$H$ is a subgroup of a finite group $G$ such that $|H|$ and $big([G:H]-1big)!$ are relatively prime. Prove...












3












$begingroup$



Suppose that $H$ is a subgroup of a finite group $G$ and that $|H|$ and $big([G:H]-1big)!$ are relatively prime. Prove that $H$ is normal in G




Let $[G:H]=m$



Let $G$ act on set $A$ of left cosets of $H$ in $G$ under left multiplication, then we have



$$phi : G to S_Asimeq S_m$$



where $$K= kerphi subseteq H$$
$$frac{G}{K} simeq phi(G)leq S_m$$



So $$[G:K] mid m!$$



$$Rightarrow [G:H][H:K]mid m!$$



$$Rightarrow [H:K]|(m-1)!$$



But $gcdbig(|H|,(m-1)!big) = 1 Rightarrow |K|>1 $



How do I proceed to prove $K=H lhd G$?



Or is this approach wrong










share|cite|improve this question











$endgroup$












  • $begingroup$
    This is a more general version of the fact that a subgroup of index $2$ is always normal
    $endgroup$
    – So Lo
    Dec 11 '18 at 15:33










  • $begingroup$
    Instead you can note that the condition implies that any prime divisor in $|H|$ is greater than the index of $H$. Then if $K$ is any subgroup of the same order as $H$ you can see from the possible order that $HK = H$.
    $endgroup$
    – Tobias Kildetoft
    Dec 11 '18 at 15:39
















3












$begingroup$



Suppose that $H$ is a subgroup of a finite group $G$ and that $|H|$ and $big([G:H]-1big)!$ are relatively prime. Prove that $H$ is normal in G




Let $[G:H]=m$



Let $G$ act on set $A$ of left cosets of $H$ in $G$ under left multiplication, then we have



$$phi : G to S_Asimeq S_m$$



where $$K= kerphi subseteq H$$
$$frac{G}{K} simeq phi(G)leq S_m$$



So $$[G:K] mid m!$$



$$Rightarrow [G:H][H:K]mid m!$$



$$Rightarrow [H:K]|(m-1)!$$



But $gcdbig(|H|,(m-1)!big) = 1 Rightarrow |K|>1 $



How do I proceed to prove $K=H lhd G$?



Or is this approach wrong










share|cite|improve this question











$endgroup$












  • $begingroup$
    This is a more general version of the fact that a subgroup of index $2$ is always normal
    $endgroup$
    – So Lo
    Dec 11 '18 at 15:33










  • $begingroup$
    Instead you can note that the condition implies that any prime divisor in $|H|$ is greater than the index of $H$. Then if $K$ is any subgroup of the same order as $H$ you can see from the possible order that $HK = H$.
    $endgroup$
    – Tobias Kildetoft
    Dec 11 '18 at 15:39














3












3








3


2



$begingroup$



Suppose that $H$ is a subgroup of a finite group $G$ and that $|H|$ and $big([G:H]-1big)!$ are relatively prime. Prove that $H$ is normal in G




Let $[G:H]=m$



Let $G$ act on set $A$ of left cosets of $H$ in $G$ under left multiplication, then we have



$$phi : G to S_Asimeq S_m$$



where $$K= kerphi subseteq H$$
$$frac{G}{K} simeq phi(G)leq S_m$$



So $$[G:K] mid m!$$



$$Rightarrow [G:H][H:K]mid m!$$



$$Rightarrow [H:K]|(m-1)!$$



But $gcdbig(|H|,(m-1)!big) = 1 Rightarrow |K|>1 $



How do I proceed to prove $K=H lhd G$?



Or is this approach wrong










share|cite|improve this question











$endgroup$





Suppose that $H$ is a subgroup of a finite group $G$ and that $|H|$ and $big([G:H]-1big)!$ are relatively prime. Prove that $H$ is normal in G




Let $[G:H]=m$



Let $G$ act on set $A$ of left cosets of $H$ in $G$ under left multiplication, then we have



$$phi : G to S_Asimeq S_m$$



where $$K= kerphi subseteq H$$
$$frac{G}{K} simeq phi(G)leq S_m$$



So $$[G:K] mid m!$$



$$Rightarrow [G:H][H:K]mid m!$$



$$Rightarrow [H:K]|(m-1)!$$



But $gcdbig(|H|,(m-1)!big) = 1 Rightarrow |K|>1 $



How do I proceed to prove $K=H lhd G$?



