$H$ is a subgroup of a finite group $G$ such that $|H|$ and $big([G:H]-1big)!$ are relatively prime. Prove...
$begingroup$
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
abstract-algebra group-theory group-actions normal-subgroups quotient-group
edited Dec 15 '18 at 21:45
Batominovski
1
asked Dec 11 '18 at 15:17
So LoSo Lo
62729
$endgroup$
add a comment |
$begingroup$
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
abstract-algebra group-theory group-actions normal-subgroups quotient-group
edited Dec 15 '18 at 21:45
Batominovski
1
asked Dec 11 '18 at 15:17
So LoSo Lo
62729
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
abstract-algebra group-theory group-actions normal-subgroups quotient-group
edited Dec 15 '18 at 21:45
Batominovski
1
asked Dec 11 '18 at 15:17
So LoSo Lo
62729
$endgroup$
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
abstract-algebra group-theory group-actions normal-subgroups quotient-group
edited Dec 15 '18 at 21:45
Batominovski
1
asked Dec 11 '18 at 15:17
So LoSo Lo
62729
edited Dec 15 '18 at 21:45
Batominovski
1
asked Dec 11 '18 at 15:17
So LoSo Lo
62729
edited Dec 15 '18 at 21:45
Batominovski
1
edited Dec 15 '18 at 21:45
Batominovski
1
edited Dec 15 '18 at 21:45
Batominovski
1
1
asked Dec 11 '18 at 15:17
So LoSo Lo
62729
asked Dec 11 '18 at 15:17
So LoSo Lo
62729
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
add a comment |
$begingroup$
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
add a comment |
1 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 $K=H$.
answered Dec 11 '18 at 15:51
BatominovskiBatominovski
1
$endgroup$
add a comment |
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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 $K=H$.
answered Dec 11 '18 at 15:51
BatominovskiBatominovski
1
$endgroup$
add a comment |
$begingroup$
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 $K=H$.
answered Dec 11 '18 at 15:51
BatominovskiBatominovski
1
$endgroup$
add a comment |
$begingroup$
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 $K=H$.
answered Dec 11 '18 at 15:51
BatominovskiBatominovski
1
$endgroup$
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 $K=H$.
answered Dec 11 '18 at 15:51
BatominovskiBatominovski
1
answered Dec 11 '18 at 15:51
BatominovskiBatominovski
1
answered Dec 11 '18 at 15:51
BatominovskiBatominovski
1
answered Dec 11 '18 at 15:51
BatominovskiBatominovski
1
1
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$begingroup$
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