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I know there is a lot that a character table can tell me but while I am figuring out a lot there are a few points that I just can not seem to figure out and any help and/or general reasoning is appreciated.

\begin{array}{rrrrrrrrrrr} & C_1 & C_2 & C_3 & C_4 & C_5 \\ \chi_0 & 1 & 1 & 1 & 1 & 1 \\ \chi_1 & 1 & 1 & 1 & -1 & -1 \\ \chi_2 & 1 & -1 & 1 & 1 & -1 \\ \chi_3 & 1 & -1 & 1 & -1 & 1 \\ \chi_3 & 2 & 0 & -2 & 0 & 0 \\ \end{array}

Question: Can I tell the order of the elements of the conjugacy classes?

I, for example can tell the order of the group $1^2 + 1^2 + 1^2 + 1^2 + 2^2 =8$ and by Schurs Lemma the columns and rows are orthogonal to one-another. I am concerned particularily with the linear characters as I know they are trying tell me something. I know that as there are 4 linear characters then the index of the derived subgroup $G:[G,G]$ is 4 which would make the order of the derived subgroup 2 as $|G|/|[G,G]| = 8/|[G,G]| = 4$.

Now some patterns that I am trying to put together with what I know: I know that the order of the group divided by sum of the square of elements in a conjugacy class give me the number of elements in the conjugacy class (correct?). That the sum of the squares of the elements in the conjugacy class is the index of the centralizer. I see that, for example the sum of the elements in the conjugacy class $C_2 = C_4= C_5 = 0$ but that for $C_3$ we have 1 + 1+ 1+ 1 + -2 =2. Does this mean anything?

Now I know that the 5 characters listed are the irredcible ones but

Question: what is the value of all the irreducible characters on the elements of $[G,G]$? I know that the characters are just numbers (traces of the matrix representation of which the character was derived), but don't know what it means to multiply numbers by elements of the derived subgroup.

Thanks again.

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On your first question, I don't know of any simple connections, but see Order of group elements from a character table for a less simple connection.

Yes, you are right that the size of the conjugacy class is the order of the group divided by the sum of squares of the corresponding column. This is related to something you glossed over in saying that "the columns and rows are orthogonal". They aren't strictly orthogonal, only if you weight the conjugacy classes by their sizes. To obtain an orthogonal matrix (one in which the rows and columns are orthonormal), you need to multiply each column by the square root of the size of the corresponding conjugacy class, and divide the entire matrix by the square root of the order of the group. It follows that the sum of squares of a column is the order of the group divided by the size of the conjugacy class.

The sums of the elements in a column don't have any significance (as far as I'm aware). Every column but the first is orthogonal to the first, and that means that the sum of the elements, weighted by the dimensions of the corresponding representations, is zero for those columns. Where the unweighted sum is zero, this is just because those conjugacy classes have character $0$ in the non-linear character, which must be weighted by $2$ to get the overall sum $0$.

On your last question, I don't understand why you talk about multiplying numbers by elements of the derived subgroup. The values of the irreducible characters on the elements of $[G,G]$ are simply their values in columns $C_1$ and $C_3$, since those are the two conjugacy classes that make up $[G,G]$.

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  • $\begingroup$ Thanks joriki, insightful indeed! I think that a point I was missing is that the two conjugacy classes that make up $[G,G]$ are $C_1$ and $C_3$. Please correct me if I am wrong, but this is because these are the only two groups with one element in their conjugacy class (as $1^2 + 1^2 + 1^2 + 1^2 + -2^2 =8$) and 8/8 = 1) - but it seems there is much more to it and I am a bit unsure as $[G,G]$ contains all the elements which are not Abelian hence $G/[G,G]$ is the Abelianization of $G$. $\endgroup$ – ThinkingInnovate Sep 16 '15 at 21:44
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    $\begingroup$ @ThinkingInnovate: No, elements that are in a conjugacy class of their own form the centre of the group; it just so happens in this case that the centre is the same as the derived group. For instance in the symmetric groups $S_n$, for $n\gt2$, the centre consists only of the identity element whereas the derived group is all of $A_n$. The adjective "abelian" is usually reserved for groups; elements that commute with all other elements (i.e. which lie in the centre) can be called "central". $\endgroup$ – joriki Sep 16 '15 at 21:54
  • $\begingroup$ I think part of the issue is that I am stuck a bit on what it means to be a "value of an irreducible character" as a character is the trace of some matrix representation - hence we have that the character is some number - this is where I got that something was supposed to happen multiplication on elements of $[G,G]$. I was figuring that maybe I was somehow supposed to find the character values of elements of the derived group, multiply them together and something magical happens. Perhaps you could help enlighten me on what is really going on? Thanks much again. $\endgroup$ – ThinkingInnovate Sep 16 '15 at 21:58
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    $\begingroup$ @ThinkingInnovate: I'm slightly guessing here where the confusion might lie. Yes, a character is the trace of a matrix representation, and since the matrix representation associates a matrix with every group element, the character associates a trace (i.e. a number) with every group element. Due to the cyclic invariance of the trace, we have $\operatorname{Tr}A^{-1}BA=\operatorname{Tr}BAA^{-1}=\operatorname{Tr}B$, which means that the character is constant on each conjugacy class. Thus it can also be regarded as a function on the conjugacy classes, as indeed the character table does. $\endgroup$ – joriki Sep 16 '15 at 22:08

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