The condition between $chi(1)$ and $[G:H]$ which gives us a normal subgroup.












8












$begingroup$


$textbf{The question is as follows:}$





Let $H le G$ with $|G : H| = n$ and suppose that $chi in Char(G)$.



$rm (a)$ Show that $[chi ; chi] ge frac{[chi_H; chi_H]}{n}$.



$rm (b)$ Show that, if $H$ is Abelian and $chi in Irr(G)$ then $chi(1) le n$.



$rm (c)$ Show that, if the equality holds in part (b), then $H vartriangleleft G$.




I can show the first part as follows:



$rm (a)$ We have
$$|H| [chi_H; chi_H] = sum_{h in H}^{} |chi(h)|^2 le sum_{g in G}^{}|chi(g)|^2 = |G|[chi, chi] $$



$rm (b)$ For second part I can write



$$chi(1)=chi|_H(1)le [chi|_H,chi|_G]le [G:H][chi,chi]=[G:H]$$
But I am not sure if it is correct?



Can someone correct me please?




$rm (c)$ For the third part I still have no idea so far!



Can someone help me to show this, please?



Thanks!










share|cite|improve this question









$endgroup$

















    8












    $begingroup$


    $textbf{The question is as follows:}$





    Let $H le G$ with $|G : H| = n$ and suppose that $chi in Char(G)$.



    $rm (a)$ Show that $[chi ; chi] ge frac{[chi_H; chi_H]}{n}$.



    $rm (b)$ Show that, if $H$ is Abelian and $chi in Irr(G)$ then $chi(1) le n$.



    $rm (c)$ Show that, if the equality holds in part (b), then $H vartriangleleft G$.




    I can show the first part as follows:



    $rm (a)$ We have
    $$|H| [chi_H; chi_H] = sum_{h in H}^{} |chi(h)|^2 le sum_{g in G}^{}|chi(g)|^2 = |G|[chi, chi] $$



    $rm (b)$ For second part I can write



    $$chi(1)=chi|_H(1)le [chi|_H,chi|_G]le [G:H][chi,chi]=[G:H]$$
    But I am not sure if it is correct?



    Can someone correct me please?




    $rm (c)$ For the third part I still have no idea so far!



    Can someone help me to show this, please?



    Thanks!










    share|cite|improve this question









    $endgroup$















      8












      8








      8


      3



      $begingroup$


      $textbf{The question is as follows:}$





      Let $H le G$ with $|G : H| = n$ and suppose that $chi in Char(G)$.



      $rm (a)$ Show that $[chi ; chi] ge frac{[chi_H; chi_H]}{n}$.



      $rm (b)$ Show that, if $H$ is Abelian and $chi in Irr(G)$ then $chi(1) le n$.



      $rm (c)$ Show that, if the equality holds in part (b), then $H vartriangleleft G$.




      I can show the first part as follows:



      $rm (a)$ We have
      $$|H| [chi_H; chi_H] = sum_{h in H}^{} |chi(h)|^2 le sum_{g in G}^{}|chi(g)|^2 = |G|[chi, chi] $$



      $rm (b)$ For second part I can write



      $$chi(1)=chi|_H(1)le [chi|_H,chi|_G]le [G:H][chi,chi]=[G:H]$$
      But I am not sure if it is correct?



      Can someone correct me please?




      $rm (c)$ For the third part I still have no idea so far!



      Can someone help me to show this, please?



      Thanks!










      share|cite|improve this question









      $endgroup$




      $textbf{The question is as follows:}$





      Let $H le G$ with $|G : H| = n$ and suppose that $chi in Char(G)$.



      $rm (a)$ Show that $[chi ; chi] ge frac{[chi_H; chi_H]}{n}$.



      $rm (b)$ Show that, if $H$ is Abelian and $chi in Irr(G)$ then $chi(1) le n$.



      $rm (c)$ Show that, if the equality holds in part (b), then $H vartriangleleft G$.




      I can show the first part as follows:



      $rm (a)$ We have
      $$|H| [chi_H; chi_H] = sum_{h in H}^{} |chi(h)|^2 le sum_{g in G}^{}|chi(g)|^2 = |G|[chi, chi] $$



      $rm (b)$ For second part I can write



      $$chi(1)=chi|_H(1)le [chi|_H,chi|_G]le [G:H][chi,chi]=[G:H]$$
      But I am not sure if it is correct?



      Can someone correct me please?




      $rm (c)$ For the third part I still have no idea so far!



      Can someone help me to show this, please?



      Thanks!







      abstract-algebra group-theory finite-groups characters






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 17 '18 at 20:47









      NikitaNikita

      435312




      435312






















          1 Answer
          1






          active

          oldest

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          6












          $begingroup$

          Any reference here is to the book of Marty Isaacs, Character Theory of Finite Groups. Now for (b) you must be a bit more precise: if $H$ is abelian, and $chi$ is an irreducible character of $G$, then $chi_H=sum_{lambda in Irr(H)}a_{lambda}lambda$, where the $lambda$'s are linear and the integers $a_lambda geq 0$. Hence, $[chi_H,chi_H]=sum_{lambda in Irr(H)}a_{lambda}^2 geq sum_{lambda in Irr(H)}a_{lambda}=chi(1).$ By (a) and the fact that $chi$ is irreducible so $[chi,chi]=1$, you get indeed $chi(1) leq [chi_H,chi_H] leq |G:H|. text{ } (*)$



          For (c) we are going to use the Lemma (2.29) of Isaacs' book, which says that if you have a $chi in Irr(G)$ and a subgroup $K$ then $[chi_K,chi_K]=|G:K|$ if and only if $chi$ vanishes outside $K$.



