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Derek is right, let $G$ act transitively on a set $\Omega$ and put $H=G_\alpha$, for some $\alpha \in \Omega$, then the permutation character $\pi=1_H^G$, the trivial character of $H$ induced to $G$. This is a standard textbook theorem. And indeed, by Frobenius Reciprocity, if $\chi \in Irr(G)$, then $[\chi, \pi]=[\chi, 1_H^G]=[\chi_H,1_H] \leq \chi(1)$. ...


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The hypothesis of the statement is that $G$ acts transitively on $M$, meaning that for any $p,q\in M$ there exists $g\in G$ such that $g\centerdot p=q$. In your example, $S^1$ does not act transitively on $S^2$.


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Since the matrix is square, it is enough to show that the rows are linearly independent. But since they are orthogonal with respect to an inner product, and each has inner product $1$ with itself, this is a standard exercise in linear algebra (we just need that for each row $v$ we have $\langle v,v\rangle\neq 0$). To elaborate a bit on the exercise, one ...



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