I'm working through problem 4.4 of Boyd's CVX Optim book.
Suppose $G=\{Q_{1},...Q_{k}\}\subset \mathbb{R}^{n \times n}$ is a group. We define $\bar{x} = (1/k)\sum_{i=1}^{k}Q_{i}x$ and the fixed subspace $\mathcal{F} = \{x| Q_{i}x = x, i=1,...k\}$.
Show that for any $x \in \mathbb{R}^{n}$ we have $\bar{x} \in \mathcal{F}$.
From the solutions, I understand the first two steps but not the last one where the permutation of $Q_{i}$ by $Q_{j}$ denoted $Q_{\sigma(i)}$ becomes $Q_{i}$ again.
$Q_{j}\bar{x} = (1/k) \sum_{i=1}^{k}Q_{j}Q_{i}x = (1/k) \sum_{i=1}^{k} Q_{\sigma(i)}x = (1/k) \sum_{i=1}^{k} Q_{i}x$
My naive guess is that there is some permutation of the set of $G$ by $Q_{j}$ such that $Q_{\sigma(i)} = Q_{i}$ but I have no idea.
$\sigma$ is a permutation.
Consider the example where $\sigma(1)=2, \sigma(2)=3, \sigma(3)=1$.
\begin{align}\sum_{i=1}^3Q_{\sigma(i)}(x)&=Q_2x + Q_3x + Q_1x \\ &= Q_1 x+Q_2x+Q_3x\\ &= \sum_{i=1}^3 Q_ix\end{align}
Being a permutation, $\sigma$ would map $\{1, \ldots, n\}$ to $\{1, \ldots, n\}$. that is the only difference is the order of summation but that would not change the sum.
$$\{1, \ldots, n\}=\{ \sigma(1), \ldots, \sigma(n)\}$$