          Now assume $H$ is abelian, $chi in Irr(G)$ with $chi(1)=|G:H|$. By $(*)$ we must have that $chi$ vanishes off $H$. Look at the set ${g in G: chi(g) neq 0 }$. This set is normal (since $chi(g)=chi(g^x)$ for any $x in G$) and is contained in $H$. Hence $N=langle g in G: chi(g) neq 0 rangle$ is an abelian normal subgroup of $G$ contained in $H$. Observe that if $h in H-N$, then we must have $chi(h)=0$. So $chi$ vanishes even off $N$, and by the Lemma (2.29) in the other direction and applying (b) to the subgroup $N$, we must have $|G:N|=[chi_N,chi_N]leq chi(1)=|G:H|$. Hence $|H| leq |N|$ and we must have $H=N$ and you are done.



          Note (February 6th) Returning to my answer above I discovered a flaw in the argument: it is in the last equation, after “we must have” - specifically $[chi_N,chi_N] leq chi(1)$ is not correct. After some thoughts, I suspected that $(c)$ above is not true at all. As a matter of fact Prof. Derek Holt was very kind to provide me with a counterexample: there is a group of order $32$, SmallGroup(32,6) (notation in Magma or in GAP). It has an abelian subgroup $H cong C_2 times C_4$, hence $|G:H|=4$. This subgroup is not normal and $G$ has a $chi in Irr(G)$ with $chi(1)=4$. In this counterexample $|N|=2$.



          My fault analysis led me to the following.



          Theorem A Let $H leq G$ be abelian, $chi in Irr(G)$ with $|G:H|=chi(1)$ a prime number, then $H unlhd G$ and $chi$ is monomial (induced by a linear character).



          Proof Since $H$ is abelian, $chi(1) leq [chi_H,chi_H] leq |G:H|$ and the last inequality is equality if and only if $chi$ vanishes outside $H$. What about the first inequality? Well, $chi_H=sum_{lambda in Irr(H)}a_lambda lambda$, where the $lambda$'s are all linear and the integers $a_lambda geq 0$. $chi(1)=sum_{lambda in Irr(H)}a_lambda$ and $[chi_H,chi_H]=sum_{lambda in Irr(H)}a^2_lambda$. So we see that $chi(1)=[chi_H,chi_H]$ if and only if $a_lambda in {0,1}$. In that case, $chi_H=lambda_1 + cdots + lambda _{|G:H|}$. Observe that for these $lambda$'s by Frobenius Reciprocity, $[chi_H,lambda_i]=[chi,lambda_i^G]=1$ and it follows that $chi=lambda_i^G$ for all $i=1, cdots ,|G:H|$. Hence, $chi$ is a monomial character.



          Next we concentrate on the normal subgroup $N=langle g in G: chi(g) neq 0 rangle$ as defined above, and we apply Clifford Theory to this:
          $chi_N=esum_{i=1}^tmu_i$, where $t=|G:I_G(mu)|$, the index of the inertia subgroup of the (linear) $mu_1=mu$ and $e$ is a positive integer. We already remarked above that $chi$ vanishes outside $N$, hence $[chi_N,chi_N]=e^2t=|G:N|$. And also, $chi(1)=et=|G:H|$. It follows that $|H:N|=e$.



          Now if $|G:H|=p$, $p$ a prime, then $p=et$, hence we have two cases: $e=1$ and $t=p$, or $e=p$ and $t=1$. In the first case we get $|H:N|=1$, that is $H=N$ and hence $H$ is normal. The latter case yields $|G:N|=p^2$, so $G/N$ is a group of order $p^2$ whence abelian, and again $H$ must be normal.$square$



          Note (February 12th) There is a generalization of the above theorem, that I can prove now. It also follows from Udo Riese's results.



          Theorem B Let $H leq G$ be an abelian maximal subgroup, $chi in Irr(G)$ with $|G:H|=chi(1)$, then $H unlhd G$ and $chi$ is monomial (induced by a linear character).



          Proof The fact that $chi$ is monomial runs exactly in the same way as in the proof of Theorem A.
          Now I use the 2.11 Theorem from Marty Isaacs' book Finite Group Theory. The proof of this Theorem is based on the famous Zipper Lemma of Helmut Wielandt. The subgroup $H$ satisfies the conditions of this Theorem (caution: in the aforementioned book our $H$ here is called $A$), namely $H$ and $G$ are the only two subgroups between $H$ and $G$ by the maximality of $H$, and we need to check that $|G:H|^2 leq |G:Z(G)|$. But this trivially true since $|G:H|^2=chi(1)^2 leq |G:Z(G)|$ (for the last inequality, this is easy to prove or see (2.30) Lemma in Character Theory of Finite Groups above). By the 2.11 Theorem, it follows $H subseteq F(G)$, the Fitting subgroup of $G$. Hence either $H=F(G)$, and then $H$ is normal, even characteristic. Or, $G=F(G)$, meaning $G$ is nilpotent and it is well-known that maximal subgroups of nilpotent groups are normal. $square$



          Final remark A famous theorem of I.N. Herstein (1957) asserts that if $G$ has an abelian maximal subgroup, $G$ must be solvable, see here.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Many thanks for your complete answer!
            $endgroup$
            – Nikita
            Jan 18 '18 at 13:29










