Dimension of an irreducible representation and the index of the group's center












1












$begingroup$


Let $(pi,V_pi)$ be an irreducible representation of finite group $G$, over algebraically closed field $F$ s.t. $char(F)$ is coprime to $|G|$ (for example $mathbb{C}$).



Prove that $dim(pi)^2 le [G:Z(G)]$ where $Z(G)$ is the group's center.




My thoughts so far: let us recall that there exists a character of G (i.e. 1-dim representation) of the center: $omega _pi : Z(G) to F^*$ such that for every $zin Z(G): pi(z) = omega_pi (z) Id_{V_pi}$. also we know $dim(Ind_{Z(G)}^G (omega _pi )) = [G:Z(G)]$, i'm not sure how to continue or even if my intuition so far is correct.




thanks ahead






finite-groups representation-theory







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$endgroup$












  • $begingroup$
    I know a way to prove that $dim(pi)^2 mid [G: Z(G)]|G|$ but I don't know the result you mention; I'd be interested in seeing a proof
    $endgroup$
    – Max
    Jan 12 at 12:59
















1












$begingroup$


Let $(pi,V_pi)$ be an irreducible representation of finite group $G$, over algebraically closed field $F$ s.t. $char(F)$ is coprime to $|G|$ (for example $mathbb{C}$).



Prove that $dim(pi)^2 le [G:Z(G)]$ where $Z(G)$ is the group's center.




My thoughts so far: let us recall that there exists a character of G (i.e. 1-dim representation) of the center: $omega _pi : Z(G) to F^*$ such that for every $zin Z(G): pi(z) = omega_pi (z) Id_{V_pi}$. also we know $dim(Ind_{Z(G)}^G (omega _pi )) = [G:Z(G)]$, i'm not sure how to continue or even if my intuition so far is correct.




thanks ahead






finite-groups representation-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    I know a way to prove that $dim(pi)^2 mid [G: Z(G)]|G|$ but I don't know the result you mention; I'd be interested in seeing a proof
    $endgroup$
    – Max
    Jan 12 at 12:59














1












1








1





$begingroup$


Let $(pi,V_pi)$ be an irreducible representation of finite group $G$, over algebraically closed field $F$ s.t. $char(F)$ is coprime to $|G|$ (for example $mathbb{C}$).



Prove that $dim(pi)^2 le [G:Z(G)]$ where $Z(G)$ is the group's center.




My thoughts so far: let us recall that there exists a character of G (i.e. 1-dim representation) of the center: $omega _pi : Z(G) to F^*$ such that for every $zin Z(G): pi(z) = omega_pi (z) Id_{V_pi}$. also we know $dim(Ind_{Z(G)}^G (omega _pi )) = [G:Z(G)]$, i'm not sure how to continue or even if my intuition so far is correct.




thanks ahead






finite-groups representation-theory







share|cite|improve this question









$endgroup$




Let $(pi,V_pi)$ be an irreducible representation of finite group $G$, over algebraically closed field $F$ s.t. $char(F)$ is coprime to $|G|$ (for example $mathbb{C}$).



Prove that $dim(pi)^2 le [G:Z(G)]$ where $Z(G)$ is the group's center.




My thoughts so far: let us recall that there exists a character of G (i.e. 1-dim representation) of the center: $omega _pi : Z(G) to F^*$ such that for every $zin Z(G): pi(z) = omega_pi (z) Id_{V_pi}$. also we know $dim(Ind_{Z(G)}^G (omega _pi )) = [G:Z(G)]$, i'm not sure how to continue or even if my intuition so far is correct.




thanks ahead






finite-groups representation-theory




finite-groups representation-theory






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asked Jan 12 at 9:05









ned grekerzbergned grekerzberg

494318




494318












  • $begingroup$
    I know a way to prove that $dim(pi)^2 mid [G: Z(G)]|G|$ but I don't know the result you mention; I'd be interested in seeing a proof
    $endgroup$
    – Max
    Jan 12 at 12:59


















  • $begingroup$
    I know a way to prove that $dim(pi)^2 mid [G: Z(G)]|G|$ but I don't know the result you mention; I'd be interested in seeing a proof
    $endgroup$
    – Max
    Jan 12 at 12:59
















$begingroup$
I know a way to prove that $dim(pi)^2 mid [G: Z(G)]|G|$ but I don't know the result you mention; I'd be interested in seeing a proof
$endgroup$
– Max
Jan 12 at 12:59




$begingroup$
I know a way to prove that $dim(pi)^2 mid [G: Z(G)]|G|$ but I don't know the result you mention; I'd be interested in seeing a proof
$endgroup$
– Max
Jan 12 at 12:59










1 Answer
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1












$begingroup$

First, note that $pi|_{Z(G)}=dim(pi)omega_pi$ and $pi$ is a summand of $mathrm{Ind}_{Z(G)}^Gomega_pi$. The inequality now follows since $dimmathrm{Ind}_{Z(G)}^Gomega_pi=[G:Z(G)]$.



