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Number of irreducible representations of a group over a field
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Let $G$ be a finite group and $K$ a field with $mathbbQ subseteq K subseteq mathbbC$.
For $K=mathbbC$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of $G$.
For $K=mathbbR$ the number of irreducible representations of $KG$ is equal to $fracr+s2$, where $r$ denotes the number of conjugacy classes of $G$ and $s$ the number of classes stable under inversion.
For $K=mathbbQ$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of cyclic subgroups of $G$.
Question:
Are there such nice closed forumlas for other fields $K$? For example quadratic, cubic or cyclotomic field extensions of $mathbbQ$.
co.combinatorics rt.representation-theory finite-groups
asked 1 hour ago
Mare
3,19121029
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up vote
8
down vote
favorite
Let $G$ be a finite group and $K$ a field with $mathbbQ subseteq K subseteq mathbbC$.
For $K=mathbbC$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of $G$.
For $K=mathbbR$ the number of irreducible representations of $KG$ is equal to $fracr+s2$, where $r$ denotes the number of conjugacy classes of $G$ and $s$ the number of classes stable under inversion.
For $K=mathbbQ$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of cyclic subgroups of $G$.
Question:
Are there such nice closed forumlas for other fields $K$? For example quadratic, cubic or cyclotomic field extensions of $mathbbQ$.
co.combinatorics rt.representation-theory finite-groups
asked 1 hour ago
Mare
3,19121029
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
Let $G$ be a finite group and $K$ a field with $mathbbQ subseteq K subseteq mathbbC$.
For $K=mathbbC$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of $G$.
For $K=mathbbR$ the number of irreducible representations of $KG$ is equal to $fracr+s2$, where $r$ denotes the number of conjugacy classes of $G$ and $s$ the number of classes stable under inversion.
For $K=mathbbQ$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of cyclic subgroups of $G$.
Question:
Are there such nice closed forumlas for other fields $K$? For example quadratic, cubic or cyclotomic field extensions of $mathbbQ$.
co.combinatorics rt.representation-theory finite-groups
asked 1 hour ago
Mare
3,19121029
Let $G$ be a finite group and $K$ a field with $mathbbQ subseteq K subseteq mathbbC$.
For $K=mathbbC$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of $G$.
For $K=mathbbR$ the number of irreducible representations of $KG$ is equal to $fracr+s2$, where $r$ denotes the number of conjugacy classes of $G$ and $s$ the number of classes stable under inversion.
For $K=mathbbQ$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of cyclic subgroups of $G$.
Question:
Are there such nice closed forumlas for other fields $K$? For example quadratic, cubic or cyclotomic field extensions of $mathbbQ$.
co.combinatorics rt.representation-theory finite-groups
co.combinatorics rt.representation-theory finite-groups
asked 1 hour ago
Mare
3,19121029
asked 1 hour ago
Mare
3,19121029
asked 1 hour ago
Mare
3,19121029
asked 1 hour ago
Mare
3,19121029
asked 1 hour ago
Mare
3,19121029
3,19121029
add a comment |Â
add a comment |Â
1 Answer
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There is a characterization due to Berman for any field. In your case let $n$ be the least common multiple of the orders of elements of $G$. Let $zeta$ be a primitive $n^th$ root of unity. Let $H=Gal(K(zeta)/K)$. We can identify $H$ with a subgroup of $(mathbb Z/nmathbb Z)^*$. Call $a,bin G$ $K$-conjugate if $b^j=gag^-1$ for some $gin G$ and $jin H$. This is an equivalence relation and the number of irreducible $KG$-modules is the number of $K$-conjugacy classes.
In characteristic $p$ you do the analogous thing but only look at this equivalence relation on $p$-regular elements.
answered 19 mins ago
Benjamin Steinberg
22.5k263122
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1 Answer
1
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
There is a characterization due to Berman for any field. In your case let $n$ be the least common multiple of the orders of elements of $G$. Let $zeta$ be a primitive $n^th$ root of unity. Let $H=Gal(K(zeta)/K)$. We can identify $H$ with a subgroup of $(mathbb Z/nmathbb Z)^*$. Call $a,bin G$ $K$-conjugate if $b^j=gag^-1$ for some $gin G$ and $jin H$. This is an equivalence relation and the number of irreducible $KG$-modules is the number of $K$-conjugacy classes.
In characteristic $p$ you do the analogous thing but only look at this equivalence relation on $p$-regular elements.
answered 19 mins ago
Benjamin Steinberg
22.5k263122
add a comment |Â
up vote
4
down vote
There is a characterization due to Berman for any field. In your case let $n$ be the least common multiple of the orders of elements of $G$. Let $zeta$ be a primitive $n^th$ root of unity. Let $H=Gal(K(zeta)/K)$. We can identify $H$ with a subgroup of $(mathbb Z/nmathbb Z)^*$. Call $a,bin G$ $K$-conjugate if $b^j=gag^-1$ for some $gin G$ and $jin H$. This is an equivalence relation and the number of irreducible $KG$-modules is the number of $K$-conjugacy classes.
In characteristic $p$ you do the analogous thing but only look at this equivalence relation on $p$-regular elements.
answered 19 mins ago
Benjamin Steinberg
22.5k263122
add a comment |Â
up vote
4
down vote
up vote
4
down vote
There is a characterization due to Berman for any field. In your case let $n$ be the least common multiple of the orders of elements of $G$. Let $zeta$ be a primitive $n^th$ root of unity. Let $H=Gal(K(zeta)/K)$. We can identify $H$ with a subgroup of $(mathbb Z/nmathbb Z)^*$. Call $a,bin G$ $K$-conjugate if $b^j=gag^-1$ for some $gin G$ and $jin H$. This is an equivalence relation and the number of irreducible $KG$-modules is the number of $K$-conjugacy classes.
In characteristic $p$ you do the analogous thing but only look at this equivalence relation on $p$-regular elements.
answered 19 mins ago
Benjamin Steinberg
22.5k263122
There is a characterization due to Berman for any field. In your case let $n$ be the least common multiple of the orders of elements of $G$. Let $zeta$ be a primitive $n^th$ root of unity. Let $H=Gal(K(zeta)/K)$. We can identify $H$ with a subgroup of $(mathbb Z/nmathbb Z)^*$. Call $a,bin G$ $K$-conjugate if $b^j=gag^-1$ for some $gin G$ and $jin H$. This is an equivalence relation and the number of irreducible $KG$-modules is the number of $K$-conjugacy classes.
In characteristic $p$ you do the analogous thing but only look at this equivalence relation on $p$-regular elements.
answered 19 mins ago
Benjamin Steinberg
22.5k263122
answered 19 mins ago
Benjamin Steinberg
22.5k263122
answered 19 mins ago
Benjamin Steinberg
22.5k263122
answered 19 mins ago
Benjamin Steinberg
22.5k263122
22.5k263122
add a comment |Â
add a comment |Â
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