Basis for Center of Group Ring












1














There is already a similar post, that has an answer to my question, but it is quite short and I don't get it... So I am trying to show that
begin{align}
{e_K |~ K subset G~ conjugacy~ class }
end{align} is a basis for $Z(mathbb{Z}[G])$. Serre even writes "one immediately checks that the $e_K$ form a basis", but even reading the other post I don't see that...



My approach to this is:



Be $x in Z(mathbb{Z}[G])$, which is equivalent to: $x = gxg^{-1}$ for all $g in G$. Writing
$x = sumlimits_{h in G} lambda_h h$, we get $x = sumlimits_{h in G} lambda_h ghg^{-1}$. I don't really know how to go on at this point. I mean, I know that when $a,b in K$ (same conjugacy class) you get $sumlimits_{s in K} mu_s s = musumlimits_{s in K} s$, since
begin{align}
a = hbh^{-1} Rightarrow mu_a = mu_b ~forall a,bin K
end{align}
So all elements in the same conjugacy class have the same scalar.



But from here I don't really know how I can generate the $x = sumlimits_{h in G} lambda_h ghg^{-1}$ with the $e_K$ ...



I would be really thankful, if someone could give me quite a basic proof that even I understand...






abstract-algebra representation-theory







share|cite|improve this question






















  • Could you link to the "similar post" you mention? This would help us in understanding your problem.
    – Pierre-Guy Plamondon
    Jun 3 at 14:24












  • Ah yes, of course... math.stackexchange.com/questions/2803614/…
    – Gilligans
    Jun 3 at 16:53
















1














There is already a similar post, that has an answer to my question, but it is quite short and I don't get it... So I am trying to show that
begin{align}
{e_K |~ K subset G~ conjugacy~ class }
end{align} is a basis for $Z(mathbb{Z}[G])$. Serre even writes "one immediately checks that the $e_K$ form a basis", but even reading the other post I don't see that...



My approach to this is:



Be $x in Z(mathbb{Z}[G])$, which is equivalent to: $x = gxg^{-1}$ for all $g in G$. Writing
$x = sumlimits_{h in G} lambda_h h$, we get $x = sumlimits_{h in G} lambda_h ghg^{-1}$. I don't really know how to go on at this point. I mean, I know that when $a,b in K$ (same conjugacy class) you get $sumlimits_{s in K} mu_s s = musumlimits_{s in K} s$, since
begin{align}
a = hbh^{-1} Rightarrow mu_a = mu_b ~forall a,bin K
end{align}
So all elements in the same conjugacy class have the same scalar.



But from here I don't really know how I can generate the $x = sumlimits_{h in G} lambda_h ghg^{-1}$ with the $e_K$ ...



I would be really thankful, if someone could give me quite a basic proof that even I understand...






abstract-algebra representation-theory







share|cite|improve this question






















  • Could you link to the "similar post" you mention? This would help us in understanding your problem.
    – Pierre-Guy Plamondon
    Jun 3 at 14:24












  • Ah yes, of course... math.stackexchange.com/questions/2803614/…
    – Gilligans
    Jun 3 at 16:53














1












1








1







There is already a similar post, that has an answer to my question, but it is quite short and I don't get it... So I am trying to show that
begin{align}
{e_K |~ K subset G~ conjugacy~ class }
end{align} is a basis for $Z(mathbb{Z}[G])$. Serre even writes "one immediately checks that the $e_K$ form a basis", but even reading the other post I don't see that...



My approach to this is:



Be $x in Z(mathbb{Z}[G])$, which is equivalent to: $x = gxg^{-1}$ for all $g in G$. Writing
$x = sumlimits_{h in G} lambda_h h$, we get $x = sumlimits_{h in G} lambda_h ghg^{-1}$. I don't really know how to go on at this point. I mean, I know that when $a,b in K$ (same conjugacy class) you get $sumlimits_{s in K} mu_s s = musumlimits_{s in K} s$, since
begin{align}
a = hbh^{-1} Rightarrow mu_a = mu_b ~forall a,bin K
end{align}
So all elements in the same conjugacy class have the same scalar.



But from here I don't really know how I can generate the $x = sumlimits_{h in G} lambda_h ghg^{-1}$ with the $e_K$ ...



