Mixing
finite groups: class constants relation
It is stated in Jansen, finite groups:
$c_l C_{j,k}^l=c_j C_{k(-l)}^{-j}$
where $c_l$ is the number of elements in class $l$ and where a fronting negative sign denotes the inverse class which is same as the class for ambivalent classes only and $c_l=c_{-l}$ always regardless. I understand and can reconcile the expression in the special case for $l=1$ meaning the class of identity in which case $c_l=1$ and the class of inverse $-l$ is same as for $l$ in this case since class of identity is ambivalent. In this case the only nonzero nontrivial result is for $j=-k,c_j=c_k$ being the number of elements in the class $k$. But cannot derive the general case and cannot find it anywhere on internet search nor other group theory texts.
It is also given :
$c_j c_k=sum C_{j,k}^l c_l$ sum over classes.
And
$C_{j,k}^l=C_{(-j),(-k)}^{-l}$
both of which I can reconcile but cannot derive the general case of the first expression above from these.
group-theory
asked Nov 22 '18 at 10:55
user158293user158293
61
add a comment |
It is stated in Jansen, finite groups:
$c_l C_{j,k}^l=c_j C_{k(-l)}^{-j}$
where $c_l$ is the number of elements in class $l$ and where a fronting negative sign denotes the inverse class which is same as the class for ambivalent classes only and $c_l=c_{-l}$ always regardless. I understand and can reconcile the expression in the special case for $l=1$ meaning the class of identity in which case $c_l=1$ and the class of inverse $-l$ is same as for $l$ in this case since class of identity is ambivalent. In this case the only nonzero nontrivial result is for $j=-k,c_j=c_k$ being the number of elements in the class $k$. But cannot derive the general case and cannot find it anywhere on internet search nor other group theory texts.
It is also given :
$c_j c_k=sum C_{j,k}^l c_l$ sum over classes.
And
$C_{j,k}^l=C_{(-j),(-k)}^{-l}$
both of which I can reconcile but cannot derive the general case of the first expression above from these.
group-theory
asked Nov 22 '18 at 10:55
user158293user158293
61
add a comment |
It is stated in Jansen, finite groups:
$c_l C_{j,k}^l=c_j C_{k(-l)}^{-j}$
where $c_l$ is the number of elements in class $l$ and where a fronting negative sign denotes the inverse class which is same as the class for ambivalent classes only and $c_l=c_{-l}$ always regardless. I understand and can reconcile the expression in the special case for $l=1$ meaning the class of identity in which case $c_l=1$ and the class of inverse $-l$ is same as for $l$ in this case since class of identity is ambivalent. In this case the only nonzero nontrivial result is for $j=-k,c_j=c_k$ being the number of elements in the class $k$. But cannot derive the general case and cannot find it anywhere on internet search nor other group theory texts.
It is also given :
$c_j c_k=sum C_{j,k}^l c_l$ sum over classes.
And
$C_{j,k}^l=C_{(-j),(-k)}^{-l}$
both of which I can reconcile but cannot derive the general case of the first expression above from these.
group-theory
asked Nov 22 '18 at 10:55
user158293user158293
61
It is stated in Jansen, finite groups:
$c_l C_{j,k}^l=c_j C_{k(-l)}^{-j}$
where $c_l$ is the number of elements in class $l$ and where a fronting negative sign denotes the inverse class which is same as the class for ambivalent classes only and $c_l=c_{-l}$ always regardless. I understand and can reconcile the expression in the special case for $l=1$ meaning the class of identity in which case $c_l=1$ and the class of inverse $-l$ is same as for $l$ in this case since class of identity is ambivalent. In this case the only nonzero nontrivial result is for $j=-k,c_j=c_k$ being the number of elements in the class $k$. But cannot derive the general case and cannot find it anywhere on internet search nor other group theory texts.
It is also given :
$c_j c_k=sum C_{j,k}^l c_l$ sum over classes.
And
$C_{j,k}^l=C_{(-j),(-k)}^{-l}$
both of which I can reconcile but cannot derive the general case of the first expression above from these.
group-theory
group-theory
asked Nov 22 '18 at 10:55
user158293user158293
61
asked Nov 22 '18 at 10:55
user158293user158293
61
asked Nov 22 '18 at 10:55
user158293user158293
61
asked Nov 22 '18 at 10:55
user158293user158293
61
asked Nov 22 '18 at 10:55
user158293user158293
61
61
add a comment |
add a comment |
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