          • $begingroup$
            @Nikita, if you deem my answer to be the right one please tick it as such, follwing the habit here at this StackExchange site, thanks.
            $endgroup$
            – Nicky Hekster
            Feb 6 '18 at 10:25










          • $begingroup$
            I am so sorry! Many thanks for your answer!
            $endgroup$
            – Nikita
            Feb 6 '18 at 12:21










          • $begingroup$
            Digging into (c) somewhat deeper I found a paper of Udo Riese, A Subnormality Criterion in Finite Groups Related to Character Degrees (1998), J. of Algebra 201 which proves that $H$ must be subnormal. Moreover, $H^G=langle H^g: g in G rangle$, the normal closure of $H$ is nilpotent! The results are far from trivial, the first uses the famous Zipper Lemma of Helmut Wielandt.
            $endgroup$
            – Nicky Hekster
            Feb 9 '18 at 13:38












          • $begingroup$
            @Nikita I would like to thank YOU for your post and hence inspiration. Nicky
            $endgroup$
            – Nicky Hekster
            Feb 12 '18 at 12:43











          Your Answer





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          1 Answer
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          active

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          $begingroup$

          Any reference here is to the book of Marty Isaacs, Character Theory of Finite Groups. Now for (b) you must be a bit more precise: if $H$ is abelian, and $chi$ is an irreducible character of $G$, then $chi_H=sum_{lambda in Irr(H)}a_{lambda}lambda$, where the $lambda$'s are linear and the integers $a_lambda geq 0$. Hence, $[chi_H,chi_H]=sum_{lambda in Irr(H)}a_{lambda}^2 geq sum_{lambda in Irr(H)}a_{lambda}=chi(1).$ By (a) and the fact that $chi$ is irreducible so $[chi,chi]=1$, you get indeed $chi(1) leq [chi_H,chi_H] leq |G:H|. text{ } (*)$



          For (c) we are going to use the Lemma (2.29) of Isaacs' book, which says that if you have a $chi in Irr(G)$ and a subgroup $K$ then $[chi_K,chi_K]=|G:K|$ if and only if $chi$ vanishes outside $K$.



          Now assume $H$ is abelian, $chi in Irr(G)$ with $chi(1)=|G:H|$. By $(*)$ we must have that $chi$ vanishes off $H$. Look at the set ${g in G: chi(g) neq 0 }$. This set is normal (since $chi(g)=chi(g^x)$ for any $x in G$) and is contained in $H$. Hence $N=langle g in G: chi(g) neq 0 rangle$ is an abelian normal subgroup of $G$ contained in $H$. Observe that if $h in H-N$, then we must have $chi(h)=0$. So $chi$ vanishes even off $N$, and by the Lemma (2.29) in the other direction and applying (b) to the subgroup $N$, we must have $|G:N|=[chi_N,chi_N]leq chi(1)=|G:H|$. Hence $|H| leq |N|$ and we must have $H=N$ and you are done.



          Note (February 6th) Returning to my answer above I discovered a flaw in the argument: it is in the last equation, after “we must have” - specifically $[chi_N,chi_N] leq chi(1)$ is not correct. After some thoughts, I suspected that $(c)$ above is not true at all. As a matter of fact Prof. Derek Holt was very kind to provide me with a counterexample: there is a group of order $32$, SmallGroup(32,6) (notation in Magma or in GAP). It has an abelian subgroup $H cong C_2 times C_4$, hence $|G:H|=4$. This subgroup is not normal and $G$ has a $chi in Irr(G)$ with $chi(1)=4$. In this counterexample $|N|=2$.



          My fault analysis led me to the following.



          Theorem A Let $H leq G$ be abelian, $chi in Irr(G)$ with $|G:H|=chi(1)$ a prime number, then $H unlhd G$ and $chi$ is monomial (induced by a linear character).



          Proof Since $H$ is abelian, $chi(1) leq [chi_H,chi_H] leq |G:H|$ and the last inequality is equality if and only if $chi$ vanishes outside $H$. What about the first inequality? Well, $chi_H=sum_{lambda in Irr(H)}a_lambda lambda$, where the $lambda$'s are all linear and the integers $a_lambda geq 0$. $chi(1)=sum_{lambda in Irr(H)}a_lambda$ and $[chi_H,chi_H]=sum_{lambda in Irr(H)}a^2_lambda$. So we see that $chi(1)=[chi_H,chi_H]$ if and only if $a_lambda in {0,1}$. In that case, $chi_H=lambda_1 + cdots + lambda _{|G:H|}$. Observe that for these $lambda$'s by Frobenius Reciprocity, $[chi_H,lambda_i]=[chi,lambda_i^G]=1$ and it follows that $chi=lambda_i^G$ for all $i=1, cdots ,|G:H|$. Hence, $chi$ is a monomial character.



          Next we concentrate on the normal subgroup $N=langle g in G: chi(g) neq 0 rangle$ as defined above, and we apply Clifford Theory to this:
          $chi_N=esum_{i=1}^tmu_i$, where $t=|G:I_G(mu)|$, the index of the inertia subgroup of the (linear) $mu_1=mu$ and $e$ is a positive integer. We already remarked above that $chi$ vanishes outside $N$, hence $[chi_N,chi_N]=e^2t=|G:N|$. And also, $chi(1)=et=|G:H|$. It follows that $|H:N|=e$.