I would be really thankful, if someone could give me quite a basic proof that even I understand...






abstract-algebra representation-theory







share|cite|improve this question













There is already a similar post, that has an answer to my question, but it is quite short and I don't get it... So I am trying to show that
begin{align}
{e_K |~ K subset G~ conjugacy~ class }
end{align} is a basis for $Z(mathbb{Z}[G])$. Serre even writes "one immediately checks that the $e_K$ form a basis", but even reading the other post I don't see that...



My approach to this is:



Be $x in Z(mathbb{Z}[G])$, which is equivalent to: $x = gxg^{-1}$ for all $g in G$. Writing
$x = sumlimits_{h in G} lambda_h h$, we get $x = sumlimits_{h in G} lambda_h ghg^{-1}$. I don't really know how to go on at this point. I mean, I know that when $a,b in K$ (same conjugacy class) you get $sumlimits_{s in K} mu_s s = musumlimits_{s in K} s$, since
begin{align}
a = hbh^{-1} Rightarrow mu_a = mu_b ~forall a,bin K
end{align}
So all elements in the same conjugacy class have the same scalar.



But from here I don't really know how I can generate the $x = sumlimits_{h in G} lambda_h ghg^{-1}$ with the $e_K$ ...



I would be really thankful, if someone could give me quite a basic proof that even I understand...






abstract-algebra representation-theory




abstract-algebra representation-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jun 3 at 7:51









Gilligans

184




184












  • Could you link to the "similar post" you mention? This would help us in understanding your problem.
    – Pierre-Guy Plamondon
    Jun 3 at 14:24












  • Ah yes, of course... math.stackexchange.com/questions/2803614/…
    – Gilligans
    Jun 3 at 16:53


















  • Could you link to the "similar post" you mention? This would help us in understanding your problem.
    – Pierre-Guy Plamondon
    Jun 3 at 14:24












  • Ah yes, of course... math.stackexchange.com/questions/2803614/…
    – Gilligans
    Jun 3 at 16:53
















Could you link to the "similar post" you mention? This would help us in understanding your problem.
– Pierre-Guy Plamondon
Jun 3 at 14:24






Could you link to the "similar post" you mention? This would help us in understanding your problem.
– Pierre-Guy Plamondon
Jun 3 at 14:24














Ah yes, of course... math.stackexchange.com/questions/2803614/…
– Gilligans
Jun 3 at 16:53




Ah yes, of course... math.stackexchange.com/questions/2803614/…
– Gilligans
Jun 3 at 16:53










1 Answer
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Notice that $sum lambda_h h = sum lambda_h ghg^{-1}$ implies that $lambda_h = lambda_{ghg^{-1}}$ for all $g$, i.e $lambda_h$ is constant along conjugacy classes. It follows that an element is the center can be written $ sum lambda_r c_r$ where $r$ runs along the conjugacy classes and $c_r = sum_{h in r}h$. Since the $c_r$ are obviously linearly independant the claim follows.






share|cite|improve this answer





















  • But why do you need $x = sumlimits_{h in G} lambda_h h = sumlimits_{h in G} lambda_h ghg^{-1}$ ? I mean $lambda_h = lambda_{ghg^{-1}}$ does follow for all elements of the same conjugacy class anyway, doesn't it ?
    – Gilligans
    Jun 3 at 19:30












  • I don't understand your objection. We want to show that a basis of $Z(Bbb Z[G])$ is given by the $c_r$ (or $e_K$ with your notation). Essentially we need to show that any $x$ in the center can be written as linear combinaison of the $e_K$. This is exactly what my answer does.
    – Nicolas Hemelsoet
    Jun 3 at 19:36










  • Ohh, am I right in the thought that the $e_K$ even form a basis for the whole $mathbb{Z}[G]$ ?
    – Gilligans
    Jun 3 at 19:44










  • @Gilligans : certainly not ! For example, if $e_K$ is a conjugacy class with more than one element and $g in e_K$, then $x=g$ can't be expressed as linear combinaisons of the $e_K$.
    – Nicolas Hemelsoet
    Jun 3 at 19:46










  • Ok thank you very much, I think I got it now :)
    – Gilligans
    Jun 3 at 20:06











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Notice that $sum lambda_h h = sum lambda_h ghg^{-1}$ implies that $lambda_h = lambda_{ghg^{-1}}$ for all $g$, i.e $lambda_h$ is constant along conjugacy classes. It follows that an element is the center can be written $ sum lambda_r c_r$ where $r$ runs along the conjugacy classes and $c_r = sum_{h in r}h$. Since the $c_r$ are obviously linearly independant the claim follows.






