Mixing
Does a nondegenerate alternating map determines the exponent of the group?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Let $G$ be a finite abelian group, i.e., $G$ is a $mathbbZ$-module. Consider the second component of the exterior algebra of $G$ over $mathbbZ$, $wedge^2 G=Gwedge G$. Suppose that $a:Gwedge GrightarrowmathbbQ/mathbbZ$ is a nonzero, nondegenerate alternating form.
Does it follow that the exponent of the group $G$ and exponent of the image of $a$ is same?
Note that since $a$ is a nondegenerate form, the group $G$ is not cyclic.
Thank you for your time!
linear-algebra abstract-algebra finite-groups
asked Jul 15 at 16:57
user219197
294
add a comment |Â
up vote
1
down vote
favorite
Let $G$ be a finite abelian group, i.e., $G$ is a $mathbbZ$-module. Consider the second component of the exterior algebra of $G$ over $mathbbZ$, $wedge^2 G=Gwedge G$. Suppose that $a:Gwedge GrightarrowmathbbQ/mathbbZ$ is a nonzero, nondegenerate alternating form.
Does it follow that the exponent of the group $G$ and exponent of the image of $a$ is same?
Note that since $a$ is a nondegenerate form, the group $G$ is not cyclic.
Thank you for your time!
linear-algebra abstract-algebra finite-groups
asked Jul 15 at 16:57
user219197
294
What are your thoughts regarding the question and how you might address it? Have you tried to answer this?
– amWhy
Jul 15 at 17:17
I was thinking to use structure theorem for finite abelian groups. But do not know how to proceed. Also if there is a 'symplectic basis' as in vector space case then it might follow easily.
– user219197
Jul 16 at 4:50
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $G$ be a finite abelian group, i.e., $G$ is a $mathbbZ$-module. Consider the second component of the exterior algebra of $G$ over $mathbbZ$, $wedge^2 G=Gwedge G$. Suppose that $a:Gwedge GrightarrowmathbbQ/mathbbZ$ is a nonzero, nondegenerate alternating form.
Does it follow that the exponent of the group $G$ and exponent of the image of $a$ is same?
Note that since $a$ is a nondegenerate form, the group $G$ is not cyclic.
Thank you for your time!
linear-algebra abstract-algebra finite-groups
asked Jul 15 at 16:57
user219197
294
Let $G$ be a finite abelian group, i.e., $G$ is a $mathbbZ$-module. Consider the second component of the exterior algebra of $G$ over $mathbbZ$, $wedge^2 G=Gwedge G$. Suppose that $a:Gwedge GrightarrowmathbbQ/mathbbZ$ is a nonzero, nondegenerate alternating form.
Does it follow that the exponent of the group $G$ and exponent of the image of $a$ is same?
Note that since $a$ is a nondegenerate form, the group $G$ is not cyclic.
Thank you for your time!
linear-algebra abstract-algebra finite-groups
asked Jul 15 at 16:57
user219197
294
asked Jul 15 at 16:57
user219197
294
asked Jul 15 at 16:57
user219197
294
asked Jul 15 at 16:57
user219197
294
294
What are your thoughts regarding the question and how you might address it? Have you tried to answer this?
– amWhy
Jul 15 at 17:17
I was thinking to use structure theorem for finite abelian groups. But do not know how to proceed. Also if there is a 'symplectic basis' as in vector space case then it might follow easily.
– user219197
Jul 16 at 4:50
add a comment |Â
What are your thoughts regarding the question and how you might address it? Have you tried to answer this?
– amWhy
Jul 15 at 17:17
I was thinking to use structure theorem for finite abelian groups. But do not know how to proceed. Also if there is a 'symplectic basis' as in vector space case then it might follow easily.
– user219197
Jul 16 at 4:50
What are your thoughts regarding the question and how you might address it? Have you tried to answer this?
– amWhy
Jul 15 at 17:17
What are your thoughts regarding the question and how you might address it? Have you tried to answer this?
– amWhy
Jul 15 at 17:17
I was thinking to use structure theorem for finite abelian groups. But do not know how to proceed. Also if there is a 'symplectic basis' as in vector space case then it might follow easily.
– user219197
Jul 16 at 4:50
I was thinking to use structure theorem for finite abelian groups. But do not know how to proceed. Also if there is a 'symplectic basis' as in vector space case then it might follow easily.
– user219197
Jul 16 at 4:50
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%2f2852673%2fdoes-a-nondegenerate-alternating-map-determines-the-exponent-of-the-group%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
What are your thoughts regarding the question and how you might address it? Have you tried to answer this?
– amWhy
Jul 15 at 17:17
I was thinking to use structure theorem for finite abelian groups. But do not know how to proceed. Also if there is a 'symplectic basis' as in vector space case then it might follow easily.
– user219197
Jul 16 at 4:50