Isomorphism between two groups
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Prove that $langle Bbb Z/nBbb Z, +rangle$ and $langle U_{n}, {}cdot{}rangle$ are isomorphic binary structure where $U_n$ is roots of unity and $Bbb Z/nBbb Z$ is integers modulo $n$.
I know that for isomorphic binary structure, we define a function between groups and we should check homomorphism property and bijection. But I can not define a function. Please help me, if you have any good idea.
abstract-algebra
$endgroup$
|
show 1 more comment
$begingroup$
Prove that $langle Bbb Z/nBbb Z, +rangle$ and $langle U_{n}, {}cdot{}rangle$ are isomorphic binary structure where $U_n$ is roots of unity and $Bbb Z/nBbb Z$ is integers modulo $n$.
I know that for isomorphic binary structure, we define a function between groups and we should check homomorphism property and bijection. But I can not define a function. Please help me, if you have any good idea.
abstract-algebra
$endgroup$
$begingroup$
That's better. Now, if you're stuck on a general problem (like here, "show that for all $n$, something something"), it usually helps to check a few small examples and see if you can get some insight. What about $Bbb Z/Bbb Z_2$ and $U_2$? What about $Bbb Z/3Bbb Z$ and $U_3$? Can you find isomorhisms there? What about $4$ and $5$? Does that generalize in any way?
$endgroup$
– Arthur
Dec 4 '18 at 9:16
$begingroup$
You are right. Thank you. I can define a map between $Z/3Z$ and $U3$ for example : $f(3k)$ = $(3k)^i$ or can not ?
$endgroup$
– mathsstudent
Dec 4 '18 at 9:20
$begingroup$
What are the elements of $Bbb Z/3Bbb Z$, and what are the elements of $U_3$?
$endgroup$
– Arthur
Dec 4 '18 at 9:22
$begingroup$
$Z/3Z= {0, 3, 6, 9, 12,....}$ and $U3={1,t, t^{2}}$ where $t=e^{2pi(i)/3}$
$endgroup$
– mathsstudent
Dec 4 '18 at 9:27
$begingroup$
No, $0, 3, 6, 9, ldots$ is $3Bbb Z$, not $Bbb Z/3Bbb Z$. The three elements of $Bbb Z/3Bbb Z$ are ${ldots, -3, 0, 3, 6, ldots}$ and ${ldots,-2, 1, 4, 7, ldots}$ and ${ldots,-1, 2, 5, 8, ldots}$, usually called $[0], [1]$ and $[2]$.
$endgroup$
– Arthur
Dec 4 '18 at 9:29
|
show 1 more comment
$begingroup$
Prove that $langle Bbb Z/nBbb Z, +rangle$ and $langle U_{n}, {}cdot{}rangle$ are isomorphic binary structure where $U_n$ is roots of unity and $Bbb Z/nBbb Z$ is integers modulo $n$.
I know that for isomorphic binary structure, we define a function between groups and we should check homomorphism property and bijection. But I can not define a function. Please help me, if you have any good idea.
abstract-algebra
$endgroup$
Prove that $langle Bbb Z/nBbb Z, +rangle$ and $langle U_{n}, {}cdot{}rangle$ are isomorphic binary structure where $U_n$ is roots of unity and $Bbb Z/nBbb Z$ is integers modulo $n$.
I know that for isomorphic binary structure, we define a function between groups and we should check homomorphism property and bijection. But I can not define a function. Please help me, if you have any good idea.
abstract-algebra
abstract-algebra
edited Dec 4 '18 at 9:50
Chinnapparaj R
5,4942928
5,4942928
asked Dec 4 '18 at 9:07
mathsstudentmathsstudent
386
386
$begingroup$
That's better. Now, if you're stuck on a general problem (like here, "show that for all $n$, something something"), it usually helps to check a few small examples and see if you can get some insight. What about $Bbb Z/Bbb Z_2$ and $U_2$? What about $Bbb Z/3Bbb Z$ and $U_3$? Can you find isomorhisms there? What about $4$ and $5$? Does that generalize in any way?
$endgroup$
– Arthur
Dec 4 '18 at 9:16
$begingroup$
You are right. Thank you. I can define a map between $Z/3Z$ and $U3$ for example : $f(3k)$ = $(3k)^i$ or can not ?
$endgroup$
– mathsstudent
Dec 4 '18 at 9:20
$begingroup$
What are the elements of $Bbb Z/3Bbb Z$, and what are the elements of $U_3$?
$endgroup$
– Arthur
Dec 4 '18 at 9:22
$begingroup$
$Z/3Z= {0, 3, 6, 9, 12,....}$ and $U3={1,t, t^{2}}$ where $t=e^{2pi(i)/3}$
$endgroup$
– mathsstudent
Dec 4 '18 at 9:27
$begingroup$
No, $0, 3, 6, 9, ldots$ is $3Bbb Z$, not $Bbb Z/3Bbb Z$. The three elements of $Bbb Z/3Bbb Z$ are ${ldots, -3, 0, 3, 6, ldots}$ and ${ldots,-2, 1, 4, 7, ldots}$ and ${ldots,-1, 2, 5, 8, ldots}$, usually called $[0], [1]$ and $[2]$.