share|cite|improve this answer





















  • But why do you need $x = sumlimits_{h in G} lambda_h h = sumlimits_{h in G} lambda_h ghg^{-1}$ ? I mean $lambda_h = lambda_{ghg^{-1}}$ does follow for all elements of the same conjugacy class anyway, doesn't it ?
    – Gilligans
    Jun 3 at 19:30












  • I don't understand your objection. We want to show that a basis of $Z(Bbb Z[G])$ is given by the $c_r$ (or $e_K$ with your notation). Essentially we need to show that any $x$ in the center can be written as linear combinaison of the $e_K$. This is exactly what my answer does.
    – Nicolas Hemelsoet
    Jun 3 at 19:36










  • Ohh, am I right in the thought that the $e_K$ even form a basis for the whole $mathbb{Z}[G]$ ?
    – Gilligans
    Jun 3 at 19:44










  • @Gilligans : certainly not ! For example, if $e_K$ is a conjugacy class with more than one element and $g in e_K$, then $x=g$ can't be expressed as linear combinaisons of the $e_K$.
    – Nicolas Hemelsoet
    Jun 3 at 19:46










  • Ok thank you very much, I think I got it now :)
    – Gilligans
    Jun 3 at 20:06
















2














Notice that $sum lambda_h h = sum lambda_h ghg^{-1}$ implies that $lambda_h = lambda_{ghg^{-1}}$ for all $g$, i.e $lambda_h$ is constant along conjugacy classes. It follows that an element is the center can be written $ sum lambda_r c_r$ where $r$ runs along the conjugacy classes and $c_r = sum_{h in r}h$. Since the $c_r$ are obviously linearly independant the claim follows.






share|cite|improve this answer





















  • But why do you need $x = sumlimits_{h in G} lambda_h h = sumlimits_{h in G} lambda_h ghg^{-1}$ ? I mean $lambda_h = lambda_{ghg^{-1}}$ does follow for all elements of the same conjugacy class anyway, doesn't it ?
    – Gilligans
    Jun 3 at 19:30












  • I don't understand your objection. We want to show that a basis of $Z(Bbb Z[G])$ is given by the $c_r$ (or $e_K$ with your notation). Essentially we need to show that any $x$ in the center can be written as linear combinaison of the $e_K$. This is exactly what my answer does.
    – Nicolas Hemelsoet
    Jun 3 at 19:36










  • Ohh, am I right in the thought that the $e_K$ even form a basis for the whole $mathbb{Z}[G]$ ?
    – Gilligans
    Jun 3 at 19:44










  • @Gilligans : certainly not ! For example, if $e_K$ is a conjugacy class with more than one element and $g in e_K$, then $x=g$ can't be expressed as linear combinaisons of the $e_K$.
    – Nicolas Hemelsoet
    Jun 3 at 19:46










  • Ok thank you very much, I think I got it now :)
    – Gilligans
    Jun 3 at 20:06














2












2








2






Notice that $sum lambda_h h = sum lambda_h ghg^{-1}$ implies that $lambda_h = lambda_{ghg^{-1}}$ for all $g$, i.e $lambda_h$ is constant along conjugacy classes. It follows that an element is the center can be written $ sum lambda_r c_r$ where $r$ runs along the conjugacy classes and $c_r = sum_{h in r}h$. Since the $c_r$ are obviously linearly independant the claim follows.






share|cite|improve this answer












Notice that $sum lambda_h h = sum lambda_h ghg^{-1}$ implies that $lambda_h = lambda_{ghg^{-1}}$ for all $g$, i.e $lambda_h$ is constant along conjugacy classes. It follows that an element is the center can be written $ sum lambda_r c_r$ where $r$ runs along the conjugacy classes and $c_r = sum_{h in r}h$. Since the $c_r$ are obviously linearly independant the claim follows.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jun 3 at 14:31









Nicolas Hemelsoet

5,7452417




5,7452417












  • But why do you need $x = sumlimits_{h in G} lambda_h h = sumlimits_{h in G} lambda_h ghg^{-1}$ ? I mean $lambda_h = lambda_{ghg^{-1}}$ does follow for all elements of the same conjugacy class anyway, doesn't it ?
    – Gilligans
    Jun 3 at 19:30












