Mixing
How to compute group direct product between group of units
$begingroup$
Let $U(N)$ is a group under multiplication modulo $N$ called
the group of units of $mathbb{Z}/N$.
$U(N) = {a in Z/N: (a,N) = 1 } $, a and N are relative prime.
$U(3) = {1,2}$
$U(5) = {1,2,3,4}$
$U(15) = {1,2,4,7,8,11,13,14}$
I did not understand the group direct product between $U(3)$ and $U(5)$.
It is claimed (in a book) that $U(15) = U(3) times U(5)$.
How the group direct product is calculated?
abstract-algebra group-theory finite-groups
edited Jan 14 at 11:43
Cosmin
1,4441527
asked Jan 14 at 11:20
VinodVinod
95941630
$endgroup$
add a comment |
$begingroup$
Let $U(N)$ is a group under multiplication modulo $N$ called
the group of units of $mathbb{Z}/N$.
$U(N) = {a in Z/N: (a,N) = 1 } $, a and N are relative prime.
$U(3) = {1,2}$
$U(5) = {1,2,3,4}$
$U(15) = {1,2,4,7,8,11,13,14}$
I did not understand the group direct product between $U(3)$ and $U(5)$.
It is claimed (in a book) that $U(15) = U(3) times U(5)$.
How the group direct product is calculated?
abstract-algebra group-theory finite-groups
edited Jan 14 at 11:43
Cosmin
1,4441527
asked Jan 14 at 11:20
VinodVinod
95941630
$endgroup$
add a comment |
$begingroup$
Let $U(N)$ is a group under multiplication modulo $N$ called
the group of units of $mathbb{Z}/N$.
$U(N) = {a in Z/N: (a,N) = 1 } $, a and N are relative prime.
$U(3) = {1,2}$
$U(5) = {1,2,3,4}$
$U(15) = {1,2,4,7,8,11,13,14}$
I did not understand the group direct product between $U(3)$ and $U(5)$.
It is claimed (in a book) that $U(15) = U(3) times U(5)$.
How the group direct product is calculated?
abstract-algebra group-theory finite-groups
edited Jan 14 at 11:43
Cosmin
1,4441527
asked Jan 14 at 11:20
VinodVinod
95941630
$endgroup$
Let $U(N)$ is a group under multiplication modulo $N$ called
the group of units of $mathbb{Z}/N$.
$U(N) = {a in Z/N: (a,N) = 1 } $, a and N are relative prime.
$U(3) = {1,2}$
$U(5) = {1,2,3,4}$
$U(15) = {1,2,4,7,8,11,13,14}$
I did not understand the group direct product between $U(3)$ and $U(5)$.
It is claimed (in a book) that $U(15) = U(3) times U(5)$.
How the group direct product is calculated?
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
edited Jan 14 at 11:43
Cosmin
1,4441527
asked Jan 14 at 11:20
VinodVinod
95941630
edited Jan 14 at 11:43
Cosmin
1,4441527
asked Jan 14 at 11:20
VinodVinod
95941630
edited Jan 14 at 11:43
Cosmin
1,4441527
edited Jan 14 at 11:43
Cosmin
1,4441527
edited Jan 14 at 11:43
Cosmin
1,4441527
1,4441527
asked Jan 14 at 11:20
VinodVinod
95941630
asked Jan 14 at 11:20
VinodVinod
95941630
asked Jan 14 at 11:20
VinodVinod
95941630
95941630
add a comment |
add a comment |
1 Answer
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$begingroup$
For sure your book says that $$U(mathbb{Z}/ 15mathbb{Z}) simeq U(mathbb{Z}/ 3mathbb{Z}) times U(mathbb{Z}/ 5mathbb{Z}), $$ where $simeq$ means "isomorphic", because they are not equal.
That is, you can find a function $f:U(mathbb{Z}/ 15mathbb{Z}) to U(mathbb{Z}/ 3mathbb{Z}) times U(mathbb{Z}/ 5mathbb{Z})$ such that $f$ is bijective and $f$ is a homomorphism i.e. $f(xy) = f(x)f(y), forall x ,yin U(mathbb{Z}/ 15mathbb{Z})$ (where the LHS multiplication is in $U(mathbb{Z}/ 15mathbb{Z})$ and the RHS multiplication is in $ U(mathbb{Z}/ 3mathbb{Z}) times U(mathbb{Z}/ 5mathbb{Z}))$. Here, a function that satisfies the conditions mentioned is $f(x pmod {15}) = (x pmod 3, x pmod 5).$
The direct product of two groups $(G,*)$ and $(H, @)$ is just the cartesian product $G times H$ with the operation $cdot: (G times H) times (G times H) to G times H$, that makes it a group, defined as following: $$(g_1,h_1) cdot (g_2,h_2) = (g_1 * g_2, h_1 @ h_2), forall g_1,g_2 in G text{ and } forall h_1,h_2 in H, $$ I'll leave you to check the group axioms.
answered Jan 14 at 11:41
CosminCosmin
1,4441527
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add a comment |
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1 Answer
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$begingroup$
For sure your book says that $$U(mathbb{Z}/ 15mathbb{Z}) simeq U(mathbb{Z}/ 3mathbb{Z}) times U(mathbb{Z}/ 5mathbb{Z}), $$ where $simeq$ means "isomorphic", because they are not equal.
