Let $C_{j,k}^l$ ,usually called class structure constants, eg Jansen and Boon and/or JQ Chen, be the number of times the class $l$ is generated from the product of classes $j,k$ and $c_j=c_{j}$ (a fronting minus sign just means the inverse class which is the same class for ambivalent classes) is the order or number of elements in class $j$. It is given (i) $c_j c_k=\sum C_{j,k}^l c_l$ , (ii) $C_{j,k}^l=C_{(j),(k)}^{l}$ both of which I can understand. It is also stated in Jansen and Boon (iii) $c_l C_{j,k}^l=c_j C_{k(l)}^{j}$ which I can't prove in the general case. I can reconcile it in the special identity case $l=1$ , say, where $c_l=c_{l}=1$ meaning the class of the single identity element in which case the only non zero result in (iii) is $j=k$ and $C_{j,j}^1=c_j , C_{j,1}^j=C_{j,1}^j=1$. Years ago I think I did reconcile the general case but now cannot and have tried in vain to search for the proof which has to be given in some texts likely quite a few decades ago with no luck. I can reconcile, eg by $g_k$ I mean a group element of group G from class $k$, that for some triplet of elements $g_j, g_k,g_l$ lhs. of (iii) says $g_j g_k=g_l$ which means $g_k g_l^{1}=g_j^{1}$ which corresponds to rhs. But this isn't saying much nor any quantitative one to one correspondance. I could go on and on in a round about way with examples etc. showing it is true but need a direct concise proof which seems should not be all that difficult.

4$\begingroup$ You recently posted a closely related question (mathoverflow.net/questions/316162), which at the moment has 3 closing votes as "unclear what you're asking". Could you clarify the link between the two questions? $\endgroup$ – YCor Nov 30 '18 at 18:51
My derivation: conjugate both of the last two pertinent expressions, $g_j g_k=g_l$ and $g_k g_l^{1}=g_j^{1}$ which follows from the first. Sum over all $\circ_{G} $(order or number of elemets of group G) elements of G and obtain \begin{eqnarray} & \sum_g g^{1}g_j g_k\, g= \sum_g g^{1}g_j \,g g^{1}g_k\, g=\sum_g g^{1}g_l \,g \label{class_struct_1} \\ & \sum_g g^{1}g_k g_l^{1} g=\sum_g g^{1}g_k\,g g^{1} g_l^{1} g=\sum_g g^{1}g_j^{1} g \label{class_struct_2}\end{eqnarray} The right hand sides of both contain total $\circ_{G}$ elements from classes $g_l,g_j^{1}$ respectively each generally being repeated more than once with each member of the classes being repeated the same number of times  that "number of times" is not generally the same for the two different expressions. By looking at the form of the left hand sides then by definition that same $\circ_{G}$ number is from some multiple of $C_{j,k}^l$ in first expression and some generally different multiple of $C_{k(l)}^{j}$ in the 2nd. And since by definition $C_{j,k}^l$ is the number of times class $l$ appears or is repeated from the product of classes $j,k$ then that multiple must be inversely proportional to the number $c_l,c_{j}=c_j$ of elements in classes $l,j$ from right hand side of first,2nd respectively. So $c_l C_{j,k}^l=c_j C_{k(l)}^{j}$ which is what to have been proved. Note I have not shown the prior statement that the same number of elements from each member of the class on rhs's are repeated the same number of times but this was shown prior in the text by Jansen and Boon and already understand that. Ok back again to I have shown to my satisfaction the proof of the statement with additional comment of showing the same number of repeats of a certain pair from each class occurs the same number of times in both cases which is the number of times the same element from class k is repeated. All this is kind of 'wordy' and feel there is a more concise direct way of putting it that the average reader can understand if someone can answer or inform of in what paper or text the proof can be found.
P.S. This has nothing to do with my question in the prior post a few days ago using some of the same symbols and which still stands.

$\begingroup$ above regarding the statement "...$g_l,g_j^{1}$ respectively..." is meant the classes that contain the two elements. $\endgroup$ – Reza Vino Zmth Dec 1 '18 at 4:59

1$\begingroup$ There should also be a close vote for answers, with explanation "unclear what you're answering" :P $\endgroup$ – მამუკა ჯიბლაძე Dec 4 '18 at 22:55

$\begingroup$ close vote ? that would be ridiculous as the question has never been answered in any suitable way. Again prove $c_l C_{j,k}^l=c_j C_{k(l)}^{j}$ is the question. $\endgroup$ – Reza Vino Zmth Dec 5 '18 at 17:56

$\begingroup$ This was a joke, sorry. But then, there was a reason behind it. I realize that it is only natural to primarily address specialists for an answer, but still, there are users (like me) who, although nonspecialists, would like to learn something from the answers, and from the question as well. For such people, both the question and answer are extremely hard to follow since you provide the absolute minimum of background. You provide a reference, yes, but it would be way easier for a reader if you could just say few more words in the beginning. It took me quite a while to reach the second book $\endgroup$ – მამუკა ჯიბლაძე Dec 5 '18 at 20:32

$\begingroup$ (could not get the first) to find out that you work with sums of all elements in a conjugacy class of some finite group. But I still cannot quite figure out what does it mean "the number of times the class $l$ is generated from the product of classes $j$, $k$". Is this the coefficient at $l$ of the product of $j$ and $k$ in the standard basis of the class algebra? $\endgroup$ – მამუკა ჯიბლაძე Dec 5 '18 at 20:36