  • I don't understand your objection. We want to show that a basis of $Z(Bbb Z[G])$ is given by the $c_r$ (or $e_K$ with your notation). Essentially we need to show that any $x$ in the center can be written as linear combinaison of the $e_K$. This is exactly what my answer does.
    – Nicolas Hemelsoet
    Jun 3 at 19:36










  • Ohh, am I right in the thought that the $e_K$ even form a basis for the whole $mathbb{Z}[G]$ ?
    – Gilligans
    Jun 3 at 19:44










  • @Gilligans : certainly not ! For example, if $e_K$ is a conjugacy class with more than one element and $g in e_K$, then $x=g$ can't be expressed as linear combinaisons of the $e_K$.
    – Nicolas Hemelsoet
    Jun 3 at 19:46










  • Ok thank you very much, I think I got it now :)
    – Gilligans
    Jun 3 at 20:06


















  • But why do you need $x = sumlimits_{h in G} lambda_h h = sumlimits_{h in G} lambda_h ghg^{-1}$ ? I mean $lambda_h = lambda_{ghg^{-1}}$ does follow for all elements of the same conjugacy class anyway, doesn't it ?
    – Gilligans
    Jun 3 at 19:30












  • I don't understand your objection. We want to show that a basis of $Z(Bbb Z[G])$ is given by the $c_r$ (or $e_K$ with your notation). Essentially we need to show that any $x$ in the center can be written as linear combinaison of the $e_K$. This is exactly what my answer does.
    – Nicolas Hemelsoet
    Jun 3 at 19:36










  • Ohh, am I right in the thought that the $e_K$ even form a basis for the whole $mathbb{Z}[G]$ ?
    – Gilligans
    Jun 3 at 19:44










  • @Gilligans : certainly not ! For example, if $e_K$ is a conjugacy class with more than one element and $g in e_K$, then $x=g$ can't be expressed as linear combinaisons of the $e_K$.
    – Nicolas Hemelsoet
    Jun 3 at 19:46










  • Ok thank you very much, I think I got it now :)
    – Gilligans
    Jun 3 at 20:06
















But why do you need $x = sumlimits_{h in G} lambda_h h = sumlimits_{h in G} lambda_h ghg^{-1}$ ? I mean $lambda_h = lambda_{ghg^{-1}}$ does follow for all elements of the same conjugacy class anyway, doesn't it ?
– Gilligans
Jun 3 at 19:30






But why do you need $x = sumlimits_{h in G} lambda_h h = sumlimits_{h in G} lambda_h ghg^{-1}$ ? I mean $lambda_h = lambda_{ghg^{-1}}$ does follow for all elements of the same conjugacy class anyway, doesn't it ?
– Gilligans
Jun 3 at 19:30














I don't understand your objection. We want to show that a basis of $Z(Bbb Z[G])$ is given by the $c_r$ (or $e_K$ with your notation). Essentially we need to show that any $x$ in the center can be written as linear combinaison of the $e_K$. This is exactly what my answer does.
– Nicolas Hemelsoet
Jun 3 at 19:36




I don't understand your objection. We want to show that a basis of $Z(Bbb Z[G])$ is given by the $c_r$ (or $e_K$ with your notation). Essentially we need to show that any $x$ in the center can be written as linear combinaison of the $e_K$. This is exactly what my answer does.
– Nicolas Hemelsoet
Jun 3 at 19:36












Ohh, am I right in the thought that the $e_K$ even form a basis for the whole $mathbb{Z}[G]$ ?
– Gilligans
Jun 3 at 19:44




Ohh, am I right in the thought that the $e_K$ even form a basis for the whole $mathbb{Z}[G]$ ?
– Gilligans
Jun 3 at 19:44












@Gilligans : certainly not ! For example, if $e_K$ is a conjugacy class with more than one element and $g in e_K$, then $x=g$ can't be expressed as linear combinaisons of the $e_K$.
– Nicolas Hemelsoet
Jun 3 at 19:46




@Gilligans : certainly not ! For example, if $e_K$ is a conjugacy class with more than one element and $g in e_K$, then $x=g$ can't be expressed as linear combinaisons of the $e_K$.
– Nicolas Hemelsoet
Jun 3 at 19:46












Ok thank you very much, I think I got it now :)
– Gilligans
Jun 3 at 20:06




Ok thank you very much, I think I got it now :)
– Gilligans
Jun 3 at 20:06


















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