          Now if $|G:H|=p$, $p$ a prime, then $p=et$, hence we have two cases: $e=1$ and $t=p$, or $e=p$ and $t=1$. In the first case we get $|H:N|=1$, that is $H=N$ and hence $H$ is normal. The latter case yields $|G:N|=p^2$, so $G/N$ is a group of order $p^2$ whence abelian, and again $H$ must be normal.$square$



          Note (February 12th) There is a generalization of the above theorem, that I can prove now. It also follows from Udo Riese's results.



          Theorem B Let $H leq G$ be an abelian maximal subgroup, $chi in Irr(G)$ with $|G:H|=chi(1)$, then $H unlhd G$ and $chi$ is monomial (induced by a linear character).



          Proof The fact that $chi$ is monomial runs exactly in the same way as in the proof of Theorem A.
          Now I use the 2.11 Theorem from Marty Isaacs' book Finite Group Theory. The proof of this Theorem is based on the famous Zipper Lemma of Helmut Wielandt. The subgroup $H$ satisfies the conditions of this Theorem (caution: in the aforementioned book our $H$ here is called $A$), namely $H$ and $G$ are the only two subgroups between $H$ and $G$ by the maximality of $H$, and we need to check that $|G:H|^2 leq |G:Z(G)|$. But this trivially true since $|G:H|^2=chi(1)^2 leq |G:Z(G)|$ (for the last inequality, this is easy to prove or see (2.30) Lemma in Character Theory of Finite Groups above). By the 2.11 Theorem, it follows $H subseteq F(G)$, the Fitting subgroup of $G$. Hence either $H=F(G)$, and then $H$ is normal, even characteristic. Or, $G=F(G)$, meaning $G$ is nilpotent and it is well-known that maximal subgroups of nilpotent groups are normal. $square$



          Final remark A famous theorem of I.N. Herstein (1957) asserts that if $G$ has an abelian maximal subgroup, $G$ must be solvable, see here.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Many thanks for your complete answer!
            $endgroup$
            – Nikita
            Jan 18 '18 at 13:29










          • $begingroup$
            @Nikita, if you deem my answer to be the right one please tick it as such, follwing the habit here at this StackExchange site, thanks.
            $endgroup$
            – Nicky Hekster
            Feb 6 '18 at 10:25










          • $begingroup$
            I am so sorry! Many thanks for your answer!
            $endgroup$
            – Nikita
            Feb 6 '18 at 12:21










          • $begingroup$
            Digging into (c) somewhat deeper I found a paper of Udo Riese, A Subnormality Criterion in Finite Groups Related to Character Degrees (1998), J. of Algebra 201 which proves that $H$ must be subnormal. Moreover, $H^G=langle H^g: g in G rangle$, the normal closure of $H$ is nilpotent! The results are far from trivial, the first uses the famous Zipper Lemma of Helmut Wielandt.
            $endgroup$
            – Nicky Hekster
            Feb 9 '18 at 13:38












          • $begingroup$
            @Nikita I would like to thank YOU for your post and hence inspiration. Nicky
            $endgroup$
            – Nicky Hekster
            Feb 12 '18 at 12:43
















          6












          $begingroup$

          Any reference here is to the book of Marty Isaacs, Character Theory of Finite Groups. Now for (b) you must be a bit more precise: if $H$ is abelian, and $chi$ is an irreducible character of $G$, then $chi_H=sum_{lambda in Irr(H)}a_{lambda}lambda$, where the $lambda$'s are linear and the integers $a_lambda geq 0$. Hence, $[chi_H,chi_H]=sum_{lambda in Irr(H)}a_{lambda}^2 geq sum_{lambda in Irr(H)}a_{lambda}=chi(1).$ By (a) and the fact that $chi$ is irreducible so $[chi,chi]=1$, you get indeed $chi(1) leq [chi_H,chi_H] leq |G:H|. text{ } (*)$



          For (c) we are going to use the Lemma (2.29) of Isaacs' book, which says that if you have a $chi in Irr(G)$ and a subgroup $K$ then $[chi_K,chi_K]=|G:K|$ if and only if $chi$ vanishes outside $K$.



          Now assume $H$ is abelian, $chi in Irr(G)$ with $chi(1)=|G:H|$. By $(*)$ we must have that $chi$ vanishes off $H$. Look at the set ${g in G: chi(g) neq 0 }$. This set is normal (since $chi(g)=chi(g^x)$ for any $x in G$) and is contained in $H$. Hence $N=langle g in G: chi(g) neq 0 rangle$ is an abelian normal subgroup of $G$ contained in $H$. Observe that if $h in H-N$, then we must have $chi(h)=0$. So $chi$ vanishes even off $N$, and by the Lemma (2.29) in the other direction and applying (b) to the subgroup $N$, we must have $|G:N|=[chi_N,chi_N]leq chi(1)=|G:H|$. Hence $|H| leq |N|$ and we must have $H=N$ and you are done.



          Note (February 6th) Returning to my answer above I discovered a flaw in the argument: it is in the last equation, after “we must have” - specifically $[chi_N,chi_N] leq chi(1)$ is not correct. After some thoughts, I suspected that $(c)$ above is not true at all. As a matter of fact Prof. Derek Holt was very kind to provide me with a counterexample: there is a group of order $32$, SmallGroup(32,6) (notation in Magma or in GAP). It has an abelian subgroup $H cong C_2 times C_4$, hence $|G:H|=4$. This subgroup is not normal and $G$ has a $chi in Irr(G)$ with $chi(1)=4$. In this counterexample $|N|=2$.



          My fault analysis led me to the following.



          Theorem A Let $H leq G$ be abelian, $chi in Irr(G)$ with $|G:H|=chi(1)$ a prime number, then $H unlhd G$ and $chi$ is monomial (induced by a linear character).



