I'm reading the representation theory book by Gordon James. There is a part involving filling out the character table using column orthogonality. And he went by using the first relation onto the first column with itself, and then every other columns. He then made the following remark:
Notice that the orgonality relations hold between all pairs of columns, although our calculation used only those relation involve the first column...
Is this true in general? When using this relation, am I supposed to test the relation between all the column? Or they just conveniently become orthogonal to each other? As I am trying to complete the following character table:
\begin{array}{cccccc } g & 1 & a & a^2 & b & ab \\ |C_G(g)| & 8 & 2 & 8 & 4 & 4 \\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1\\ \chi_2 & 1 & 1 & 1 & -1 & -1\\ \chi_3 & 1 & -1 & 1 & 1 & -1\\ \chi_4 & 1 & -1 & 1 & -1 & 1\\ \hline \end{array}
Surely if I was doing the same thing as James Gordon, I would still have 5 column orthogonal to each other and get the following table: \begin{array}{cccccc } g & 1 & a & a^2 & b & ab \\ |C_G(g)| & 8 & 2 & 8 & 4 & 4 \\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1\\ \chi_2 & 1 & 1 & 1 & -1 & -1\\ \chi_3 & 1 & -1 & 1 & 1 & -1\\ \chi_4 & 1 & -1 & 1 & -1 & 1\\ \chi_5 & 2 & 0 & -2 & 0 & 0\\ \hline \end{array}
However I can't help but notice that if I had used the column orthogonality relation on the $a^2$ with its own, then I'd have $\chi_5(a^2)=2$. That's when the $1$ column and the $a^2$ column are no longer orthogonal. Did I misread the author's remark? Should I check the orthogonality between every pair of column? Or the algorithm of using exactly one column to complete the whole table would do the job just fine?