Mixing
Prove that $BbbZ×BbbZ$ and $BbbZ×BbbZ×BbbZ$ is not isomorphic groups. [closed]
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Considering them as rings, I can prove that they are not isomorphic as they have different nos of idempotent elements , but in case of group isomorphism I am unable to prove it.
abstract-algebra group-isomorphism
edited Jul 28 at 13:32
Sou
2,6792820
asked Jul 28 at 13:20
A. B.
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closed as off-topic by amWhy, anomaly, Dietrich Burde, Xander Henderson, Alexander Gruber♦ Jul 29 at 5:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, anomaly, Dietrich Burde, Xander Henderson, Alexander Gruber
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up vote
-3
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Considering them as rings, I can prove that they are not isomorphic as they have different nos of idempotent elements , but in case of group isomorphism I am unable to prove it.
abstract-algebra group-isomorphism
edited Jul 28 at 13:32
Sou
2,6792820
asked Jul 28 at 13:20
A. B.
1
closed as off-topic by amWhy, anomaly, Dietrich Burde, Xander Henderson, Alexander Gruber♦ Jul 29 at 5:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, anomaly, Dietrich Burde, Xander Henderson, Alexander Gruber
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 28 at 13:26
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up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
Considering them as rings, I can prove that they are not isomorphic as they have different nos of idempotent elements , but in case of group isomorphism I am unable to prove it.
abstract-algebra group-isomorphism
edited Jul 28 at 13:32
Sou
2,6792820
asked Jul 28 at 13:20
A. B.
1
Considering them as rings, I can prove that they are not isomorphic as they have different nos of idempotent elements , but in case of group isomorphism I am unable to prove it.
abstract-algebra group-isomorphism
edited Jul 28 at 13:32
Sou
2,6792820
asked Jul 28 at 13:20
A. B.
1
edited Jul 28 at 13:32
Sou
2,6792820
edited Jul 28 at 13:32
Sou
2,6792820
edited Jul 28 at 13:32
Sou
2,6792820
2,6792820
asked Jul 28 at 13:20
A. B.
1
asked Jul 28 at 13:20
A. B.
1
asked Jul 28 at 13:20
A. B.
1
1
closed as off-topic by amWhy, anomaly, Dietrich Burde, Xander Henderson, Alexander Gruber♦ Jul 29 at 5:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, anomaly, Dietrich Burde, Xander Henderson, Alexander Gruber
closed as off-topic by amWhy, anomaly, Dietrich Burde, Xander Henderson, Alexander Gruber♦ Jul 29 at 5:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, anomaly, Dietrich Burde, Xander Henderson, Alexander Gruber
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 28 at 13:26
add a comment |Â
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 28 at 13:26
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 28 at 13:26
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 28 at 13:26
add a comment |Â
3 Answers
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Hint: consider the possible cardinalities of sets of odd elements such that the sum of any two is odd. Alternatively, show that if $G_1$, $G_2$ are isomorphic abelian groups, then $G_1/2G_1$ and $G_2/2G_2$ are isomorphic.
answered Jul 28 at 13:36
tomasz
22.7k22975
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Hint:
$mathbb Ztimesmathbb Z$ can be generated by $2$ elements.
$mathbb Ztimesmathbb Ztimesmathbb Z$ cannot be generated by $2$ elements.
(Prove that for two elements $(a,b,c)$ and $(u,v,w)$ the set $n(a,b,c)+m(u,v,w)mid n,minmathbb Z$ is Always a proper subset of $mathbb Ztimesmathbb Ztimesmathbb Z$)
answered Jul 28 at 13:26
Vera
1,696313
1
Of course the second line requires proof.
– GEdgar
Jul 28 at 13:37
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That results from the Invariant Basis Number property in commutative rings: they're frr $bf Z$-modules, with a basis of $2$ and $3$ vectors respectively.
answered Jul 28 at 13:27
Bernard
110k635102
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Hint: consider the possible cardinalities of sets of odd elements such that the sum of any two is odd. Alternatively, show that if $G_1$, $G_2$ are isomorphic abelian groups, then $G_1/2G_1$ and $G_2/2G_2$ are isomorphic.
answered Jul 28 at 13:36
tomasz
22.7k22975
add a comment |Â
up vote
3
down vote
Hint: consider the possible cardinalities of sets of odd elements such that the sum of any two is odd. Alternatively, show that if $G_1$, $G_2$ are isomorphic abelian groups, then $G_1/2G_1$ and $G_2/2G_2$ are isomorphic.
answered Jul 28 at 13:36
tomasz
22.7k22975
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Hint: consider the possible cardinalities of sets of odd elements such that the sum of any two is odd. Alternatively, show that if $G_1$, $G_2$ are isomorphic abelian groups, then $G_1/2G_1$ and $G_2/2G_2$ are isomorphic.
