Units in group rings.
up vote
6
down vote
favorite
Let $G$ be a finite solvable group of order $n$, and let $g_1 ... g_n$ be an enumeration of its elements. Let $a_1 ... a_n$ be a sequence of integers, such that $sum a_i$ is relatively prime to $n$.
Consider $mathbb{C}[G]$, the group ring of $G$ with complex coefficients. Does the element $sum a_i g_i$ necessarily have to be a unit in the group ring? (I believe that the element does have to be a unit, and have a proof in the cyclic and abelian case, but was hoping for a reference in greater generality, at least in the case when $G$ is solvable.)
gr.group-theory rt.representation-theory
asked 4 hours ago
Fedex
511
add a comment |
up vote
6
down vote
favorite
Let $G$ be a finite solvable group of order $n$, and let $g_1 ... g_n$ be an enumeration of its elements. Let $a_1 ... a_n$ be a sequence of integers, such that $sum a_i$ is relatively prime to $n$.
Consider $mathbb{C}[G]$, the group ring of $G$ with complex coefficients. Does the element $sum a_i g_i$ necessarily have to be a unit in the group ring? (I believe that the element does have to be a unit, and have a proof in the cyclic and abelian case, but was hoping for a reference in greater generality, at least in the case when $G$ is solvable.)
gr.group-theory rt.representation-theory
asked 4 hours ago
Fedex
511
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Let $G$ be a finite solvable group of order $n$, and let $g_1 ... g_n$ be an enumeration of its elements. Let $a_1 ... a_n$ be a sequence of integers, such that $sum a_i$ is relatively prime to $n$.
Consider $mathbb{C}[G]$, the group ring of $G$ with complex coefficients. Does the element $sum a_i g_i$ necessarily have to be a unit in the group ring? (I believe that the element does have to be a unit, and have a proof in the cyclic and abelian case, but was hoping for a reference in greater generality, at least in the case when $G$ is solvable.)
gr.group-theory rt.representation-theory
asked 4 hours ago
Fedex
511
Let $G$ be a finite solvable group of order $n$, and let $g_1 ... g_n$ be an enumeration of its elements. Let $a_1 ... a_n$ be a sequence of integers, such that $sum a_i$ is relatively prime to $n$.
Consider $mathbb{C}[G]$, the group ring of $G$ with complex coefficients. Does the element $sum a_i g_i$ necessarily have to be a unit in the group ring? (I believe that the element does have to be a unit, and have a proof in the cyclic and abelian case, but was hoping for a reference in greater generality, at least in the case when $G$ is solvable.)
gr.group-theory rt.representation-theory
gr.group-theory rt.representation-theory
asked 4 hours ago
Fedex
511
asked 4 hours ago
Fedex
511
asked 4 hours ago
Fedex
511
asked 4 hours ago
Fedex
511
asked 4 hours ago
Fedex
511
511
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
5
down vote
This is false for the cyclic group of order $6$. Let $g$ be a generator. Then $g^2-g+1$ acts by $0$ on the representations where $g$ acts by a primitive $6$-th root of $1$, and hence is not a unit in the group ring, but $1-1+1=1$ is relatively prime to $6$.
answered 1 hour ago
David E Speyer
105k8271533
1
Alternatively, the product of this element with $(g + 1)(g^3 - 1)$ is $0$. (That's probably effectively the same proof in a minor disguise.)
– LSpice
17 mins ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
This is false for the cyclic group of order $6$. Let $g$ be a generator. Then $g^2-g+1$ acts by $0$ on the representations where $g$ acts by a primitive $6$-th root of $1$, and hence is not a unit in the group ring, but $1-1+1=1$ is relatively prime to $6$.
answered 1 hour ago
David E Speyer
105k8271533
1
Alternatively, the product of this element with $(g + 1)(g^3 - 1)$ is $0$. (That's probably effectively the same proof in a minor disguise.)
– LSpice
17 mins ago
add a comment |
up vote
5
down vote
This is false for the cyclic group of order $6$. Let $g$ be a generator. Then $g^2-g+1$ acts by $0$ on the representations where $g$ acts by a primitive $6$-th root of $1$, and hence is not a unit in the group ring, but $1-1+1=1$ is relatively prime to $6$.
answered 1 hour ago
David E Speyer
105k8271533
1
Alternatively, the product of this element with $(g + 1)(g^3 - 1)$ is $0$. (That's probably effectively the same proof in a minor disguise.)
– LSpice
17 mins ago
add a comment |
up vote
5
down vote
up vote
5
down vote
This is false for the cyclic group of order $6$. Let $g$ be a generator. Then $g^2-g+1$ acts by $0$ on the representations where $g$ acts by a primitive $6$-th root of $1$, and hence is not a unit in the group ring, but $1-1+1=1$ is relatively prime to $6$.
answered 1 hour ago
David E Speyer
105k8271533
This is false for the cyclic group of order $6$. Let $g$ be a generator. Then $g^2-g+1$ acts by $0$ on the representations where $g$ acts by a primitive $6$-th root of $1$, and hence is not a unit in the group ring, but $1-1+1=1$ is relatively prime to $6$.
answered 1 hour ago
David E Speyer
105k8271533
edited 45 mins ago
answered 1 hour ago
David E Speyer
105k8271533
answered 1 hour ago
David E Speyer
105k8271533
answered 1 hour ago
David E Speyer
105k8271533
105k8271533
1
Alternatively, the product of this element with $(g + 1)(g^3 - 1)$ is $0$. (That's probably effectively the same proof in a minor disguise.)
– LSpice
17 mins ago
add a comment |
1
Alternatively, the product of this element with $(g + 1)(g^3 - 1)$ is $0$. (That's probably effectively the same proof in a minor disguise.)
– LSpice
17 mins ago
1
1
Alternatively, the product of this element with $(g + 1)(g^3 - 1)$ is $0$. (That's probably effectively the same proof in a minor disguise.)
– LSpice
17 mins ago
Alternatively, the product of this element with $(g + 1)(g^3 - 1)$ is $0$. (That's probably effectively the same proof in a minor disguise.)
– LSpice
17 mins ago
add a comment |
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f317376%2funits-in-group-rings%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