          Proof Since $H$ is abelian, $chi(1) leq [chi_H,chi_H] leq |G:H|$ and the last inequality is equality if and only if $chi$ vanishes outside $H$. What about the first inequality? Well, $chi_H=sum_{lambda in Irr(H)}a_lambda lambda$, where the $lambda$'s are all linear and the integers $a_lambda geq 0$. $chi(1)=sum_{lambda in Irr(H)}a_lambda$ and $[chi_H,chi_H]=sum_{lambda in Irr(H)}a^2_lambda$. So we see that $chi(1)=[chi_H,chi_H]$ if and only if $a_lambda in {0,1}$. In that case, $chi_H=lambda_1 + cdots + lambda _{|G:H|}$. Observe that for these $lambda$'s by Frobenius Reciprocity, $[chi_H,lambda_i]=[chi,lambda_i^G]=1$ and it follows that $chi=lambda_i^G$ for all $i=1, cdots ,|G:H|$. Hence, $chi$ is a monomial character.



          Next we concentrate on the normal subgroup $N=langle g in G: chi(g) neq 0 rangle$ as defined above, and we apply Clifford Theory to this:
          $chi_N=esum_{i=1}^tmu_i$, where $t=|G:I_G(mu)|$, the index of the inertia subgroup of the (linear) $mu_1=mu$ and $e$ is a positive integer. We already remarked above that $chi$ vanishes outside $N$, hence $[chi_N,chi_N]=e^2t=|G:N|$. And also, $chi(1)=et=|G:H|$. It follows that $|H:N|=e$.



          Now if $|G:H|=p$, $p$ a prime, then $p=et$, hence we have two cases: $e=1$ and $t=p$, or $e=p$ and $t=1$. In the first case we get $|H:N|=1$, that is $H=N$ and hence $H$ is normal. The latter case yields $|G:N|=p^2$, so $G/N$ is a group of order $p^2$ whence abelian, and again $H$ must be normal.$square$



          Note (February 12th) There is a generalization of the above theorem, that I can prove now. It also follows from Udo Riese's results.



          Theorem B Let $H leq G$ be an abelian maximal subgroup, $chi in Irr(G)$ with $|G:H|=chi(1)$, then $H unlhd G$ and $chi$ is monomial (induced by a linear character).



          Proof The fact that $chi$ is monomial runs exactly in the same way as in the proof of Theorem A.
          Now I use the 2.11 Theorem from Marty Isaacs' book Finite Group Theory. The proof of this Theorem is based on the famous Zipper Lemma of Helmut Wielandt. The subgroup $H$ satisfies the conditions of this Theorem (caution: in the aforementioned book our $H$ here is called $A$), namely $H$ and $G$ are the only two subgroups between $H$ and $G$ by the maximality of $H$, and we need to check that $|G:H|^2 leq |G:Z(G)|$. But this trivially true since $|G:H|^2=chi(1)^2 leq |G:Z(G)|$ (for the last inequality, this is easy to prove or see (2.30) Lemma in Character Theory of Finite Groups above). By the 2.11 Theorem, it follows $H subseteq F(G)$, the Fitting subgroup of $G$. Hence either $H=F(G)$, and then $H$ is normal, even characteristic. Or, $G=F(G)$, meaning $G$ is nilpotent and it is well-known that maximal subgroups of nilpotent groups are normal. $square$



          Final remark A famous theorem of I.N. Herstein (1957) asserts that if $G$ has an abelian maximal subgroup, $G$ must be solvable, see here.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Many thanks for your complete answer!
            $endgroup$
            – Nikita
            Jan 18 '18 at 13:29










          • $begingroup$
            @Nikita, if you deem my answer to be the right one please tick it as such, follwing the habit here at this StackExchange site, thanks.
            $endgroup$
            – Nicky Hekster
            Feb 6 '18 at 10:25










          • $begingroup$
            I am so sorry! Many thanks for your answer!
            $endgroup$
            – Nikita
            Feb 6 '18 at 12:21










          • $begingroup$
            Digging into (c) somewhat deeper I found a paper of Udo Riese, A Subnormality Criterion in Finite Groups Related to Character Degrees (1998), J. of Algebra 201 which proves that $H$ must be subnormal. Moreover, $H^G=langle H^g: g in G rangle$, the normal closure of $H$ is nilpotent! The results are far from trivial, the first uses the famous Zipper Lemma of Helmut Wielandt.
            $endgroup$
            – Nicky Hekster
            Feb 9 '18 at 13:38












          • $begingroup$
            @Nikita I would like to thank YOU for your post and hence inspiration. Nicky
            $endgroup$
            – Nicky Hekster
            Feb 12 '18 at 12:43














          6












          6








          6





          $begingroup$

          Any reference here is to the book of Marty Isaacs, Character Theory of Finite Groups. Now for (b) you must be a bit more precise: if $H$ is abelian, and $chi$ is an irreducible character of $G$, then $chi_H=sum_{lambda in Irr(H)}a_{lambda}lambda$, where the $lambda$'s are linear and the integers $a_lambda geq 0$. Hence, $[chi_H,chi_H]=sum_{lambda in Irr(H)}a_{lambda}^2 geq sum_{lambda in Irr(H)}a_{lambda}=chi(1).$ By (a) and the fact that $chi$ is irreducible so $[chi,chi]=1$, you get indeed $chi(1) leq [chi_H,chi_H] leq |G:H|. text{ } (*)$



          For (c) we are going to use the Lemma (2.29) of Isaacs' book, which says that if you have a $chi in Irr(G)$ and a subgroup $K$ then $[chi_K,chi_K]=|G:K|$ if and only if $chi$ vanishes outside $K$.



