Suppose G is a finite group, H and K normal subgroups, gcd(|H|, |K|) = 1, and |G| = |H| |K|. Prove $G cong H times K$

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Suppose G is a finite group, H and K normal subgroups, gcd(|H|, |K|) = 1, and |G| = |H| |K|. Prove $G cong H times K$.

I have a proposition that say if H and K are normal subgroups of G, $H cap K = e$, and G = HK, then $G cong H times K$. My question is if this is the proposition that I need to use and how to go about showing that $H cap K = e$, and G = HK, or how else can I go about proving this.

abstract-algebra group-theory

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asked Jul 19 at 16:55

frostyfeet

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  The order of $Hcap K$ divides the order of $H$ and the order of $K$...
  – David C. Ullrich
  Jul 19 at 16:57
  

  
  
  
  

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up vote
1
down vote

favorite

Suppose G is a finite group, H and K normal subgroups, gcd(|H|, |K|) = 1, and |G| = |H| |K|. Prove $G cong H times K$.

I have a proposition that say if H and K are normal subgroups of G, $H cap K = e$, and G = HK, then $G cong H times K$. My question is if this is the proposition that I need to use and how to go about showing that $H cap K = e$, and G = HK, or how else can I go about proving this.

abstract-algebra group-theory

share|cite|improve this question

asked Jul 19 at 16:55

frostyfeet

1248

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  The order of $Hcap K$ divides the order of $H$ and the order of $K$...
  – David C. Ullrich
  Jul 19 at 16:57
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Suppose G is a finite group, H and K normal subgroups, gcd(|H|, |K|) = 1, and |G| = |H| |K|. Prove $G cong H times K$.

I have a proposition that say if H and K are normal subgroups of G, $H cap K = e$, and G = HK, then $G cong H times K$. My question is if this is the proposition that I need to use and how to go about showing that $H cap K = e$, and G = HK, or how else can I go about proving this.

abstract-algebra group-theory

share|cite|improve this question

asked Jul 19 at 16:55

frostyfeet

1248

Suppose G is a finite group, H and K normal subgroups, gcd(|H|, |K|) = 1, and |G| = |H| |K|. Prove $G cong H times K$.

I have a proposition that say if H and K are normal subgroups of G, $H cap K = e$, and G = HK, then $G cong H times K$. My question is if this is the proposition that I need to use and how to go about showing that $H cap K = e$, and G = HK, or how else can I go about proving this.

abstract-algebra group-theory

share|cite|improve this question

asked Jul 19 at 16:55

frostyfeet

1248

share|cite|improve this question

share|cite|improve this question

asked Jul 19 at 16:55

frostyfeet

1248

asked Jul 19 at 16:55

frostyfeet

1248

asked Jul 19 at 16:55

frostyfeet

1248

1248

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  The order of $Hcap K$ divides the order of $H$ and the order of $K$...
  – David C. Ullrich
  Jul 19 at 16:57
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  The order of $Hcap K$ divides the order of $H$ and the order of $K$...
  – David C. Ullrich
  Jul 19 at 16:57
  

  
  
  
  

6

6

The order of $Hcap K$ divides the order of $H$ and the order of $K$...
– David C. Ullrich
Jul 19 at 16:57

The order of $Hcap K$ divides the order of $H$ and the order of $K$...
– David C. Ullrich
Jul 19 at 16:57

add a comment | 

2 Answers
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As in the David's comment , you will get that since $|H cap K| | H;K$ and $gcd(|H|;|K|)=1$, their intersection is the identity elementy. Since the intersection of two normal subgroups $H$ and $K$ is $aba^-1b^-1$ we get that the elements of $H$ and $K$ commute and using the fact that $|G|=|H||K|=fracH1=fracH=|HK|$, we have that $G=HK$ and hence $G= H rtimes K $ and from the commutativity of the elements of $H$ and $K$, we have that $G cong H times K$.

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edited Jul 19 at 17:18

answered Jul 19 at 17:08

Mario 04

6013

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For the first part see @David C. Ullrich's comment...

For the second part, of course $HKsubset G$, and then $mid HKmid=mid Gmid$, since $mid HKmid=fracmid Hmid mid Kmidmid Hcap Kmid$. See this... So $HK=G$.

