Units in group rings.
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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 5 hours ago
Fedex
511
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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 5 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 5 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 5 hours ago
Fedex
511
asked 5 hours ago
Fedex
511
asked 5 hours ago
Fedex
511
asked 5 hours ago
Fedex
511
asked 5 hours ago
Fedex
511
511
add a comment |
add a comment |
1 Answer
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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 2 hours 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
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
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 2 hours 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
1 hour ago
add a comment |
up vote
6
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 2 hours 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
1 hour ago
add a comment |
up vote
6
down vote
up vote
6
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 2 hours 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 2 hours ago
David E Speyer
105k8271533
edited 2 hours ago
answered 2 hours ago
David E Speyer
105k8271533
answered 2 hours ago
David E Speyer
105k8271533
answered 2 hours 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
add a comment |
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