Mixing
Additive bijection concept. [closed]
Could someone help me understand what is meant by additive bijection ?
Especially between the sets $(Z^*_7,cdot)$ and $(Z_6,+)$
where $Z^*_7 = Z_7/{[0]}$.
Also is $(Z^*_7,cdot)$ the set of module 7 where only scalar multiplication is defined?
Thank you in advance.
linear-algebra abstract-algebra
edited Nov 21 '18 at 15:04
Andrés E. Caicedo
64.8k8158246
asked Nov 21 '18 at 14:59
Ricouello
1355
closed as unclear what you're asking by Andrés E. Caicedo, Frpzzd, amWhy, Mostafa Ayaz, Don Thousand Nov 21 '18 at 21:32
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Could someone help me understand what is meant by additive bijection ?
Especially between the sets $(Z^*_7,cdot)$ and $(Z_6,+)$
where $Z^*_7 = Z_7/{[0]}$.
Also is $(Z^*_7,cdot)$ the set of module 7 where only scalar multiplication is defined?
Thank you in advance.
linear-algebra abstract-algebra
edited Nov 21 '18 at 15:04
Andrés E. Caicedo
64.8k8158246
asked Nov 21 '18 at 14:59
Ricouello
1355
closed as unclear what you're asking by Andrés E. Caicedo, Frpzzd, amWhy, Mostafa Ayaz, Don Thousand Nov 21 '18 at 21:32
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Could someone help me understand what is meant by additive bijection ?
Especially between the sets $(Z^*_7,cdot)$ and $(Z_6,+)$
where $Z^*_7 = Z_7/{[0]}$.
Also is $(Z^*_7,cdot)$ the set of module 7 where only scalar multiplication is defined?
Thank you in advance.
linear-algebra abstract-algebra
edited Nov 21 '18 at 15:04
Andrés E. Caicedo
64.8k8158246
asked Nov 21 '18 at 14:59
Ricouello
1355
Could someone help me understand what is meant by additive bijection ?
Especially between the sets $(Z^*_7,cdot)$ and $(Z_6,+)$
where $Z^*_7 = Z_7/{[0]}$.
Also is $(Z^*_7,cdot)$ the set of module 7 where only scalar multiplication is defined?
Thank you in advance.
linear-algebra abstract-algebra
linear-algebra abstract-algebra
edited Nov 21 '18 at 15:04
Andrés E. Caicedo
64.8k8158246
asked Nov 21 '18 at 14:59
Ricouello
1355
edited Nov 21 '18 at 15:04
Andrés E. Caicedo
64.8k8158246
asked Nov 21 '18 at 14:59
Ricouello
1355
edited Nov 21 '18 at 15:04
Andrés E. Caicedo
64.8k8158246
edited Nov 21 '18 at 15:04
Andrés E. Caicedo
64.8k8158246
edited Nov 21 '18 at 15:04
Andrés E. Caicedo
64.8k8158246
64.8k8158246
asked Nov 21 '18 at 14:59
Ricouello
1355
asked Nov 21 '18 at 14:59
Ricouello
1355
asked Nov 21 '18 at 14:59
Ricouello
1355
1355
closed as unclear what you're asking by Andrés E. Caicedo, Frpzzd, amWhy, Mostafa Ayaz, Don Thousand Nov 21 '18 at 21:32
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Andrés E. Caicedo, Frpzzd, amWhy, Mostafa Ayaz, Don Thousand Nov 21 '18 at 21:32
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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$(Z^{*}_7,.)$ is a multiplicative group and $(Z_6,+)$ is an additive group. But both are cyclic group of order $6$.
$Z_7^*=<[3]_7>$ and $Z_6=<[1]_6>$.
One of the isomorphism $f:(Z^{*}_7,.)to(Z_6,+)$ is defined by $f([3]_7^n)=[n]_6$.
I think, isomorphism is meant by "additive bijection".
answered Nov 21 '18 at 15:13
Offlaw
2649
Thank you, also what do you mean by multiplicative group? And what is the reasoning that brought you to use $3^n$ out of nowhere?That's what I can't understand
– Ricouello
Nov 21 '18 at 15:17
Multiplicative group means notations related with the operation in the group looks like "natural" multiplication (Which we usually do in daily life).
