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







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    up vote
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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







    share|cite|improve this question
























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      favorite
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      up vote
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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







      share|cite|improve this question













      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






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      asked 6 hours ago









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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 $6$.






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            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
            2 hours ago











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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 $6$.






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          • 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
            2 hours ago















          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 $6$.






          share|cite|improve this answer



















          • 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
            2 hours ago













          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 $6$.






          share|cite|improve this answer














          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 $6$.







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          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 hours ago

























          answered 3 hours ago









          David E Speyer

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          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
            2 hours ago














          • 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
            2 hours 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
          2 hours 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
          2 hours ago


















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