Prove that $G_2=B_1,B_2,…,B_p$ is a multiplicative group isomorphic with $G_1=A_1,A_2,…,A_p$How to compute in $BbbZ_6 rtimes BbbZ_2$?Prove that $f$ is a group isomorphism, where $f$ interchanges the primes.The central product of two cyclic subgroups of prime power order for one $p$ is isomorphic to direct product of two cyclic groups of prime power orderIs this a feasible way to show that a group $G$ is isomorphic to its opposite group $G^op$?Are these subgroups?Where is the error in this proof? (possibly bad surjection assumption)$mathbbC^3$ as a pointwise alegbra has only one simple module: true or false?Isomorphic subgroups of $S_n$ and bijection of setsOn a certain bijection of the multiplicative group modulo prime $p=2^k+1$.Problem : Normal form of product of matrices over a PID
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Prove that $G_2=B_1,B_2,…,B_p$ is a multiplicative group isomorphic with $G_1=A_1,A_2,…,A_p$
How to compute in $BbbZ_6 rtimes BbbZ_2$?Prove that $f$ is a group isomorphism, where $f$ interchanges the primes.The central product of two cyclic subgroups of prime power order for one $p$ is isomorphic to direct product of two cyclic groups of prime power orderIs this a feasible way to show that a group $G$ is isomorphic to its opposite group $G^op$?Are these subgroups?Where is the error in this proof? (possibly bad surjection assumption)$mathbbC^3$ as a pointwise alegbra has only one simple module: true or false?Isomorphic subgroups of $S_n$ and bijection of setsOn a certain bijection of the multiplicative group modulo prime $p=2^k+1$.Problem : Normal form of product of matrices over a PID
$begingroup$
Let $m,n,pgeq3$ be natural numbers and $G_1=A_1,A_2,...,A_p$ a multiplicative group of order $p$ with elements from $M_2(mathbbZ)$. For every $A_i$ from $G$ we attach $B_i$ such that if $A=(a_ij), i,j=1,n$, then $B=(a_ij) pmod m$. Prove that the set $G_2=B_1,B_2,...,B_p$ is a multiplicative group isomorphic with $G_1$, where the multiplication is that of $M_n(mathbbZ_m)$.
I think $f:G_1rightarrow G_2$, $f(A_i)=B_i$ is an isomorphism and thus the problem can be easily solved, but I am unsure it's so basic. Any help to finish it?
abstract-algebra group-theory finite-groups group-isomorphism
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asked 2 days ago
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add a comment |
$begingroup$
Let $m,n,pgeq3$ be natural numbers and $G_1=A_1,A_2,...,A_p$ a multiplicative group of order $p$ with elements from $M_2(mathbbZ)$. For every $A_i$ from $G$ we attach $B_i$ such that if $A=(a_ij), i,j=1,n$, then $B=(a_ij) pmod m$. Prove that the set $G_2=B_1,B_2,...,B_p$ is a multiplicative group isomorphic with $G_1$, where the multiplication is that of $M_n(mathbbZ_m)$.
I think $f:G_1rightarrow G_2$, $f(A_i)=B_i$ is an isomorphism and thus the problem can be easily solved, but I am unsure it's so basic. Any help to finish it?
abstract-algebra group-theory finite-groups group-isomorphism
edited yesterday
user26857
39.4k124183
asked 2 days ago
user651754user651754
454
New contributor
user651754 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let $m,n,pgeq3$ be natural numbers and $G_1=A_1,A_2,...,A_p$ a multiplicative group of order $p$ with elements from $M_2(mathbbZ)$. For every $A_i$ from $G$ we attach $B_i$ such that if $A=(a_ij), i,j=1,n$, then $B=(a_ij) pmod m$. Prove that the set $G_2=B_1,B_2,...,B_p$ is a multiplicative group isomorphic with $G_1$, where the multiplication is that of $M_n(mathbbZ_m)$.