          Now assume $H$ is abelian, $chi in Irr(G)$ with $chi(1)=|G:H|$. By $(*)$ we must have that $chi$ vanishes off $H$. Look at the set ${g in G: chi(g) neq 0 }$. This set is normal (since $chi(g)=chi(g^x)$ for any $x in G$) and is contained in $H$. Hence $N=langle g in G: chi(g) neq 0 rangle$ is an abelian normal subgroup of $G$ contained in $H$. Observe that if $h in H-N$, then we must have $chi(h)=0$. So $chi$ vanishes even off $N$, and by the Lemma (2.29) in the other direction and applying (b) to the subgroup $N$, we must have $|G:N|=[chi_N,chi_N]leq chi(1)=|G:H|$. Hence $|H| leq |N|$ and we must have $H=N$ and you are done.



          Note (February 6th) Returning to my answer above I discovered a flaw in the argument: it is in the last equation, after “we must have” - specifically $[chi_N,chi_N] leq chi(1)$ is not correct. After some thoughts, I suspected that $(c)$ above is not true at all. As a matter of fact Prof. Derek Holt was very kind to provide me with a counterexample: there is a group of order $32$, SmallGroup(32,6) (notation in Magma or in GAP). It has an abelian subgroup $H cong C_2 times C_4$, hence $|G:H|=4$. This subgroup is not normal and $G$ has a $chi in Irr(G)$ with $chi(1)=4$. In this counterexample $|N|=2$.



          My fault analysis led me to the following.



          Theorem A Let $H leq G$ be abelian, $chi in Irr(G)$ with $|G:H|=chi(1)$ a prime number, then $H unlhd G$ and $chi$ is monomial (induced by a linear character).



          Proof Since $H$ is abelian, $chi(1) leq [chi_H,chi_H] leq |G:H|$ and the last inequality is equality if and only if $chi$ vanishes outside $H$. What about the first inequality? Well, $chi_H=sum_{lambda in Irr(H)}a_lambda lambda$, where the $lambda$'s are all linear and the integers $a_lambda geq 0$. $chi(1)=sum_{lambda in Irr(H)}a_lambda$ and $[chi_H,chi_H]=sum_{lambda in Irr(H)}a^2_lambda$. So we see that $chi(1)=[chi_H,chi_H]$ if and only if $a_lambda in {0,1}$. In that case, $chi_H=lambda_1 + cdots + lambda _{|G:H|}$. Observe that for these $lambda$'s by Frobenius Reciprocity, $[chi_H,lambda_i]=[chi,lambda_i^G]=1$ and it follows that $chi=lambda_i^G$ for all $i=1, cdots ,|G:H|$. Hence, $chi$ is a monomial character.



          Next we concentrate on the normal subgroup $N=langle g in G: chi(g) neq 0 rangle$ as defined above, and we apply Clifford Theory to this:
          $chi_N=esum_{i=1}^tmu_i$, where $t=|G:I_G(mu)|$, the index of the inertia subgroup of the (linear) $mu_1=mu$ and $e$ is a positive integer. We already remarked above that $chi$ vanishes outside $N$, hence $[chi_N,chi_N]=e^2t=|G:N|$. And also, $chi(1)=et=|G:H|$. It follows that $|H:N|=e$.



          Now if $|G:H|=p$, $p$ a prime, then $p=et$, hence we have two cases: $e=1$ and $t=p$, or $e=p$ and $t=1$. In the first case we get $|H:N|=1$, that is $H=N$ and hence $H$ is normal. The latter case yields $|G:N|=p^2$, so $G/N$ is a group of order $p^2$ whence abelian, and again $H$ must be normal.$square$



          Note (February 12th) There is a generalization of the above theorem, that I can prove now. It also follows from Udo Riese's results.



          Theorem B Let $H leq G$ be an abelian maximal subgroup, $chi in Irr(G)$ with $|G:H|=chi(1)$, then $H unlhd G$ and $chi$ is monomial (induced by a linear character).



          Proof The fact that $chi$ is monomial runs exactly in the same way as in the proof of Theorem A.
          Now I use the 2.11 Theorem from Marty Isaacs' book Finite Group Theory. The proof of this Theorem is based on the famous Zipper Lemma of Helmut Wielandt. The subgroup $H$ satisfies the conditions of this Theorem (caution: in the aforementioned book our $H$ here is called $A$), namely $H$ and $G$ are the only two subgroups between $H$ and $G$ by the maximality of $H$, and we need to check that $|G:H|^2 leq |G:Z(G)|$. But this trivially true since $|G:H|^2=chi(1)^2 leq |G:Z(G)|$ (for the last inequality, this is easy to prove or see (2.30) Lemma in Character Theory of Finite Groups above). By the 2.11 Theorem, it follows $H subseteq F(G)$, the Fitting subgroup of $G$. Hence either $H=F(G)$, and then $H$ is normal, even characteristic. Or, $G=F(G)$, meaning $G$ is nilpotent and it is well-known that maximal subgroups of nilpotent groups are normal. $square$



          Final remark A famous theorem of I.N. Herstein (1957) asserts that if $G$ has an abelian maximal subgroup, $G$ must be solvable, see here.






