How to find normal subgroups from a character table?

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4












$begingroup$


I know that normal subgroups are the union of some conjugacy classes



Conjugacy classes are represented by the the columns in a matrix



How could we use character values in the table to determine normal subgroups?






abstract-algebra representation-theory normal-subgroups characters







share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    I know that normal subgroups are the union of some conjugacy classes



    Conjugacy classes are represented by the the columns in a matrix



    How could we use character values in the table to determine normal subgroups?






    abstract-algebra representation-theory normal-subgroups characters







    share|cite|improve this question









    $endgroup$















      4












      4








      4


      5



      $begingroup$


      I know that normal subgroups are the union of some conjugacy classes



      Conjugacy classes are represented by the the columns in a matrix



      How could we use character values in the table to determine normal subgroups?






      abstract-algebra representation-theory normal-subgroups characters







      share|cite|improve this question









      $endgroup$




      I know that normal subgroups are the union of some conjugacy classes



      Conjugacy classes are represented by the the columns in a matrix



      How could we use character values in the table to determine normal subgroups?






      abstract-algebra representation-theory normal-subgroups characters




      abstract-algebra representation-theory normal-subgroups characters






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked May 17 '16 at 6:57









      thinkerthinker

      632518




      632518






















          1 Answer
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          9












          $begingroup$

          This is quite well-known and can be found in books on representation theory. Here is an explanation, which is far from being original.



          First fact : $N$ is a normal subgroup of a finite group $G$ if and only if there exists a character $chi$ of $G$ such that $N = ker chi := {g in G | chi(g)=chi(1)}$. Indeed, if $N$ is normal then $G$ acts on the complex algebra $mathbf{C}[G/N] = displaystyle bigoplus_{gN in G/N} mathbf{C} e_{gN}$ by $h cdot e_{gN}=e_{hgN}$. This is a linear representation of $G$ (coming from the regular representation of $G/N$). Let $chi$ be its character. It is easy to check that $chi(h) = 0$ if $h notin N$ and $chi(h) = mathrm{Card}(G/N) = chi(1)$ if $h in N$. So $N = ker chi$. Conversely, using the fact that a character is constant on every conjugacy class, any subgroup of the form $ker chi$ is normal.



          Second fact : if $rho : G to mathrm{GL}(V)$ is the representation associated to the character $chi$ then $ker rho = ker chi$. The inclusion $subseteq$ is trivial. Conversely, assume $chi(g) = chi(1) = dim V$. Since the eigenvalues of $rho(g)$ are roots of $1$ and $chi(g)$ is the sum of the eigenvalues (with multiplicities), these eigenvalues are forced to be all equal to $1$. So $rho(g) = mathrm{id}_V$, that is to say $g in ker rho$.



          Third fact : if $chi = displaystyle sum_{i=1}^r n_i chi_i$ (where the $chi_i$ are pairwise distinct irreducible characters and $n_i geq 1$) then $ker chi = displaystyle bigcap_{i=1}^r ker chi_i$. Writing $rho, rho_1,ldots,rho_r$ for the corresponding representations, $rho$ is the direct sum of copies of $rho_1,ldots,rho_r$ so $ker rho = displaystyle bigcap_{i=1}^r ker rho_i$. Then apply the second fact.



          Conclusion : with your character table, you can read the subgroups $N_i:=ker chi_i$. Then the normal subgroups of $G$ are exactly the intersections of some $N_i$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Good summary! I've always had trouble retaining this, even after a couple readings of texts. This makes it easier.
            $endgroup$
            – rschwieb
            May 17 '16 at 16:29













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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          9












          $begingroup$

          This is quite well-known and can be found in books on representation theory. Here is an explanation, which is far from being original.



          First fact : $N$ is a normal subgroup of a finite group $G$ if and only if there exists a character $chi$ of $G$ such that $N = ker chi := {g in G | chi(g)=chi(1)}$. Indeed, if $N$ is normal then $G$ acts on the complex algebra $mathbf{C}[G/N] = displaystyle bigoplus_{gN in G/N} mathbf{C} e_{gN}$ by $h cdot e_{gN}=e_{hgN}$. This is a linear representation of $G$ (coming from the regular representation of $G/N$). Let $chi$ be its character. It is easy to check that $chi(h) = 0$ if $h notin N$ and $chi(h) = mathrm{Card}(G/N) = chi(1)$ if $h in N$. So $N = ker chi$. Conversely, using the fact that a character is constant on every conjugacy class, any subgroup of the form $ker chi$ is normal.