Or is this approach wrong







abstract-algebra group-theory group-actions normal-subgroups quotient-group






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edited Dec 15 '18 at 21:45









Batominovski

33k33293




33k33293










asked Dec 11 '18 at 15:17









So LoSo Lo

62729




62729












  • $begingroup$
    This is a more general version of the fact that a subgroup of index $2$ is always normal
    $endgroup$
    – So Lo
    Dec 11 '18 at 15:33










  • $begingroup$
    Instead you can note that the condition implies that any prime divisor in $|H|$ is greater than the index of $H$. Then if $K$ is any subgroup of the same order as $H$ you can see from the possible order that $HK = H$.
    $endgroup$
    – Tobias Kildetoft
    Dec 11 '18 at 15:39


















  • $begingroup$
    This is a more general version of the fact that a subgroup of index $2$ is always normal
    $endgroup$
    – So Lo
    Dec 11 '18 at 15:33










  • $begingroup$
    Instead you can note that the condition implies that any prime divisor in $|H|$ is greater than the index of $H$. Then if $K$ is any subgroup of the same order as $H$ you can see from the possible order that $HK = H$.
    $endgroup$
    – Tobias Kildetoft
    Dec 11 '18 at 15:39
















$begingroup$
This is a more general version of the fact that a subgroup of index $2$ is always normal
$endgroup$
– So Lo
Dec 11 '18 at 15:33




$begingroup$
This is a more general version of the fact that a subgroup of index $2$ is always normal
$endgroup$
– So Lo
Dec 11 '18 at 15:33












$begingroup$
Instead you can note that the condition implies that any prime divisor in $|H|$ is greater than the index of $H$. Then if $K$ is any subgroup of the same order as $H$ you can see from the possible order that $HK = H$.
$endgroup$
– Tobias Kildetoft
Dec 11 '18 at 15:39




$begingroup$
Instead you can note that the condition implies that any prime divisor in $|H|$ is greater than the index of $H$. Then if $K$ is any subgroup of the same order as $H$ you can see from the possible order that $HK = H$.
$endgroup$
– Tobias Kildetoft
Dec 11 '18 at 15:39










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$begingroup$

Note that $[H:K]$ divides both $|H|$ and $big([G:H]-1big)!=(m-1)!$. Therefore, it divides the greatest common divisor of $|H|$ and $big([G:H]-1big)!$, which is $1$. Hence, $[H:K]=1$, making $K=H$.






share|cite|improve this answer









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    $begingroup$

    Note that $[H:K]$ divides both $|H|$ and $big([G:H]-1big)!=(m-1)!$. Therefore, it divides the greatest common divisor of $|H|$ and $big([G:H]-1big)!$, which is $1$. Hence, $[H:K]=1$, making $K=H$.






    share|cite|improve this answer









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      4












      $begingroup$

      Note that $[H:K]$ divides both $|H|$ and $big([G:H]-1big)!=(m-1)!$. Therefore, it divides the greatest common divisor of $|H|$ and $big([G:H]-1big)!$, which is $1$. Hence, $[H:K]=1$, making $K=H$.






      share|cite|improve this answer









      $endgroup$
















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        4








        4





        $begingroup$

        Note that $[H:K]$ divides both $|H|$ and $big([G:H]-1big)!=(m-1)!$. Therefore, it divides the greatest common divisor of $|H|$ and $big([G:H]-1big)!$, which is $1$. Hence, $[H:K]=1$, making $K=H$.






        share|cite|improve this answer









        $endgroup$



        Note that $[H:K]$ divides both $|H|$ and $big([G:H]-1big)!=(m-1)!$. Therefore, it divides the greatest common divisor of $|H|$ and $big([G:H]-1big)!$, which is $1$. Hence, $[H:K]=1$, making $K=H$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 15:51









        BatominovskiBatominovski

        33k33293




        33k33293






























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