That is, you can find a function $f:U(mathbb{Z}/ 15mathbb{Z}) to U(mathbb{Z}/ 3mathbb{Z}) times U(mathbb{Z}/ 5mathbb{Z})$ such that $f$ is bijective and $f$ is a homomorphism i.e. $f(xy) = f(x)f(y), forall x ,yin U(mathbb{Z}/ 15mathbb{Z})$ (where the LHS multiplication is in $U(mathbb{Z}/ 15mathbb{Z})$ and the RHS multiplication is in $ U(mathbb{Z}/ 3mathbb{Z}) times U(mathbb{Z}/ 5mathbb{Z}))$. Here, a function that satisfies the conditions mentioned is $f(x pmod {15}) = (x pmod 3, x pmod 5).$
The direct product of two groups $(G,*)$ and $(H, @)$ is just the cartesian product $G times H$ with the operation $cdot: (G times H) times (G times H) to G times H$, that makes it a group, defined as following: $$(g_1,h_1) cdot (g_2,h_2) = (g_1 * g_2, h_1 @ h_2), forall g_1,g_2 in G text{ and } forall h_1,h_2 in H, $$ I'll leave you to check the group axioms.
answered Jan 14 at 11:41
CosminCosmin
1,4441527
$endgroup$
add a comment |
$begingroup$
For sure your book says that $$U(mathbb{Z}/ 15mathbb{Z}) simeq U(mathbb{Z}/ 3mathbb{Z}) times U(mathbb{Z}/ 5mathbb{Z}), $$ where $simeq$ means "isomorphic", because they are not equal.
That is, you can find a function $f:U(mathbb{Z}/ 15mathbb{Z}) to U(mathbb{Z}/ 3mathbb{Z}) times U(mathbb{Z}/ 5mathbb{Z})$ such that $f$ is bijective and $f$ is a homomorphism i.e. $f(xy) = f(x)f(y), forall x ,yin U(mathbb{Z}/ 15mathbb{Z})$ (where the LHS multiplication is in $U(mathbb{Z}/ 15mathbb{Z})$ and the RHS multiplication is in $ U(mathbb{Z}/ 3mathbb{Z}) times U(mathbb{Z}/ 5mathbb{Z}))$. Here, a function that satisfies the conditions mentioned is $f(x pmod {15}) = (x pmod 3, x pmod 5).$
The direct product of two groups $(G,*)$ and $(H, @)$ is just the cartesian product $G times H$ with the operation $cdot: (G times H) times (G times H) to G times H$, that makes it a group, defined as following: $$(g_1,h_1) cdot (g_2,h_2) = (g_1 * g_2, h_1 @ h_2), forall g_1,g_2 in G text{ and } forall h_1,h_2 in H, $$ I'll leave you to check the group axioms.
answered Jan 14 at 11:41
CosminCosmin
1,4441527
$endgroup$
add a comment |
$begingroup$
For sure your book says that $$U(mathbb{Z}/ 15mathbb{Z}) simeq U(mathbb{Z}/ 3mathbb{Z}) times U(mathbb{Z}/ 5mathbb{Z}), $$ where $simeq$ means "isomorphic", because they are not equal.
That is, you can find a function $f:U(mathbb{Z}/ 15mathbb{Z}) to U(mathbb{Z}/ 3mathbb{Z}) times U(mathbb{Z}/ 5mathbb{Z})$ such that $f$ is bijective and $f$ is a homomorphism i.e. $f(xy) = f(x)f(y), forall x ,yin U(mathbb{Z}/ 15mathbb{Z})$ (where the LHS multiplication is in $U(mathbb{Z}/ 15mathbb{Z})$ and the RHS multiplication is in $ U(mathbb{Z}/ 3mathbb{Z}) times U(mathbb{Z}/ 5mathbb{Z}))$. Here, a function that satisfies the conditions mentioned is $f(x pmod {15}) = (x pmod 3, x pmod 5).$
The direct product of two groups $(G,*)$ and $(H, @)$ is just the cartesian product $G times H$ with the operation $cdot: (G times H) times (G times H) to G times H$, that makes it a group, defined as following: $$(g_1,h_1) cdot (g_2,h_2) = (g_1 * g_2, h_1 @ h_2), forall g_1,g_2 in G text{ and } forall h_1,h_2 in H, $$ I'll leave you to check the group axioms.
answered Jan 14 at 11:41
CosminCosmin
1,4441527
$endgroup$
For sure your book says that $$U(mathbb{Z}/ 15mathbb{Z}) simeq U(mathbb{Z}/ 3mathbb{Z}) times U(mathbb{Z}/ 5mathbb{Z}), $$ where $simeq$ means "isomorphic", because they are not equal.
That is, you can find a function $f:U(mathbb{Z}/ 15mathbb{Z}) to U(mathbb{Z}/ 3mathbb{Z}) times U(mathbb{Z}/ 5mathbb{Z})$ such that $f$ is bijective and $f$ is a homomorphism i.e. $f(xy) = f(x)f(y), forall x ,yin U(mathbb{Z}/ 15mathbb{Z})$ (where the LHS multiplication is in $U(mathbb{Z}/ 15mathbb{Z})$ and the RHS multiplication is in $ U(mathbb{Z}/ 3mathbb{Z}) times U(mathbb{Z}/ 5mathbb{Z}))$. Here, a function that satisfies the conditions mentioned is $f(x pmod {15}) = (x pmod 3, x pmod 5).$
The direct product of two groups $(G,*)$ and $(H, @)$ is just the cartesian product $G times H$ with the operation $cdot: (G times H) times (G times H) to G times H$, that makes it a group, defined as following: $$(g_1,h_1) cdot (g_2,h_2) = (g_1 * g_2, h_1 @ h_2), forall g_1,g_2 in G text{ and } forall h_1,h_2 in H, $$ I'll leave you to check the group axioms.
answered Jan 14 at 11:41
CosminCosmin
1,4441527
edited Feb 1 at 15:19
answered Jan 14 at 11:41
CosminCosmin
1,4441527
answered Jan 14 at 11:41
CosminCosmin
1,4441527
answered Jan 14 at 11:41
CosminCosmin
1,4441527
1,4441527
add a comment |
add a comment |
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