Mixing
Isomorphism of representation induced by a morphism of groups
$begingroup$
Let $G_1$,$G_2$ be groups. The question is to prove that if $$u: G_1 rightarrow G_2$$ induces an isomorphism
$$overline{u}: operatorname{Hom}(G_2,Gl(M)) rightarrow operatorname{Hom}(G_1,Gl(M))$$ for all $M$ module over $mathbb{Z}$, then $u$ is an isomorphism.
I proved that is true for $G_1$, $G_2$ abelian, but I'm in trouble proving it is true in general.
Thanks.
abstract-algebra group-theory representation-theory
edited Jan 12 at 18:14
Shaun
9,082113683
asked Jan 12 at 15:30
Matvey TizovskyMatvey Tizovsky
314
$endgroup$
add a comment |
$begingroup$
Let $G_1$,$G_2$ be groups. The question is to prove that if $$u: G_1 rightarrow G_2$$ induces an isomorphism
$$overline{u}: operatorname{Hom}(G_2,Gl(M)) rightarrow operatorname{Hom}(G_1,Gl(M))$$ for all $M$ module over $mathbb{Z}$, then $u$ is an isomorphism.
I proved that is true for $G_1$, $G_2$ abelian, but I'm in trouble proving it is true in general.
Thanks.
abstract-algebra group-theory representation-theory
edited Jan 12 at 18:14
Shaun
9,082113683
asked Jan 12 at 15:30
Matvey TizovskyMatvey Tizovsky
314
$endgroup$
add a comment |
$begingroup$
Let $G_1$,$G_2$ be groups. The question is to prove that if $$u: G_1 rightarrow G_2$$ induces an isomorphism
$$overline{u}: operatorname{Hom}(G_2,Gl(M)) rightarrow operatorname{Hom}(G_1,Gl(M))$$ for all $M$ module over $mathbb{Z}$, then $u$ is an isomorphism.
I proved that is true for $G_1$, $G_2$ abelian, but I'm in trouble proving it is true in general.
Thanks.
abstract-algebra group-theory representation-theory
edited Jan 12 at 18:14
Shaun
9,082113683
asked Jan 12 at 15:30
Matvey TizovskyMatvey Tizovsky
314
$endgroup$
Let $G_1$,$G_2$ be groups. The question is to prove that if $$u: G_1 rightarrow G_2$$ induces an isomorphism
$$overline{u}: operatorname{Hom}(G_2,Gl(M)) rightarrow operatorname{Hom}(G_1,Gl(M))$$ for all $M$ module over $mathbb{Z}$, then $u$ is an isomorphism.
I proved that is true for $G_1$, $G_2$ abelian, but I'm in trouble proving it is true in general.
Thanks.
abstract-algebra group-theory representation-theory
abstract-algebra group-theory representation-theory
edited Jan 12 at 18:14
Shaun
9,082113683
asked Jan 12 at 15:30
Matvey TizovskyMatvey Tizovsky
314
edited Jan 12 at 18:14
Shaun
9,082113683
asked Jan 12 at 15:30
Matvey TizovskyMatvey Tizovsky
314
edited Jan 12 at 18:14
Shaun
9,082113683
edited Jan 12 at 18:14
Shaun
9,082113683
edited Jan 12 at 18:14
Shaun
9,082113683
9,082113683
asked Jan 12 at 15:30
Matvey TizovskyMatvey Tizovsky
314
asked Jan 12 at 15:30
Matvey TizovskyMatvey Tizovsky
314
asked Jan 12 at 15:30
Matvey TizovskyMatvey Tizovsky
314
314
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
First, we prove that $u$ is injective. Let $M$ be the free abelian group with basis indexed by $G_1$. Then there is an injective morphism $G_1to Gl(M)$ given by the action of $G_1$ on $M$ by permutation of its basis elements. But if $u$ is not injective, then no element in the image of $bar u$ is injective. Thus $u$ has to be injective, otherwise $bar u$ would not be surjective.
Next, we prove that $u$ is surjective. Assume that it is not, and consider the set $G_2/u(G_1)$ of cosets of $u(G_1)$ in $G_2$. Let $M$ be the free abelian group with basis indexed by $G_2/u(G_1)$. The group $G_2$ acts on $G_2/u(G_1)$ by translation of cosets; therefore, there is a corresponding non-trivial morphism $phi:G_2to Gl(M)$ given by the action of $G_2$ on $M$ by permutation of its basis elements. Note that $bar u (phi)$ is the trivial morphism, since the elements of $u(G_1)$ act trivially on $M$. Thus, $bar u$ is not injective, a contradiction.