EDIT: To see that $pi$ is a direct summand of $mathrm{Ind}_{Z(G)}^Gomega_pi$, use Frobenius reciprocity:
$$langlemathrm{Ind}_{Z(G)}^Gomega_pi,pirangle_G=langleomega_pi,mathrm{Res}^G_{Z(G)}pirangle_{Z(G)}=langle omega_pi,dim(pi)omega_pirangle_{Z(G)}=dimpi.$$
In fact, this means that $pi$ appears with multiplicity $dim(pi)$ in $mathrm{Ind}_{Z(G)}^Gomega_pi$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    would you be willing to explain why $pi$ is a summand of $Ind_{Z(G)}^G omega _ pi$?
    $endgroup$
    – ned grekerzberg
    Jan 15 at 9:10












  • $begingroup$
    You only get $dim (pi) leq [G:Z(G)]$, not $dim (pi)^2$; or am I missing something ?
    $endgroup$
    – Max
    Jan 16 at 18:46










  • $begingroup$
    The Frobenius Reciprocity argument implies that $pi$ appears $dimpi$ times in $Indomega_pi$.
    $endgroup$
    – David Hill
    Jan 16 at 18:59












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$begingroup$

First, note that $pi|_{Z(G)}=dim(pi)omega_pi$ and $pi$ is a summand of $mathrm{Ind}_{Z(G)}^Gomega_pi$. The inequality now follows since $dimmathrm{Ind}_{Z(G)}^Gomega_pi=[G:Z(G)]$.



EDIT: To see that $pi$ is a direct summand of $mathrm{Ind}_{Z(G)}^Gomega_pi$, use Frobenius reciprocity:
$$langlemathrm{Ind}_{Z(G)}^Gomega_pi,pirangle_G=langleomega_pi,mathrm{Res}^G_{Z(G)}pirangle_{Z(G)}=langle omega_pi,dim(pi)omega_pirangle_{Z(G)}=dimpi.$$
In fact, this means that $pi$ appears with multiplicity $dim(pi)$ in $mathrm{Ind}_{Z(G)}^Gomega_pi$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    would you be willing to explain why $pi$ is a summand of $Ind_{Z(G)}^G omega _ pi$?
    $endgroup$
    – ned grekerzberg
    Jan 15 at 9:10












  • $begingroup$
    You only get $dim (pi) leq [G:Z(G)]$, not $dim (pi)^2$; or am I missing something ?
    $endgroup$
    – Max
    Jan 16 at 18:46










  • $begingroup$
    The Frobenius Reciprocity argument implies that $pi$ appears $dimpi$ times in $Indomega_pi$.
    $endgroup$
    – David Hill
    Jan 16 at 18:59
















1












$begingroup$

First, note that $pi|_{Z(G)}=dim(pi)omega_pi$ and $pi$ is a summand of $mathrm{Ind}_{Z(G)}^Gomega_pi$. The inequality now follows since $dimmathrm{Ind}_{Z(G)}^Gomega_pi=[G:Z(G)]$.



EDIT: To see that $pi$ is a direct summand of $mathrm{Ind}_{Z(G)}^Gomega_pi$, use Frobenius reciprocity:
$$langlemathrm{Ind}_{Z(G)}^Gomega_pi,pirangle_G=langleomega_pi,mathrm{Res}^G_{Z(G)}pirangle_{Z(G)}=langle omega_pi,dim(pi)omega_pirangle_{Z(G)}=dimpi.$$
In fact, this means that $pi$ appears with multiplicity $dim(pi)$ in $mathrm{Ind}_{Z(G)}^Gomega_pi$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    would you be willing to explain why $pi$ is a summand of $Ind_{Z(G)}^G omega _ pi$?
    $endgroup$
    – ned grekerzberg
    Jan 15 at 9:10












  • $begingroup$
    You only get $dim (pi) leq [G:Z(G)]$, not $dim (pi)^2$; or am I missing something ?
    $endgroup$
    – Max
    Jan 16 at 18:46










  • $begingroup$
    The Frobenius Reciprocity argument implies that $pi$ appears $dimpi$ times in $Indomega_pi$.
    $endgroup$
    – David Hill
    Jan 16 at 18:59














1












1








1





$begingroup$

First, note that $pi|_{Z(G)}=dim(pi)omega_pi$ and $pi$ is a summand of $mathrm{Ind}_{Z(G)}^Gomega_pi$. The inequality now follows since $dimmathrm{Ind}_{Z(G)}^Gomega_pi=[G:Z(G)]$.