– Offlaw
Nov 22 '18 at 11:37
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$(Z^{*}_7,.)$ is a multiplicative group and $(Z_6,+)$ is an additive group. But both are cyclic group of order $6$.
$Z_7^*=<[3]_7>$ and $Z_6=<[1]_6>$.
One of the isomorphism $f:(Z^{*}_7,.)to(Z_6,+)$ is defined by $f([3]_7^n)=[n]_6$.
I think, isomorphism is meant by "additive bijection".
answered Nov 21 '18 at 15:13
Offlaw
2649
Thank you, also what do you mean by multiplicative group? And what is the reasoning that brought you to use $3^n$ out of nowhere?That's what I can't understand
– Ricouello
Nov 21 '18 at 15:17
Multiplicative group means notations related with the operation in the group looks like "natural" multiplication (Which we usually do in daily life).
– Offlaw
Nov 22 '18 at 11:37
add a comment |
$(Z^{*}_7,.)$ is a multiplicative group and $(Z_6,+)$ is an additive group. But both are cyclic group of order $6$.
$Z_7^*=<[3]_7>$ and $Z_6=<[1]_6>$.
One of the isomorphism $f:(Z^{*}_7,.)to(Z_6,+)$ is defined by $f([3]_7^n)=[n]_6$.
I think, isomorphism is meant by "additive bijection".
answered Nov 21 '18 at 15:13
Offlaw
2649
Thank you, also what do you mean by multiplicative group? And what is the reasoning that brought you to use $3^n$ out of nowhere?That's what I can't understand
– Ricouello
Nov 21 '18 at 15:17
Multiplicative group means notations related with the operation in the group looks like "natural" multiplication (Which we usually do in daily life).
– Offlaw
Nov 22 '18 at 11:37
add a comment |
$(Z^{*}_7,.)$ is a multiplicative group and $(Z_6,+)$ is an additive group. But both are cyclic group of order $6$.
$Z_7^*=<[3]_7>$ and $Z_6=<[1]_6>$.
One of the isomorphism $f:(Z^{*}_7,.)to(Z_6,+)$ is defined by $f([3]_7^n)=[n]_6$.
I think, isomorphism is meant by "additive bijection".
answered Nov 21 '18 at 15:13
Offlaw
2649
$(Z^{*}_7,.)$ is a multiplicative group and $(Z_6,+)$ is an additive group. But both are cyclic group of order $6$.
$Z_7^*=<[3]_7>$ and $Z_6=<[1]_6>$.
One of the isomorphism $f:(Z^{*}_7,.)to(Z_6,+)$ is defined by $f([3]_7^n)=[n]_6$.
I think, isomorphism is meant by "additive bijection".
answered Nov 21 '18 at 15:13
Offlaw
2649
answered Nov 21 '18 at 15:13
Offlaw
2649
answered Nov 21 '18 at 15:13
Offlaw
2649
answered Nov 21 '18 at 15:13
Offlaw
2649
2649
Thank you, also what do you mean by multiplicative group? And what is the reasoning that brought you to use $3^n$ out of nowhere?That's what I can't understand
– Ricouello
Nov 21 '18 at 15:17
Multiplicative group means notations related with the operation in the group looks like "natural" multiplication (Which we usually do in daily life).
– Offlaw
Nov 22 '18 at 11:37
add a comment |
Thank you, also what do you mean by multiplicative group? And what is the reasoning that brought you to use $3^n$ out of nowhere?That's what I can't understand
– Ricouello
Nov 21 '18 at 15:17
Multiplicative group means notations related with the operation in the group looks like "natural" multiplication (Which we usually do in daily life).
– Offlaw
Nov 22 '18 at 11:37
Thank you, also what do you mean by multiplicative group? And what is the reasoning that brought you to use $3^n$ out of nowhere?That's what I can't understand
– Ricouello
Nov 21 '18 at 15:17
Thank you, also what do you mean by multiplicative group? And what is the reasoning that brought you to use $3^n$ out of nowhere?That's what I can't understand
– Ricouello
Nov 21 '18 at 15:17
Multiplicative group means notations related with the operation in the group looks like "natural" multiplication (Which we usually do in daily life).
– Offlaw
Nov 22 '18 at 11:37
Multiplicative group means notations related with the operation in the group looks like "natural" multiplication (Which we usually do in daily life).
– Offlaw
Nov 22 '18 at 11:37
add a comment |