Therefore $u$ is bijective, so it is an isomorphism of groups.
answered Jan 12 at 23:24
Pierre-Guy PlamondonPierre-Guy Plamondon
8,83011639
$endgroup$
1
$begingroup$
Thanks you. Anyway I bypassed the abelian case with the fact that $Hom(G, Gl(M))$ is isomorphic to $Hom(mathbb{Z}[G], End(M))$ and so i can prove an isomorphism between $mathbb{Z}[G_1]$ and $mathbb{Z}[G_2]$ , but it is induced by $u$ and so $u$ is isomorphism.
$endgroup$
– Matvey Tizovsky
Jan 13 at 18:54
add a comment |
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1 Answer
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$begingroup$
First, we prove that $u$ is injective. Let $M$ be the free abelian group with basis indexed by $G_1$. Then there is an injective morphism $G_1to Gl(M)$ given by the action of $G_1$ on $M$ by permutation of its basis elements. But if $u$ is not injective, then no element in the image of $bar u$ is injective. Thus $u$ has to be injective, otherwise $bar u$ would not be surjective.
Next, we prove that $u$ is surjective. Assume that it is not, and consider the set $G_2/u(G_1)$ of cosets of $u(G_1)$ in $G_2$. Let $M$ be the free abelian group with basis indexed by $G_2/u(G_1)$. The group $G_2$ acts on $G_2/u(G_1)$ by translation of cosets; therefore, there is a corresponding non-trivial morphism $phi:G_2to Gl(M)$ given by the action of $G_2$ on $M$ by permutation of its basis elements. Note that $bar u (phi)$ is the trivial morphism, since the elements of $u(G_1)$ act trivially on $M$. Thus, $bar u$ is not injective, a contradiction.
Therefore $u$ is bijective, so it is an isomorphism of groups.
answered Jan 12 at 23:24
Pierre-Guy PlamondonPierre-Guy Plamondon
8,83011639
$endgroup$
1
$begingroup$
Thanks you. Anyway I bypassed the abelian case with the fact that $Hom(G, Gl(M))$ is isomorphic to $Hom(mathbb{Z}[G], End(M))$ and so i can prove an isomorphism between $mathbb{Z}[G_1]$ and $mathbb{Z}[G_2]$ , but it is induced by $u$ and so $u$ is isomorphism.
$endgroup$
– Matvey Tizovsky
Jan 13 at 18:54
add a comment |
$begingroup$
First, we prove that $u$ is injective. Let $M$ be the free abelian group with basis indexed by $G_1$. Then there is an injective morphism $G_1to Gl(M)$ given by the action of $G_1$ on $M$ by permutation of its basis elements. But if $u$ is not injective, then no element in the image of $bar u$ is injective. Thus $u$ has to be injective, otherwise $bar u$ would not be surjective.
Next, we prove that $u$ is surjective. Assume that it is not, and consider the set $G_2/u(G_1)$ of cosets of $u(G_1)$ in $G_2$. Let $M$ be the free abelian group with basis indexed by $G_2/u(G_1)$. The group $G_2$ acts on $G_2/u(G_1)$ by translation of cosets; therefore, there is a corresponding non-trivial morphism $phi:G_2to Gl(M)$ given by the action of $G_2$ on $M$ by permutation of its basis elements. Note that $bar u (phi)$ is the trivial morphism, since the elements of $u(G_1)$ act trivially on $M$. Thus, $bar u$ is not injective, a contradiction.
Therefore $u$ is bijective, so it is an isomorphism of groups.
answered Jan 12 at 23:24
Pierre-Guy PlamondonPierre-Guy Plamondon
8,83011639
$endgroup$
1
$begingroup$
Thanks you. Anyway I bypassed the abelian case with the fact that $Hom(G, Gl(M))$ is isomorphic to $Hom(mathbb{Z}[G], End(M))$ and so i can prove an isomorphism between $mathbb{Z}[G_1]$ and $mathbb{Z}[G_2]$ , but it is induced by $u$ and so $u$ is isomorphism.
$endgroup$
– Matvey Tizovsky
Jan 13 at 18:54
add a comment |
$begingroup$
First, we prove that $u$ is injective. Let $M$ be the free abelian group with basis indexed by $G_1$. Then there is an injective morphism $G_1to Gl(M)$ given by the action of $G_1$ on $M$ by permutation of its basis elements. But if $u$ is not injective, then no element in the image of $bar u$ is injective. Thus $u$ has to be injective, otherwise $bar u$ would not be surjective.