EDIT: To see that $pi$ is a direct summand of $mathrm{Ind}_{Z(G)}^Gomega_pi$, use Frobenius reciprocity:
$$langlemathrm{Ind}_{Z(G)}^Gomega_pi,pirangle_G=langleomega_pi,mathrm{Res}^G_{Z(G)}pirangle_{Z(G)}=langle omega_pi,dim(pi)omega_pirangle_{Z(G)}=dimpi.$$
In fact, this means that $pi$ appears with multiplicity $dim(pi)$ in $mathrm{Ind}_{Z(G)}^Gomega_pi$.






share|cite|improve this answer











$endgroup$



First, note that $pi|_{Z(G)}=dim(pi)omega_pi$ and $pi$ is a summand of $mathrm{Ind}_{Z(G)}^Gomega_pi$. The inequality now follows since $dimmathrm{Ind}_{Z(G)}^Gomega_pi=[G:Z(G)]$.



EDIT: To see that $pi$ is a direct summand of $mathrm{Ind}_{Z(G)}^Gomega_pi$, use Frobenius reciprocity:
$$langlemathrm{Ind}_{Z(G)}^Gomega_pi,pirangle_G=langleomega_pi,mathrm{Res}^G_{Z(G)}pirangle_{Z(G)}=langle omega_pi,dim(pi)omega_pirangle_{Z(G)}=dimpi.$$
In fact, this means that $pi$ appears with multiplicity $dim(pi)$ in $mathrm{Ind}_{Z(G)}^Gomega_pi$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 16 at 18:57

























answered Jan 14 at 21:28









David HillDavid Hill

9,5761619




9,5761619












  • $begingroup$
    would you be willing to explain why $pi$ is a summand of $Ind_{Z(G)}^G omega _ pi$?
    $endgroup$
    – ned grekerzberg
    Jan 15 at 9:10












  • $begingroup$
    You only get $dim (pi) leq [G:Z(G)]$, not $dim (pi)^2$; or am I missing something ?
    $endgroup$
    – Max
    Jan 16 at 18:46










  • $begingroup$
    The Frobenius Reciprocity argument implies that $pi$ appears $dimpi$ times in $Indomega_pi$.
    $endgroup$
    – David Hill
    Jan 16 at 18:59


















  • $begingroup$
    would you be willing to explain why $pi$ is a summand of $Ind_{Z(G)}^G omega _ pi$?
    $endgroup$
    – ned grekerzberg
    Jan 15 at 9:10












  • $begingroup$
    You only get $dim (pi) leq [G:Z(G)]$, not $dim (pi)^2$; or am I missing something ?
    $endgroup$
    – Max
    Jan 16 at 18:46










  • $begingroup$
    The Frobenius Reciprocity argument implies that $pi$ appears $dimpi$ times in $Indomega_pi$.
    $endgroup$
    – David Hill
    Jan 16 at 18:59
















$begingroup$
would you be willing to explain why $pi$ is a summand of $Ind_{Z(G)}^G omega _ pi$?
$endgroup$
– ned grekerzberg
Jan 15 at 9:10






$begingroup$
would you be willing to explain why $pi$ is a summand of $Ind_{Z(G)}^G omega _ pi$?
$endgroup$
– ned grekerzberg
Jan 15 at 9:10














$begingroup$
You only get $dim (pi) leq [G:Z(G)]$, not $dim (pi)^2$; or am I missing something ?
$endgroup$
– Max
Jan 16 at 18:46




$begingroup$
You only get $dim (pi) leq [G:Z(G)]$, not $dim (pi)^2$; or am I missing something ?
$endgroup$
– Max
Jan 16 at 18:46












$begingroup$
The Frobenius Reciprocity argument implies that $pi$ appears $dimpi$ times in $Indomega_pi$.
$endgroup$
– David Hill
Jan 16 at 18:59




$begingroup$
The Frobenius Reciprocity argument implies that $pi$ appears $dimpi$ times in $Indomega_pi$.
$endgroup$
– David Hill
Jan 16 at 18:59


















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