I think $f:G_1rightarrow G_2$, $f(A_i)=B_i$ is an isomorphism and thus the problem can be easily solved, but I am unsure it's so basic. Any help to finish it?
abstract-algebra group-theory finite-groups group-isomorphism
edited yesterday
user26857
39.4k124183
asked 2 days ago
user651754user651754
454
New contributor
user651754 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let $m,n,pgeq3$ be natural numbers and $G_1=A_1,A_2,...,A_p$ a multiplicative group of order $p$ with elements from $M_2(mathbbZ)$. For every $A_i$ from $G$ we attach $B_i$ such that if $A=(a_ij), i,j=1,n$, then $B=(a_ij) pmod m$. Prove that the set $G_2=B_1,B_2,...,B_p$ is a multiplicative group isomorphic with $G_1$, where the multiplication is that of $M_n(mathbbZ_m)$.
I think $f:G_1rightarrow G_2$, $f(A_i)=B_i$ is an isomorphism and thus the problem can be easily solved, but I am unsure it's so basic. Any help to finish it?
abstract-algebra group-theory finite-groups group-isomorphism
abstract-algebra group-theory finite-groups group-isomorphism
edited yesterday
user26857
39.4k124183
asked 2 days ago
user651754user651754
454
New contributor
user651754 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
user26857
39.4k124183
asked 2 days ago
user651754user651754
454
New contributor
user651754 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
user26857
39.4k124183
edited yesterday
user26857
39.4k124183
edited yesterday
user26857
39.4k124183
39.4k124183
asked 2 days ago
user651754user651754
454
New contributor
user651754 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
user651754user651754
454
asked 2 days ago
user651754user651754
454
454
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user651754 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user651754 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1 Answer
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$begingroup$
This is not completely trivial. The map $f$ is certainly a homomorphism, but to show that it is an isomorphism, you need to show that its kernel is trivial.
This follows from the fact that the kernel of the reduction modulo $m$ map $rm SL(n,mathbb Z) to rm SL(n,mathbb Z/mmathbb Z)$ is torsion-free for $m ge 3$. You can find a proof of this in Lemma 2.6 here, for example.
answered 2 days ago
Derek HoltDerek Holt
54.2k53571
$endgroup$
$begingroup$
Thanks! I am not familiar with the notion of torsion, so is there an easier way to show that the map $f$ is an isomorphism?
$endgroup$
– user651754
2 days ago
$begingroup$
A torsion-free group is one in which every non-identity element has infinite order. I am afraid that there is no easier way to do this, because if the kernel of the reduction mod $m$ had elements of finite order, then the result that you are trying to prove would not be true.
$endgroup$
– Derek Holt
2 days ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
This is not completely trivial. The map $f$ is certainly a homomorphism, but to show that it is an isomorphism, you need to show that its kernel is trivial.
This follows from the fact that the kernel of the reduction modulo $m$ map $rm SL(n,mathbb Z) to rm SL(n,mathbb Z/mmathbb Z)$ is torsion-free for $m ge 3$. You can find a proof of this in Lemma 2.6 here, for example.
answered 2 days ago
Derek HoltDerek Holt
54.2k53571
$endgroup$
$begingroup$
Thanks! I am not familiar with the notion of torsion, so is there an easier way to show that the map $f$ is an isomorphism?
$endgroup$
– user651754
2 days ago
$begingroup$
A torsion-free group is one in which every non-identity element has infinite order. I am afraid that there is no easier way to do this, because if the kernel of the reduction mod $m$ had elements of finite order, then the result that you are trying to prove would not be true.
$endgroup$
– Derek Holt
2 days ago
add a comment |
$begingroup$
This is not completely trivial. The map $f$ is certainly a homomorphism, but to show that it is an isomorphism, you need to show that its kernel is trivial.