$endgroup$
– Arthur
Dec 4 '18 at 9:29
|
show 1 more comment
$begingroup$
That's better. Now, if you're stuck on a general problem (like here, "show that for all $n$, something something"), it usually helps to check a few small examples and see if you can get some insight. What about $Bbb Z/Bbb Z_2$ and $U_2$? What about $Bbb Z/3Bbb Z$ and $U_3$? Can you find isomorhisms there? What about $4$ and $5$? Does that generalize in any way?
$endgroup$
– Arthur
Dec 4 '18 at 9:16
$begingroup$
You are right. Thank you. I can define a map between $Z/3Z$ and $U3$ for example : $f(3k)$ = $(3k)^i$ or can not ?
$endgroup$
– mathsstudent
Dec 4 '18 at 9:20
$begingroup$
What are the elements of $Bbb Z/3Bbb Z$, and what are the elements of $U_3$?
$endgroup$
– Arthur
Dec 4 '18 at 9:22
$begingroup$
$Z/3Z= {0, 3, 6, 9, 12,....}$ and $U3={1,t, t^{2}}$ where $t=e^{2pi(i)/3}$
$endgroup$
– mathsstudent
Dec 4 '18 at 9:27
$begingroup$
No, $0, 3, 6, 9, ldots$ is $3Bbb Z$, not $Bbb Z/3Bbb Z$. The three elements of $Bbb Z/3Bbb Z$ are ${ldots, -3, 0, 3, 6, ldots}$ and ${ldots,-2, 1, 4, 7, ldots}$ and ${ldots,-1, 2, 5, 8, ldots}$, usually called $[0], [1]$ and $[2]$.
$endgroup$
– Arthur
Dec 4 '18 at 9:29
$begingroup$
That's better. Now, if you're stuck on a general problem (like here, "show that for all $n$, something something"), it usually helps to check a few small examples and see if you can get some insight. What about $Bbb Z/Bbb Z_2$ and $U_2$? What about $Bbb Z/3Bbb Z$ and $U_3$? Can you find isomorhisms there? What about $4$ and $5$? Does that generalize in any way?
$endgroup$
– Arthur
Dec 4 '18 at 9:16
$begingroup$
That's better. Now, if you're stuck on a general problem (like here, "show that for all $n$, something something"), it usually helps to check a few small examples and see if you can get some insight. What about $Bbb Z/Bbb Z_2$ and $U_2$? What about $Bbb Z/3Bbb Z$ and $U_3$? Can you find isomorhisms there? What about $4$ and $5$? Does that generalize in any way?
$endgroup$
– Arthur
Dec 4 '18 at 9:16
$begingroup$
You are right. Thank you. I can define a map between $Z/3Z$ and $U3$ for example : $f(3k)$ = $(3k)^i$ or can not ?
$endgroup$
– mathsstudent
Dec 4 '18 at 9:20
$begingroup$
You are right. Thank you. I can define a map between $Z/3Z$ and $U3$ for example : $f(3k)$ = $(3k)^i$ or can not ?
$endgroup$
– mathsstudent
Dec 4 '18 at 9:20
$begingroup$
What are the elements of $Bbb Z/3Bbb Z$, and what are the elements of $U_3$?
$endgroup$
– Arthur
Dec 4 '18 at 9:22
$begingroup$
What are the elements of $Bbb Z/3Bbb Z$, and what are the elements of $U_3$?
$endgroup$
– Arthur
Dec 4 '18 at 9:22
$begingroup$
$Z/3Z= {0, 3, 6, 9, 12,....}$ and $U3={1,t, t^{2}}$ where $t=e^{2pi(i)/3}$
$endgroup$
– mathsstudent
Dec 4 '18 at 9:27
$begingroup$
$Z/3Z= {0, 3, 6, 9, 12,....}$ and $U3={1,t, t^{2}}$ where $t=e^{2pi(i)/3}$
$endgroup$
– mathsstudent
Dec 4 '18 at 9:27
$begingroup$
No, $0, 3, 6, 9, ldots$ is $3Bbb Z$, not $Bbb Z/3Bbb Z$. The three elements of $Bbb Z/3Bbb Z$ are ${ldots, -3, 0, 3, 6, ldots}$ and ${ldots,-2, 1, 4, 7, ldots}$ and ${ldots,-1, 2, 5, 8, ldots}$, usually called $[0], [1]$ and $[2]$.
$endgroup$
– Arthur
Dec 4 '18 at 9:29
$begingroup$
No, $0, 3, 6, 9, ldots$ is $3Bbb Z$, not $Bbb Z/3Bbb Z$. The three elements of $Bbb Z/3Bbb Z$ are ${ldots, -3, 0, 3, 6, ldots}$ and ${ldots,-2, 1, 4, 7, ldots}$ and ${ldots,-1, 2, 5, 8, ldots}$, usually called $[0], [1]$ and $[2]$.