answered Jul 28 at 13:36
tomasz
22.7k22975
Hint: consider the possible cardinalities of sets of odd elements such that the sum of any two is odd. Alternatively, show that if $G_1$, $G_2$ are isomorphic abelian groups, then $G_1/2G_1$ and $G_2/2G_2$ are isomorphic.
answered Jul 28 at 13:36
tomasz
22.7k22975
answered Jul 28 at 13:36
tomasz
22.7k22975
answered Jul 28 at 13:36
tomasz
22.7k22975
answered Jul 28 at 13:36
tomasz
22.7k22975
22.7k22975
add a comment |Â
add a comment |Â
up vote
2
down vote
Hint:
$mathbb Ztimesmathbb Z$ can be generated by $2$ elements.
$mathbb Ztimesmathbb Ztimesmathbb Z$ cannot be generated by $2$ elements.
(Prove that for two elements $(a,b,c)$ and $(u,v,w)$ the set $n(a,b,c)+m(u,v,w)mid n,minmathbb Z$ is Always a proper subset of $mathbb Ztimesmathbb Ztimesmathbb Z$)
answered Jul 28 at 13:26
Vera
1,696313
1
Of course the second line requires proof.
– GEdgar
Jul 28 at 13:37
add a comment |Â
up vote
2
down vote
Hint:
$mathbb Ztimesmathbb Z$ can be generated by $2$ elements.
$mathbb Ztimesmathbb Ztimesmathbb Z$ cannot be generated by $2$ elements.
(Prove that for two elements $(a,b,c)$ and $(u,v,w)$ the set $n(a,b,c)+m(u,v,w)mid n,minmathbb Z$ is Always a proper subset of $mathbb Ztimesmathbb Ztimesmathbb Z$)
answered Jul 28 at 13:26
Vera
1,696313
1
Of course the second line requires proof.
– GEdgar
Jul 28 at 13:37
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint:
$mathbb Ztimesmathbb Z$ can be generated by $2$ elements.
$mathbb Ztimesmathbb Ztimesmathbb Z$ cannot be generated by $2$ elements.
(Prove that for two elements $(a,b,c)$ and $(u,v,w)$ the set $n(a,b,c)+m(u,v,w)mid n,minmathbb Z$ is Always a proper subset of $mathbb Ztimesmathbb Ztimesmathbb Z$)
answered Jul 28 at 13:26
Vera
1,696313
Hint:
$mathbb Ztimesmathbb Z$ can be generated by $2$ elements.
$mathbb Ztimesmathbb Ztimesmathbb Z$ cannot be generated by $2$ elements.
(Prove that for two elements $(a,b,c)$ and $(u,v,w)$ the set $n(a,b,c)+m(u,v,w)mid n,minmathbb Z$ is Always a proper subset of $mathbb Ztimesmathbb Ztimesmathbb Z$)
answered Jul 28 at 13:26
Vera
1,696313
edited Jul 28 at 13:58
answered Jul 28 at 13:26
Vera
1,696313
answered Jul 28 at 13:26
Vera
1,696313
answered Jul 28 at 13:26
Vera
1,696313
1,696313
1
Of course the second line requires proof.
– GEdgar
Jul 28 at 13:37
add a comment |Â
1
Of course the second line requires proof.
– GEdgar
Jul 28 at 13:37
1
1
Of course the second line requires proof.
– GEdgar
Jul 28 at 13:37
Of course the second line requires proof.
– GEdgar
Jul 28 at 13:37
add a comment |Â
up vote
1
down vote
That results from the Invariant Basis Number property in commutative rings: they're frr $bf Z$-modules, with a basis of $2$ and $3$ vectors respectively.
answered Jul 28 at 13:27
Bernard
110k635102
add a comment |Â
up vote
1
down vote
That results from the Invariant Basis Number property in commutative rings: they're frr $bf Z$-modules, with a basis of $2$ and $3$ vectors respectively.
answered Jul 28 at 13:27
Bernard
110k635102
add a comment |Â
up vote
1
down vote
up vote
1
down vote
That results from the Invariant Basis Number property in commutative rings: they're frr $bf Z$-modules, with a basis of $2$ and $3$ vectors respectively.
answered Jul 28 at 13:27
Bernard
110k635102
That results from the Invariant Basis Number property in commutative rings: they're frr $bf Z$-modules, with a basis of $2$ and $3$ vectors respectively.
answered Jul 28 at 13:27
Bernard
110k635102
answered Jul 28 at 13:27
Bernard
110k635102
answered Jul 28 at 13:27
Bernard
110k635102
answered Jul 28 at 13:27
Bernard
110k635102
110k635102
add a comment |Â
add a comment |Â
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 28 at 13:26