Let $G$ be a finite group with a normal subgroup $H$ such that $G/H$ has prime order . Then precisely when $G$ is isomorphic to $ H × G/H$ .
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Let $G$ be a finite group with a normal subgroup $H$ . Then precisely when $G$ is isomorphic to $ H × G/H$ .
I was trying to directly construct an isomorphism with trial and error between $G$ and $ H × G/H$ and my hope was to find appropriate conditions on $G$ and $H$ along the way.But so far not able to make any notable progress.
I was trying different onto homomorphisms from $Gto H$ and $Gto G/H$ , and then use these homomorphisms as the first and the second component of the desired isomorphism between $G$ and $ H × G/H$. I was trying to see how those homomorphisms should relate to each other in order to do the job and what constrain they put on $G$ and $H$ ?
Any progress in this situation will be help full to me .Thanks
abstract-algebra group-theory finite-groups
add a comment |Â
up vote
2
down vote
favorite
Let $G$ be a finite group with a normal subgroup $H$ . Then precisely when $G$ is isomorphic to $ H × G/H$ .
I was trying to directly construct an isomorphism with trial and error between $G$ and $ H × G/H$ and my hope was to find appropriate conditions on $G$ and $H$ along the way.But so far not able to make any notable progress.
I was trying different onto homomorphisms from $Gto H$ and $Gto G/H$ , and then use these homomorphisms as the first and the second component of the desired isomorphism between $G$ and $ H × G/H$. I was trying to see how those homomorphisms should relate to each other in order to do the job and what constrain they put on $G$ and $H$ ?
Any progress in this situation will be help full to me .Thanks
abstract-algebra group-theory finite-groups
It is not very clear to me, what are you asking. Please give more details.
– peterh
Jul 15 at 15:44
1
This is a very broad question. One characterization is when there is a projection $pi:Gto H$ which is the identity on $H$.
– Ashwin Trisal
Jul 15 at 15:47
A necessary (but not sufficient) condition is that $H$ is in the center of $G$.
– Andreas Blass
Jul 15 at 17:45
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $G$ be a finite group with a normal subgroup $H$ . Then precisely when $G$ is isomorphic to $ H × G/H$ .
I was trying to directly construct an isomorphism with trial and error between $G$ and $ H × G/H$ and my hope was to find appropriate conditions on $G$ and $H$ along the way.But so far not able to make any notable progress.
I was trying different onto homomorphisms from $Gto H$ and $Gto G/H$ , and then use these homomorphisms as the first and the second component of the desired isomorphism between $G$ and $ H × G/H$. I was trying to see how those homomorphisms should relate to each other in order to do the job and what constrain they put on $G$ and $H$ ?
Any progress in this situation will be help full to me .Thanks
abstract-algebra group-theory finite-groups
Let $G$ be a finite group with a normal subgroup $H$ . Then precisely when $G$ is isomorphic to $ H × G/H$ .
I was trying to directly construct an isomorphism with trial and error between $G$ and $ H × G/H$ and my hope was to find appropriate conditions on $G$ and $H$ along the way.But so far not able to make any notable progress.
I was trying different onto homomorphisms from $Gto H$ and $Gto G/H$ , and then use these homomorphisms as the first and the second component of the desired isomorphism between $G$ and $ H × G/H$. I was trying to see how those homomorphisms should relate to each other in order to do the job and what constrain they put on $G$ and $H$ ?
Any progress in this situation will be help full to me .Thanks
abstract-algebra group-theory finite-groups
edited Jul 16 at 11:38
asked Jul 15 at 15:07
bumba
246
246
It is not very clear to me, what are you asking. Please give more details.
– peterh
Jul 15 at 15:44
1
This is a very broad question. One characterization is when there is a projection $pi:Gto H$ which is the identity on $H$.
– Ashwin Trisal
Jul 15 at 15:47
A necessary (but not sufficient) condition is that $H$ is in the center of $G$.
– Andreas Blass
Jul 15 at 17:45
add a comment |Â
It is not very clear to me, what are you asking. Please give more details.
– peterh
Jul 15 at 15:44
1
This is a very broad question. One characterization is when there is a projection $pi:Gto H$ which is the identity on $H$.
– Ashwin Trisal
Jul 15 at 15:47
A necessary (but not sufficient) condition is that $H$ is in the center of $G$.
– Andreas Blass
Jul 15 at 17:45
It is not very clear to me, what are you asking. Please give more details.
– peterh
Jul 15 at 15:44
It is not very clear to me, what are you asking. Please give more details.
– peterh
Jul 15 at 15:44
1
1
This is a very broad question. One characterization is when there is a projection $pi:Gto H$ which is the identity on $H$.
– Ashwin Trisal
Jul 15 at 15:47
This is a very broad question. One characterization is when there is a projection $pi:Gto H$ which is the identity on $H$.
– Ashwin Trisal
Jul 15 at 15:47
A necessary (but not sufficient) condition is that $H$ is in the center of $G$.
– Andreas Blass
Jul 15 at 17:45
A necessary (but not sufficient) condition is that $H$ is in the center of $G$.
– Andreas Blass
Jul 15 at 17:45
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2852609%2flet-g-be-a-finite-group-with-a-normal-subgroup-h-such-that-g-h-has-prime-o%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
It is not very clear to me, what are you asking. Please give more details.
– peterh
Jul 15 at 15:44
1
This is a very broad question. One characterization is when there is a projection $pi:Gto H$ which is the identity on $H$.
– Ashwin Trisal
Jul 15 at 15:47
A necessary (but not sufficient) condition is that $H$ is in the center of $G$.
– Andreas Blass
Jul 15 at 17:45