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Let $G$ be a finite group with conjugacy classes $C_1, C_2, ..., C_k$ and let $g_i \in C_i$ be an element for each $i=1, ..., k$

Part 1: State the theorems on row and column orthogonality in the character table of $G$,

Row orthogonality:

$<X_i, X_j>=0$ for $i\neq j $, $1$ for $i=j$

Column orthogonality:

$\sum_{X_i}X_i(g)\overline{X_i(h)}=|C_G(g)|$ if $g, h$ are conjugate, $0$ otherwise $i=j$ where the sum is over irreducible characters $X_i$ and $|C_G(g)|$ is the size of the centralizer.


Part 2: The following shows part of a character table of a group $G$ whose conjugacy classes are $C_1, C_2, ..., C_5$. Below each conjugacy class in the table the size of the centraliser of one of its elements is given. Find the values of $x$ and $y$ in the table. Character table

I am not sure how to apply the orthogonality relations to the table. I think the first step is to consider column $1$, to find $x$ using column orthogonality. Then we can calculate $y$ using row orthogonality of row $5$. However I am not sure how to do this in practice. Many thanks for your help.

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Let $g=h$ be in $C_1$. Then they are conjugate, so $\sum_i\chi_i(g)\overline{\chi_i(g)}=|C_G(g)|$. But that's $\sum_i|\chi_i(g)|^2=24$, and the sum is $1^1+1^2+2^2+3^2+x^2$. So that should get you $x$ (well, you also have to know that $\chi_i(g)$ is a positive integer for $g$ in $C_1$).

Now can you do the row orthogonality to get $y$?

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  • $\begingroup$ So this gives $x=3$. So since the rows are orthogonal, we know that $3\times 3+(-1) \times y+1 \times (-1) + (-1) \times 1 +0=0$ This gives $y=7$. Is that correct? Many thanks $\endgroup$ – thinker Apr 4 '16 at 17:36
  • $\begingroup$ The definition of the inner product for the rows has a factor in it coming from the size of each conjugacy class, and you have left that out. $\endgroup$ – Gerry Myerson Apr 4 '16 at 23:04
  • $\begingroup$ Making any progress, thinker? $\endgroup$ – Gerry Myerson Apr 6 '16 at 3:47
  • $\begingroup$ hi, so if $S_i$ is the size of the $ith$ conjugacy class (i.e. the number of elements in it), my formula would be: $S_1 \times 3 \times 3+S_2 \times (-1) \times y +......=0$? $\endgroup$ – thinker Apr 6 '16 at 9:52
  • $\begingroup$ Yes. ${}{}{}{}$ $\endgroup$ – Gerry Myerson Apr 6 '16 at 12:41

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