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.


1 Answer 1


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 $y$?

  • $\begingroup$ 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, 2016 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. $\endgroup$ Apr 4, 2016 at 23:04
  • $\begingroup$ Making any progress, thinker? $\endgroup$ Apr 6, 2016 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, 2016 at 9:52
  • $\begingroup$ Yes. ${}{}{}{}$ $\endgroup$ Apr 6, 2016 at 12:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.