          Second fact : if $rho : G to mathrm{GL}(V)$ is the representation associated to the character $chi$ then $ker rho = ker chi$. The inclusion $subseteq$ is trivial. Conversely, assume $chi(g) = chi(1) = dim V$. Since the eigenvalues of $rho(g)$ are roots of $1$ and $chi(g)$ is the sum of the eigenvalues (with multiplicities), these eigenvalues are forced to be all equal to $1$. So $rho(g) = mathrm{id}_V$, that is to say $g in ker rho$.



          Third fact : if $chi = displaystyle sum_{i=1}^r n_i chi_i$ (where the $chi_i$ are pairwise distinct irreducible characters and $n_i geq 1$) then $ker chi = displaystyle bigcap_{i=1}^r ker chi_i$. Writing $rho, rho_1,ldots,rho_r$ for the corresponding representations, $rho$ is the direct sum of copies of $rho_1,ldots,rho_r$ so $ker rho = displaystyle bigcap_{i=1}^r ker rho_i$. Then apply the second fact.



          Conclusion : with your character table, you can read the subgroups $N_i:=ker chi_i$. Then the normal subgroups of $G$ are exactly the intersections of some $N_i$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Good summary! I've always had trouble retaining this, even after a couple readings of texts. This makes it easier.
            $endgroup$
            – rschwieb
            May 17 '16 at 16:29


















          9












          $begingroup$

          This is quite well-known and can be found in books on representation theory. Here is an explanation, which is far from being original.



          First fact : $N$ is a normal subgroup of a finite group $G$ if and only if there exists a character $chi$ of $G$ such that $N = ker chi := {g in G | chi(g)=chi(1)}$. Indeed, if $N$ is normal then $G$ acts on the complex algebra $mathbf{C}[G/N] = displaystyle bigoplus_{gN in G/N} mathbf{C} e_{gN}$ by $h cdot e_{gN}=e_{hgN}$. This is a linear representation of $G$ (coming from the regular representation of $G/N$). Let $chi$ be its character. It is easy to check that $chi(h) = 0$ if $h notin N$ and $chi(h) = mathrm{Card}(G/N) = chi(1)$ if $h in N$. So $N = ker chi$. Conversely, using the fact that a character is constant on every conjugacy class, any subgroup of the form $ker chi$ is normal.



          Second fact : if $rho : G to mathrm{GL}(V)$ is the representation associated to the character $chi$ then $ker rho = ker chi$. The inclusion $subseteq$ is trivial. Conversely, assume $chi(g) = chi(1) = dim V$. Since the eigenvalues of $rho(g)$ are roots of $1$ and $chi(g)$ is the sum of the eigenvalues (with multiplicities), these eigenvalues are forced to be all equal to $1$. So $rho(g) = mathrm{id}_V$, that is to say $g in ker rho$.



          Third fact : if $chi = displaystyle sum_{i=1}^r n_i chi_i$ (where the $chi_i$ are pairwise distinct irreducible characters and $n_i geq 1$) then $ker chi = displaystyle bigcap_{i=1}^r ker chi_i$. Writing $rho, rho_1,ldots,rho_r$ for the corresponding representations, $rho$ is the direct sum of copies of $rho_1,ldots,rho_r$ so $ker rho = displaystyle bigcap_{i=1}^r ker rho_i$. Then apply the second fact.



          Conclusion : with your character table, you can read the subgroups $N_i:=ker chi_i$. Then the normal subgroups of $G$ are exactly the intersections of some $N_i$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Good summary! I've always had trouble retaining this, even after a couple readings of texts. This makes it easier.
            $endgroup$
            – rschwieb
            May 17 '16 at 16:29
















          9












          9








          9





          $begingroup$

          This is quite well-known and can be found in books on representation theory. Here is an explanation, which is far from being original.



          First fact : $N$ is a normal subgroup of a finite group $G$ if and only if there exists a character $chi$ of $G$ such that $N = ker chi := {g in G | chi(g)=chi(1)}$. Indeed, if $N$ is normal then $G$ acts on the complex algebra $mathbf{C}[G/N] = displaystyle bigoplus_{gN in G/N} mathbf{C} e_{gN}$ by $h cdot e_{gN}=e_{hgN}$. This is a linear representation of $G$ (coming from the regular representation of $G/N$). Let $chi$ be its character. It is easy to check that $chi(h) = 0$ if $h notin N$ and $chi(h) = mathrm{Card}(G/N) = chi(1)$ if $h in N$. So $N = ker chi$. Conversely, using the fact that a character is constant on every conjugacy class, any subgroup of the form $ker chi$ is normal.