This follows from the fact that the kernel of the reduction modulo $m$ map $rm SL(n,mathbb Z) to rm SL(n,mathbb Z/mmathbb Z)$ is torsion-free for $m ge 3$. You can find a proof of this in Lemma 2.6 here, for example.
answered 2 days ago
Derek HoltDerek Holt
54.2k53571
$endgroup$
$begingroup$
Thanks! I am not familiar with the notion of torsion, so is there an easier way to show that the map $f$ is an isomorphism?
$endgroup$
– user651754
2 days ago
$begingroup$
A torsion-free group is one in which every non-identity element has infinite order. I am afraid that there is no easier way to do this, because if the kernel of the reduction mod $m$ had elements of finite order, then the result that you are trying to prove would not be true.
$endgroup$
– Derek Holt
2 days ago
add a comment |
$begingroup$
This is not completely trivial. The map $f$ is certainly a homomorphism, but to show that it is an isomorphism, you need to show that its kernel is trivial.
This follows from the fact that the kernel of the reduction modulo $m$ map $rm SL(n,mathbb Z) to rm SL(n,mathbb Z/mmathbb Z)$ is torsion-free for $m ge 3$. You can find a proof of this in Lemma 2.6 here, for example.
answered 2 days ago
Derek HoltDerek Holt
54.2k53571
$endgroup$
This is not completely trivial. The map $f$ is certainly a homomorphism, but to show that it is an isomorphism, you need to show that its kernel is trivial.
This follows from the fact that the kernel of the reduction modulo $m$ map $rm SL(n,mathbb Z) to rm SL(n,mathbb Z/mmathbb Z)$ is torsion-free for $m ge 3$. You can find a proof of this in Lemma 2.6 here, for example.
answered 2 days ago
Derek HoltDerek Holt
54.2k53571
answered 2 days ago
Derek HoltDerek Holt
54.2k53571
answered 2 days ago
Derek HoltDerek Holt
54.2k53571
answered 2 days ago
Derek HoltDerek Holt
54.2k53571
54.2k53571
$begingroup$
Thanks! I am not familiar with the notion of torsion, so is there an easier way to show that the map $f$ is an isomorphism?
$endgroup$
– user651754
2 days ago
$begingroup$
A torsion-free group is one in which every non-identity element has infinite order. I am afraid that there is no easier way to do this, because if the kernel of the reduction mod $m$ had elements of finite order, then the result that you are trying to prove would not be true.
$endgroup$
– Derek Holt
2 days ago
add a comment |
$begingroup$
Thanks! I am not familiar with the notion of torsion, so is there an easier way to show that the map $f$ is an isomorphism?
$endgroup$
– user651754
2 days ago
$begingroup$
A torsion-free group is one in which every non-identity element has infinite order. I am afraid that there is no easier way to do this, because if the kernel of the reduction mod $m$ had elements of finite order, then the result that you are trying to prove would not be true.
$endgroup$
– Derek Holt
2 days ago
$begingroup$
Thanks! I am not familiar with the notion of torsion, so is there an easier way to show that the map $f$ is an isomorphism?
$endgroup$
– user651754
2 days ago
$begingroup$
Thanks! I am not familiar with the notion of torsion, so is there an easier way to show that the map $f$ is an isomorphism?
$endgroup$
– user651754
2 days ago
$begingroup$
A torsion-free group is one in which every non-identity element has infinite order. I am afraid that there is no easier way to do this, because if the kernel of the reduction mod $m$ had elements of finite order, then the result that you are trying to prove would not be true.
$endgroup$
– Derek Holt
2 days ago
$begingroup$
A torsion-free group is one in which every non-identity element has infinite order. I am afraid that there is no easier way to do this, because if the kernel of the reduction mod $m$ had elements of finite order, then the result that you are trying to prove would not be true.
$endgroup$
– Derek Holt
2 days ago
add a comment |
user651754 is a new contributor. Be nice, and check out our Code of Conduct.
user651754 is a new contributor. Be nice, and check out our Code of Conduct.
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