Next, we prove that $u$ is surjective. Assume that it is not, and consider the set $G_2/u(G_1)$ of cosets of $u(G_1)$ in $G_2$. Let $M$ be the free abelian group with basis indexed by $G_2/u(G_1)$. The group $G_2$ acts on $G_2/u(G_1)$ by translation of cosets; therefore, there is a corresponding non-trivial morphism $phi:G_2to Gl(M)$ given by the action of $G_2$ on $M$ by permutation of its basis elements. Note that $bar u (phi)$ is the trivial morphism, since the elements of $u(G_1)$ act trivially on $M$. Thus, $bar u$ is not injective, a contradiction.
Therefore $u$ is bijective, so it is an isomorphism of groups.
answered Jan 12 at 23:24
Pierre-Guy PlamondonPierre-Guy Plamondon
8,83011639
$endgroup$
First, we prove that $u$ is injective. Let $M$ be the free abelian group with basis indexed by $G_1$. Then there is an injective morphism $G_1to Gl(M)$ given by the action of $G_1$ on $M$ by permutation of its basis elements. But if $u$ is not injective, then no element in the image of $bar u$ is injective. Thus $u$ has to be injective, otherwise $bar u$ would not be surjective.
Next, we prove that $u$ is surjective. Assume that it is not, and consider the set $G_2/u(G_1)$ of cosets of $u(G_1)$ in $G_2$. Let $M$ be the free abelian group with basis indexed by $G_2/u(G_1)$. The group $G_2$ acts on $G_2/u(G_1)$ by translation of cosets; therefore, there is a corresponding non-trivial morphism $phi:G_2to Gl(M)$ given by the action of $G_2$ on $M$ by permutation of its basis elements. Note that $bar u (phi)$ is the trivial morphism, since the elements of $u(G_1)$ act trivially on $M$. Thus, $bar u$ is not injective, a contradiction.
Therefore $u$ is bijective, so it is an isomorphism of groups.
answered Jan 12 at 23:24
Pierre-Guy PlamondonPierre-Guy Plamondon
8,83011639
answered Jan 12 at 23:24
Pierre-Guy PlamondonPierre-Guy Plamondon
8,83011639
answered Jan 12 at 23:24
Pierre-Guy PlamondonPierre-Guy Plamondon
8,83011639
answered Jan 12 at 23:24
Pierre-Guy PlamondonPierre-Guy Plamondon
8,83011639
8,83011639
1
$begingroup$
Thanks you. Anyway I bypassed the abelian case with the fact that $Hom(G, Gl(M))$ is isomorphic to $Hom(mathbb{Z}[G], End(M))$ and so i can prove an isomorphism between $mathbb{Z}[G_1]$ and $mathbb{Z}[G_2]$ , but it is induced by $u$ and so $u$ is isomorphism.
$endgroup$
– Matvey Tizovsky
Jan 13 at 18:54
add a comment |
1
$begingroup$
Thanks you. Anyway I bypassed the abelian case with the fact that $Hom(G, Gl(M))$ is isomorphic to $Hom(mathbb{Z}[G], End(M))$ and so i can prove an isomorphism between $mathbb{Z}[G_1]$ and $mathbb{Z}[G_2]$ , but it is induced by $u$ and so $u$ is isomorphism.
$endgroup$
– Matvey Tizovsky
Jan 13 at 18:54
1
1
$begingroup$
Thanks you. Anyway I bypassed the abelian case with the fact that $Hom(G, Gl(M))$ is isomorphic to $Hom(mathbb{Z}[G], End(M))$ and so i can prove an isomorphism between $mathbb{Z}[G_1]$ and $mathbb{Z}[G_2]$ , but it is induced by $u$ and so $u$ is isomorphism.
$endgroup$
– Matvey Tizovsky
Jan 13 at 18:54
$begingroup$
Thanks you. Anyway I bypassed the abelian case with the fact that $Hom(G, Gl(M))$ is isomorphic to $Hom(mathbb{Z}[G], End(M))$ and so i can prove an isomorphism between $mathbb{Z}[G_1]$ and $mathbb{Z}[G_2]$ , but it is induced by $u$ and so $u$ is isomorphism.
$endgroup$
– Matvey Tizovsky
Jan 13 at 18:54
add a comment |
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