          share|cite|improve this answer











          $endgroup$



          Any reference here is to the book of Marty Isaacs, Character Theory of Finite Groups. Now for (b) you must be a bit more precise: if $H$ is abelian, and $chi$ is an irreducible character of $G$, then $chi_H=sum_{lambda in Irr(H)}a_{lambda}lambda$, where the $lambda$'s are linear and the integers $a_lambda geq 0$. Hence, $[chi_H,chi_H]=sum_{lambda in Irr(H)}a_{lambda}^2 geq sum_{lambda in Irr(H)}a_{lambda}=chi(1).$ By (a) and the fact that $chi$ is irreducible so $[chi,chi]=1$, you get indeed $chi(1) leq [chi_H,chi_H] leq |G:H|. text{ } (*)$



          For (c) we are going to use the Lemma (2.29) of Isaacs' book, which says that if you have a $chi in Irr(G)$ and a subgroup $K$ then $[chi_K,chi_K]=|G:K|$ if and only if $chi$ vanishes outside $K$.



          Now assume $H$ is abelian, $chi in Irr(G)$ with $chi(1)=|G:H|$. By $(*)$ we must have that $chi$ vanishes off $H$. Look at the set ${g in G: chi(g) neq 0 }$. This set is normal (since $chi(g)=chi(g^x)$ for any $x in G$) and is contained in $H$. Hence $N=langle g in G: chi(g) neq 0 rangle$ is an abelian normal subgroup of $G$ contained in $H$. Observe that if $h in H-N$, then we must have $chi(h)=0$. So $chi$ vanishes even off $N$, and by the Lemma (2.29) in the other direction and applying (b) to the subgroup $N$, we must have $|G:N|=[chi_N,chi_N]leq chi(1)=|G:H|$. Hence $|H| leq |N|$ and we must have $H=N$ and you are done.



          Note (February 6th) Returning to my answer above I discovered a flaw in the argument: it is in the last equation, after “we must have” - specifically $[chi_N,chi_N] leq chi(1)$ is not correct. After some thoughts, I suspected that $(c)$ above is not true at all. As a matter of fact Prof. Derek Holt was very kind to provide me with a counterexample: there is a group of order $32$, SmallGroup(32,6) (notation in Magma or in GAP). It has an abelian subgroup $H cong C_2 times C_4$, hence $|G:H|=4$. This subgroup is not normal and $G$ has a $chi in Irr(G)$ with $chi(1)=4$. In this counterexample $|N|=2$.



          My fault analysis led me to the following.



          Theorem A Let $H leq G$ be abelian, $chi in Irr(G)$ with $|G:H|=chi(1)$ a prime number, then $H unlhd G$ and $chi$ is monomial (induced by a linear character).



          Proof Since $H$ is abelian, $chi(1) leq [chi_H,chi_H] leq |G:H|$ and the last inequality is equality if and only if $chi$ vanishes outside $H$. What about the first inequality? Well, $chi_H=sum_{lambda in Irr(H)}a_lambda lambda$, where the $lambda$'s are all linear and the integers $a_lambda geq 0$. $chi(1)=sum_{lambda in Irr(H)}a_lambda$ and $[chi_H,chi_H]=sum_{lambda in Irr(H)}a^2_lambda$. So we see that $chi(1)=[chi_H,chi_H]$ if and only if $a_lambda in {0,1}$. In that case, $chi_H=lambda_1 + cdots + lambda _{|G:H|}$. Observe that for these $lambda$'s by Frobenius Reciprocity, $[chi_H,lambda_i]=[chi,lambda_i^G]=1$ and it follows that $chi=lambda_i^G$ for all $i=1, cdots ,|G:H|$. Hence, $chi$ is a monomial character.



          Next we concentrate on the normal subgroup $N=langle g in G: chi(g) neq 0 rangle$ as defined above, and we apply Clifford Theory to this:
          $chi_N=esum_{i=1}^tmu_i$, where $t=|G:I_G(mu)|$, the index of the inertia subgroup of the (linear) $mu_1=mu$ and $e$ is a positive integer. We already remarked above that $chi$ vanishes outside $N$, hence $[chi_N,chi_N]=e^2t=|G:N|$. And also, $chi(1)=et=|G:H|$. It follows that $|H:N|=e$.



          Now if $|G:H|=p$, $p$ a prime, then $p=et$, hence we have two cases: $e=1$ and $t=p$, or $e=p$ and $t=1$. In the first case we get $|H:N|=1$, that is $H=N$ and hence $H$ is normal. The latter case yields $|G:N|=p^2$, so $G/N$ is a group of order $p^2$ whence abelian, and again $H$ must be normal.$square$



          Note (February 12th) There is a generalization of the above theorem, that I can prove now. It also follows from Udo Riese's results.



          Theorem B Let $H leq G$ be an abelian maximal subgroup, $chi in Irr(G)$ with $|G:H|=chi(1)$, then $H unlhd G$ and $chi$ is monomial (induced by a linear character).