          Second fact : if $rho : G to mathrm{GL}(V)$ is the representation associated to the character $chi$ then $ker rho = ker chi$. The inclusion $subseteq$ is trivial. Conversely, assume $chi(g) = chi(1) = dim V$. Since the eigenvalues of $rho(g)$ are roots of $1$ and $chi(g)$ is the sum of the eigenvalues (with multiplicities), these eigenvalues are forced to be all equal to $1$. So $rho(g) = mathrm{id}_V$, that is to say $g in ker rho$.



          Third fact : if $chi = displaystyle sum_{i=1}^r n_i chi_i$ (where the $chi_i$ are pairwise distinct irreducible characters and $n_i geq 1$) then $ker chi = displaystyle bigcap_{i=1}^r ker chi_i$. Writing $rho, rho_1,ldots,rho_r$ for the corresponding representations, $rho$ is the direct sum of copies of $rho_1,ldots,rho_r$ so $ker rho = displaystyle bigcap_{i=1}^r ker rho_i$. Then apply the second fact.



          Conclusion : with your character table, you can read the subgroups $N_i:=ker chi_i$. Then the normal subgroups of $G$ are exactly the intersections of some $N_i$.






          share|cite|improve this answer











          $endgroup$



          This is quite well-known and can be found in books on representation theory. Here is an explanation, which is far from being original.



          First fact : $N$ is a normal subgroup of a finite group $G$ if and only if there exists a character $chi$ of $G$ such that $N = ker chi := {g in G | chi(g)=chi(1)}$. Indeed, if $N$ is normal then $G$ acts on the complex algebra $mathbf{C}[G/N] = displaystyle bigoplus_{gN in G/N} mathbf{C} e_{gN}$ by $h cdot e_{gN}=e_{hgN}$. This is a linear representation of $G$ (coming from the regular representation of $G/N$). Let $chi$ be its character. It is easy to check that $chi(h) = 0$ if $h notin N$ and $chi(h) = mathrm{Card}(G/N) = chi(1)$ if $h in N$. So $N = ker chi$. Conversely, using the fact that a character is constant on every conjugacy class, any subgroup of the form $ker chi$ is normal.



          Second fact : if $rho : G to mathrm{GL}(V)$ is the representation associated to the character $chi$ then $ker rho = ker chi$. The inclusion $subseteq$ is trivial. Conversely, assume $chi(g) = chi(1) = dim V$. Since the eigenvalues of $rho(g)$ are roots of $1$ and $chi(g)$ is the sum of the eigenvalues (with multiplicities), these eigenvalues are forced to be all equal to $1$. So $rho(g) = mathrm{id}_V$, that is to say $g in ker rho$.



          Third fact : if $chi = displaystyle sum_{i=1}^r n_i chi_i$ (where the $chi_i$ are pairwise distinct irreducible characters and $n_i geq 1$) then $ker chi = displaystyle bigcap_{i=1}^r ker chi_i$. Writing $rho, rho_1,ldots,rho_r$ for the corresponding representations, $rho$ is the direct sum of copies of $rho_1,ldots,rho_r$ so $ker rho = displaystyle bigcap_{i=1}^r ker rho_i$. Then apply the second fact.



          Conclusion : with your character table, you can read the subgroups $N_i:=ker chi_i$. Then the normal subgroups of $G$ are exactly the intersections of some $N_i$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 10 at 6:59









          dcw

          246




          246










          answered May 17 '16 at 13:59









          BrLBrL

          996159




          996159












          • $begingroup$
            Good summary! I've always had trouble retaining this, even after a couple readings of texts. This makes it easier.
            $endgroup$
            – rschwieb
            May 17 '16 at 16:29




















          • $begingroup$
            Good summary! I've always had trouble retaining this, even after a couple readings of texts. This makes it easier.
            $endgroup$
            – rschwieb
            May 17 '16 at 16:29


















          $begingroup$
          Good summary! I've always had trouble retaining this, even after a couple readings of texts. This makes it easier.
          $endgroup$
          – rschwieb
          May 17 '16 at 16:29






          $begingroup$
          Good summary! I've always had trouble retaining this, even after a couple readings of texts. This makes it easier.
          $endgroup$
          – rschwieb
          May 17 '16 at 16:29




















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