$endgroup$
– Arthur
Dec 4 '18 at 9:29
|
show 1 more comment
1 Answer
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$begingroup$
Outline:
$$U_n={1, zeta,zeta^2,cdots, zeta^{n-1}}=langle zeta rangle$$ where $zeta=e^frac{2 pi i}{n}$
For a sake of simplicity, identify $Bbb Z/ n Bbb Z$ with $Bbb Z_n$. Here $Bbb Z_n =langle 1 rangle$. Then the map $$ f:Bbb Z_n ni 1^i mapsto zeta^i in U_n$$ is an isomorphism (!). Here $1^n$ means $underbrace{1+1+cdots+1}_{n ;times}$
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1 Answer
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$begingroup$
Outline:
$$U_n={1, zeta,zeta^2,cdots, zeta^{n-1}}=langle zeta rangle$$ where $zeta=e^frac{2 pi i}{n}$
For a sake of simplicity, identify $Bbb Z/ n Bbb Z$ with $Bbb Z_n$. Here $Bbb Z_n =langle 1 rangle$. Then the map $$ f:Bbb Z_n ni 1^i mapsto zeta^i in U_n$$ is an isomorphism (!). Here $1^n$ means $underbrace{1+1+cdots+1}_{n ;times}$
$endgroup$
add a comment |
$begingroup$
Outline:
$$U_n={1, zeta,zeta^2,cdots, zeta^{n-1}}=langle zeta rangle$$ where $zeta=e^frac{2 pi i}{n}$
For a sake of simplicity, identify $Bbb Z/ n Bbb Z$ with $Bbb Z_n$. Here $Bbb Z_n =langle 1 rangle$. Then the map $$ f:Bbb Z_n ni 1^i mapsto zeta^i in U_n$$ is an isomorphism (!). Here $1^n$ means $underbrace{1+1+cdots+1}_{n ;times}$
$endgroup$
add a comment |
$begingroup$
Outline:
$$U_n={1, zeta,zeta^2,cdots, zeta^{n-1}}=langle zeta rangle$$ where $zeta=e^frac{2 pi i}{n}$
For a sake of simplicity, identify $Bbb Z/ n Bbb Z$ with $Bbb Z_n$. Here $Bbb Z_n =langle 1 rangle$. Then the map $$ f:Bbb Z_n ni 1^i mapsto zeta^i in U_n$$ is an isomorphism (!). Here $1^n$ means $underbrace{1+1+cdots+1}_{n ;times}$
$endgroup$
Outline:
$$U_n={1, zeta,zeta^2,cdots, zeta^{n-1}}=langle zeta rangle$$ where $zeta=e^frac{2 pi i}{n}$
For a sake of simplicity, identify $Bbb Z/ n Bbb Z$ with $Bbb Z_n$. Here $Bbb Z_n =langle 1 rangle$. Then the map $$ f:Bbb Z_n ni 1^i mapsto zeta^i in U_n$$ is an isomorphism (!). Here $1^n$ means $underbrace{1+1+cdots+1}_{n ;times}$
answered Dec 4 '18 at 9:25
Chinnapparaj RChinnapparaj R
5,4942928
5,4942928
add a comment |
add a comment |
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$begingroup$
That's better. Now, if you're stuck on a general problem (like here, "show that for all $n$, something something"), it usually helps to check a few small examples and see if you can get some insight. What about $Bbb Z/Bbb Z_2$ and $U_2$? What about $Bbb Z/3Bbb Z$ and $U_3$? Can you find isomorhisms there? What about $4$ and $5$? Does that generalize in any way?
$endgroup$
– Arthur
Dec 4 '18 at 9:16
$begingroup$
You are right. Thank you. I can define a map between $Z/3Z$ and $U3$ for example : $f(3k)$ = $(3k)^i$ or can not ?
$endgroup$
– mathsstudent
Dec 4 '18 at 9:20
$begingroup$
What are the elements of $Bbb Z/3Bbb Z$, and what are the elements of $U_3$?
$endgroup$
– Arthur
Dec 4 '18 at 9:22
$begingroup$
$Z/3Z= {0, 3, 6, 9, 12,....}$ and $U3={1,t, t^{2}}$ where $t=e^{2pi(i)/3}$
$endgroup$
– mathsstudent
Dec 4 '18 at 9:27
$begingroup$
No, $0, 3, 6, 9, ldots$ is $3Bbb Z$, not $Bbb Z/3Bbb Z$. The three elements of $Bbb Z/3Bbb Z$ are ${ldots, -3, 0, 3, 6, ldots}$ and ${ldots,-2, 1, 4, 7, ldots}$ and ${ldots,-1, 2, 5, 8, ldots}$, usually called $[0], [1]$ and $[2]$.
$endgroup$
– Arthur
Dec 4 '18 at 9:29