_n)\, | \, \alpha_i \in \mathbb{Z}_{>0}\}$. The group $G$ acts on $\Sigma$ by permuting the components of the vector $\alpha$. For $\alpha\in\Sigma$, write
$$
x_\alpha = x_{\alpha_1}x_{\alpha_2}\cdots x_{\alpha_n}.
$$
The map $\alpha\mapsto x_\alpha$ is clearly constant on each $G$-orbit of $\Sigma$, so we obtain a well defined map $\Sigma/G\to \mathbb{Q}[[x_1,x_2,\dots]]$ sending the orbit containing $\alpha$ (which we also denote by $\alpha$) to $x_\alpha$.
The aim is to prove that
$$
c(G) = \sum_{\alpha\in \Sigma/G} x_\alpha,
$$
which is analogous to Polya's enumeration theorem (at least that is what I. G. Macdonald says in his book Symmetric functions and Hall polynomials, 2nd ed., in Example 9 of Section 2 in Chapter I, and this is the only reason I'm tagging Polya's counting theory).
For this, Macdonald suggests to define
$$
X = \frac{1}{|G|}\sum_{(g,\alpha)\in\Gamma} x_\alpha,
$$
where
$$
\Gamma = \{(g,\alpha)\in G\times\Sigma \, | \, g\alpha = \alpha\}
$$
and to show that $X=c(G)$ and that
$$
X = \sum_{\alpha\in \Sigma/G} x_\alpha.
$$
I proved this last equation without any trouble, but I am unable to show that $X=c(G)$. I have tried by expanding $c(G)$ and trying to compare terms, but I have not suceed.
Any help will be appreciated.
 A: Rewrite $X$ as
$$
\sum_{\lambda\vdash n}\sum_{\substack{g\\ \text{cyc}(g)=\lambda}}\sum_{\substack{\alpha\\ g\alpha=\alpha}} x_\alpha.$$
Now, it will suffice to show that, for a given $g\in G$ of cycle type $\lambda$ we have $\sum_{\alpha\ :\ g\alpha=\alpha} x_\alpha = p_\lambda$. I will indicate how to do this by constructing a bijection. Let us fix, wlog, $g = (1, 2, 3, ..., \lambda_1) (\lambda_1+1, ..., \lambda_1 + \lambda_2) ... (\lambda_1+\cdots+\lambda_{l-1} ,..., \lambda_1+\cdots+\lambda_{l})$.
For the first part of the bijection, we send a monomial $x_{j_1}^{\lambda_1}\cdots x_{j_l}^{\lambda_l}$ of $p_\lambda$ to a vector $(j_1, \stackrel{\lambda_1 \text{ times}}{...},j_1, j_2, \stackrel{\lambda_2 \text{ times}}{...}, j_2, ..., j_l, \stackrel{\lambda_l \text{ times}}{...}, j_l)$ which is obviously invariant by $g$'s action. Note that we are not supposing $j_1, j_2, ..., j_l$ to be distinct from one another.
For the second part, we notice that every vector $\alpha$ in the summation is precisely of the form $(j_1, \stackrel{\lambda_1 \text{ times}}{...},j_1, j_2, \stackrel{\lambda_2 \text{ times}}{...}, j_2, ..., j_l, \stackrel{\lambda_l \text{ times}}{...}, j_l)$, and so we send it back to $x_{j_1}^{\lambda_1}\cdots x_{j_l}^{\lambda_l}$.