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answered Jul 19 at 17:15

Chris Custer

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2 Answers
2

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2 Answers
2

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up vote
1
down vote

As in the David's comment , you will get that since $|H cap K| | H;K$ and $gcd(|H|;|K|)=1$, their intersection is the identity elementy. Since the intersection of two normal subgroups $H$ and $K$ is $aba^-1b^-1$ we get that the elements of $H$ and $K$ commute and using the fact that $|G|=|H||K|=fracH1=fracH=|HK|$, we have that $G=HK$ and hence $G= H rtimes K $ and from the commutativity of the elements of $H$ and $K$, we have that $G cong H times K$.

share|cite|improve this answer

edited Jul 19 at 17:18

answered Jul 19 at 17:08

Mario 04

6013

add a comment | 

up vote
1
down vote

As in the David's comment , you will get that since $|H cap K| | H;K$ and $gcd(|H|;|K|)=1$, their intersection is the identity elementy. Since the intersection of two normal subgroups $H$ and $K$ is $aba^-1b^-1$ we get that the elements of $H$ and $K$ commute and using the fact that $|G|=|H||K|=fracH1=fracH=|HK|$, we have that $G=HK$ and hence $G= H rtimes K $ and from the commutativity of the elements of $H$ and $K$, we have that $G cong H times K$.

share|cite|improve this answer

edited Jul 19 at 17:18

answered Jul 19 at 17:08

Mario 04

6013

add a comment | 

up vote
1
down vote

up vote
1
down vote

As in the David's comment , you will get that since $|H cap K| | H;K$ and $gcd(|H|;|K|)=1$, their intersection is the identity elementy. Since the intersection of two normal subgroups $H$ and $K$ is $aba^-1b^-1$ we get that the elements of $H$ and $K$ commute and using the fact that $|G|=|H||K|=fracH1=fracH=|HK|$, we have that $G=HK$ and hence $G= H rtimes K $ and from the commutativity of the elements of $H$ and $K$, we have that $G cong H times K$.

share|cite|improve this answer

edited Jul 19 at 17:18

answered Jul 19 at 17:08

Mario 04

6013

As in the David's comment , you will get that since $|H cap K| | H;K$ and $gcd(|H|;|K|)=1$, their intersection is the identity elementy. Since the intersection of two normal subgroups $H$ and $K$ is $aba^-1b^-1$ we get that the elements of $H$ and $K$ commute and using the fact that $|G|=|H||K|=fracH1=fracH=|HK|$, we have that $G=HK$ and hence $G= H rtimes K $ and from the commutativity of the elements of $H$ and $K$, we have that $G cong H times K$.

share|cite|improve this answer

edited Jul 19 at 17:18

answered Jul 19 at 17:08

Mario 04

6013

share|cite|improve this answer

share|cite|improve this answer

edited Jul 19 at 17:18

edited Jul 19 at 17:18

edited Jul 19 at 17:18

answered Jul 19 at 17:08

Mario 04

6013

answered Jul 19 at 17:08

Mario 04

6013

answered Jul 19 at 17:08

Mario 04

6013

6013

add a comment | 

add a comment | 

up vote
0
down vote

For the first part see @David C. Ullrich's comment...

For the second part, of course $HKsubset G$, and then $mid HKmid=mid Gmid$, since $mid HKmid=fracmid Hmid mid Kmidmid Hcap Kmid$. See this... So $HK=G$.

share|cite|improve this answer

answered Jul 19 at 17:15

Chris Custer

5,4482622

add a comment | 

up vote
0
down vote

For the first part see @David C. Ullrich's comment...

For the second part, of course $HKsubset G$, and then $mid HKmid=mid Gmid$, since $mid HKmid=fracmid Hmid mid Kmidmid Hcap Kmid$. See this... So $HK=G$.

share|cite|improve this answer

answered Jul 19 at 17:15

Chris Custer

5,4482622

add a comment | 

up vote
0
down vote

up vote
0
down vote

For the first part see @David C. Ullrich's comment...

For the second part, of course $HKsubset G$, and then $mid HKmid=mid Gmid$, since $mid HKmid=fracmid Hmid mid Kmidmid Hcap Kmid$. See this... So $HK=G$.

share|cite|improve this answer

answered Jul 19 at 17:15

Chris Custer

5,4482622

For the first part see @David C. Ullrich's comment...

For the second part, of course $HKsubset G$, and then $mid HKmid=mid Gmid$, since $mid HKmid=fracmid Hmid mid Kmidmid Hcap Kmid$. See this... So $HK=G$.

share|cite|improve this answer

answered Jul 19 at 17:15

Chris Custer

5,4482622

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 17:15

Chris Custer

5,4482622

answered Jul 19 at 17:15

Chris Custer

5,4482622

answered Jul 19 at 17:15

Chris Custer

5,4482622

5,4482622

add a comment | 

add a comment | 

 

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