          Proof The fact that $chi$ is monomial runs exactly in the same way as in the proof of Theorem A.
          Now I use the 2.11 Theorem from Marty Isaacs' book Finite Group Theory. The proof of this Theorem is based on the famous Zipper Lemma of Helmut Wielandt. The subgroup $H$ satisfies the conditions of this Theorem (caution: in the aforementioned book our $H$ here is called $A$), namely $H$ and $G$ are the only two subgroups between $H$ and $G$ by the maximality of $H$, and we need to check that $|G:H|^2 leq |G:Z(G)|$. But this trivially true since $|G:H|^2=chi(1)^2 leq |G:Z(G)|$ (for the last inequality, this is easy to prove or see (2.30) Lemma in Character Theory of Finite Groups above). By the 2.11 Theorem, it follows $H subseteq F(G)$, the Fitting subgroup of $G$. Hence either $H=F(G)$, and then $H$ is normal, even characteristic. Or, $G=F(G)$, meaning $G$ is nilpotent and it is well-known that maximal subgroups of nilpotent groups are normal. $square$



          Final remark A famous theorem of I.N. Herstein (1957) asserts that if $G$ has an abelian maximal subgroup, $G$ must be solvable, see here.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 3 at 18:35

























          answered Jan 18 '18 at 9:44









          Nicky HeksterNicky Hekster

          28.4k53456




          28.4k53456












          • $begingroup$
            Many thanks for your complete answer!
            $endgroup$
            – Nikita
            Jan 18 '18 at 13:29










          • $begingroup$
            @Nikita, if you deem my answer to be the right one please tick it as such, follwing the habit here at this StackExchange site, thanks.
            $endgroup$
            – Nicky Hekster
            Feb 6 '18 at 10:25










          • $begingroup$
            I am so sorry! Many thanks for your answer!
            $endgroup$
            – Nikita
            Feb 6 '18 at 12:21










          • $begingroup$
            Digging into (c) somewhat deeper I found a paper of Udo Riese, A Subnormality Criterion in Finite Groups Related to Character Degrees (1998), J. of Algebra 201 which proves that $H$ must be subnormal. Moreover, $H^G=langle H^g: g in G rangle$, the normal closure of $H$ is nilpotent! The results are far from trivial, the first uses the famous Zipper Lemma of Helmut Wielandt.
            $endgroup$
            – Nicky Hekster
            Feb 9 '18 at 13:38












          • $begingroup$
            @Nikita I would like to thank YOU for your post and hence inspiration. Nicky
            $endgroup$
            – Nicky Hekster
            Feb 12 '18 at 12:43


















          • $begingroup$
            Many thanks for your complete answer!
            $endgroup$
            – Nikita
            Jan 18 '18 at 13:29










          • $begingroup$
            @Nikita, if you deem my answer to be the right one please tick it as such, follwing the habit here at this StackExchange site, thanks.
            $endgroup$
            – Nicky Hekster
            Feb 6 '18 at 10:25










          • $begingroup$
            I am so sorry! Many thanks for your answer!
            $endgroup$
            – Nikita
            Feb 6 '18 at 12:21










          • $begingroup$
            Digging into (c) somewhat deeper I found a paper of Udo Riese, A Subnormality Criterion in Finite Groups Related to Character Degrees (1998), J. of Algebra 201 which proves that $H$ must be subnormal. Moreover, $H^G=langle H^g: g in G rangle$, the normal closure of $H$ is nilpotent! The results are far from trivial, the first uses the famous Zipper Lemma of Helmut Wielandt.
            $endgroup$
            – Nicky Hekster
            Feb 9 '18 at 13:38












          • $begingroup$
            @Nikita I would like to thank YOU for your post and hence inspiration. Nicky
            $endgroup$
            – Nicky Hekster
            Feb 12 '18 at 12:43
















          $begingroup$
          Many thanks for your complete answer!
          $endgroup$
          – Nikita
          Jan 18 '18 at 13:29




          $begingroup$
          Many thanks for your complete answer!
          $endgroup$
          – Nikita
          Jan 18 '18 at 13:29












          $begingroup$
          @Nikita, if you deem my answer to be the right one please tick it as such, follwing the habit here at this StackExchange site, thanks.
          $endgroup$
          – Nicky Hekster
          Feb 6 '18 at 10:25




          $begingroup$
          @Nikita, if you deem my answer to be the right one please tick it as such, follwing the habit here at this StackExchange site, thanks.
          $endgroup$
          – Nicky Hekster
          Feb 6 '18 at 10:25












          $begingroup$
          I am so sorry! Many thanks for your answer!
          $endgroup$
          – Nikita
          Feb 6 '18 at 12:21




          $begingroup$
          I am so sorry! Many thanks for your answer!
          $endgroup$
          – Nikita
          Feb 6 '18 at 12:21












          $begingroup$
          Digging into (c) somewhat deeper I found a paper of Udo Riese, A Subnormality Criterion in Finite Groups Related to Character Degrees (1998), J. of Algebra 201 which proves that $H$ must be subnormal. Moreover, $H^G=langle H^g: g in G rangle$, the normal closure of $H$ is nilpotent! The results are far from trivial, the first uses the famous Zipper Lemma of Helmut Wielandt.
          $endgroup$
          – Nicky Hekster
          Feb 9 '18 at 13:38






          $begingroup$
          Digging into (c) somewhat deeper I found a paper of Udo Riese, A Subnormality Criterion in Finite Groups Related to Character Degrees (1998), J. of Algebra 201 which proves that $H$ must be subnormal. Moreover, $H^G=langle H^g: g in G rangle$, the normal closure of $H$ is nilpotent! The results are far from trivial, the first uses the famous Zipper Lemma of Helmut Wielandt.
          $endgroup$
          – Nicky Hekster
          Feb 9 '18 at 13:38














          $begingroup$
          @Nikita I would like to thank YOU for your post and hence inspiration. Nicky
          $endgroup$
          – Nicky Hekster
          Feb 12 '18 at 12:43




          $begingroup$
          @Nikita I would like to thank YOU for your post and hence inspiration. Nicky
          $endgroup$
          – Nicky Hekster
          Feb 12 '18 at 12:43


















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