Define an operation on the Cartesian product $G = G_1 × G_2$ by $(x_1, y_1) * (x_2, y_2) = (x_1x_2,...
$begingroup$
Let $G_1 = {1, −1, i, −i}$ and $G_2 = {1, omega, omega^2}$, where $i = sqrt{−1}$ and $omega$ is a
complex cube root of $1$. Define an operation on the Cartesian product
$G = G_1 times G_2$ by
$$(x_1, y_1) * (x_2, y_2) = (x_1x_2, y_1y_2).$$
Then choose the correct option:
(A) $(G,* )$ is not a group,
(B) $ (G, *) $ is a group but not cyclic,
(C) $(G, *)$ is a group but not commutative,
(D) $(G, *)$ is a commutative cyclic group.
My attempt : I thinks option (A) is correct because inverse properties will not hold
Any hints/solution will be appreciated.
Thank you.
abstract-algebra group-theory
edited Jan 3 at 9:39
Shaun
8,951113682
asked Jan 3 at 7:21
jasminejasmine
1,718417
$endgroup$
closed as off-topic by Saad, Paul Frost, user91500, ancientmathematician, mrtaurho Jan 3 at 16:55
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Paul Frost, user91500, ancientmathematician, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 2 more comments
$begingroup$
Let $G_1 = {1, −1, i, −i}$ and $G_2 = {1, omega, omega^2}$, where $i = sqrt{−1}$ and $omega$ is a
complex cube root of $1$. Define an operation on the Cartesian product
$G = G_1 times G_2$ by
$$(x_1, y_1) * (x_2, y_2) = (x_1x_2, y_1y_2).$$
Then choose the correct option:
(A) $(G,* )$ is not a group,
(B) $ (G, *) $ is a group but not cyclic,
(C) $(G, *)$ is a group but not commutative,
(D) $(G, *)$ is a commutative cyclic group.
My attempt : I thinks option (A) is correct because inverse properties will not hold
Any hints/solution will be appreciated.
Thank you.
abstract-algebra group-theory
edited Jan 3 at 9:39
Shaun
8,951113682
asked Jan 3 at 7:21
jasminejasmine
1,718417
$endgroup$
closed as off-topic by Saad, Paul Frost, user91500, ancientmathematician, mrtaurho Jan 3 at 16:55
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Paul Frost, user91500, ancientmathematician, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
@Arthur ya inverse will not exist
$endgroup$
– jasmine
Jan 3 at 7:24
1
$begingroup$
Given $(x,y)in G$, is $(x^{-1},y^{-1})$ in $G$? What is $(x,y)*(x^{-1},y^{-1})$? Are you still claiming that we don't have inverses?
$endgroup$
– Arthur
Jan 3 at 7:27
1
$begingroup$
ya $frac{1}{0 }$ is not exist @Arthur
$endgroup$
– jasmine
Jan 3 at 7:29
2
$begingroup$
You can't invert $0$. Sure. Is that an issue? Is $0$ in $G$?
$endgroup$
– Arthur
Jan 3 at 7:30
2
$begingroup$
okss @Arthur..$0$ is not in G ..
$endgroup$
– jasmine
Jan 3 at 7:33
|
show 2 more comments
$begingroup$
Let $G_1 = {1, −1, i, −i}$ and $G_2 = {1, omega, omega^2}$, where $i = sqrt{−1}$ and $omega$ is a
complex cube root of $1$. Define an operation on the Cartesian product
$G = G_1 times G_2$ by
$$(x_1, y_1) * (x_2, y_2) = (x_1x_2, y_1y_2).$$
Then choose the correct option:
(A) $(G,* )$ is not a group,
(B) $ (G, *) $ is a group but not cyclic,
(C) $(G, *)$ is a group but not commutative,
(D) $(G, *)$ is a commutative cyclic group.
My attempt : I thinks option (A) is correct because inverse properties will not hold
Any hints/solution will be appreciated.
Thank you.
abstract-algebra group-theory
edited Jan 3 at 9:39
Shaun
8,951113682
asked Jan 3 at 7:21
jasminejasmine
1,718417
$endgroup$
Let $G_1 = {1, −1, i, −i}$ and $G_2 = {1, omega, omega^2}$, where $i = sqrt{−1}$ and $omega$ is a
complex cube root of $1$. Define an operation on the Cartesian product
$G = G_1 times G_2$ by
$$(x_1, y_1) * (x_2, y_2) = (x_1x_2, y_1y_2).$$
Then choose the correct option:
(A) $(G,* )$ is not a group,
(B) $ (G, *) $ is a group but not cyclic,
(C) $(G, *)$ is a group but not commutative,
(D) $(G, *)$ is a commutative cyclic group.
My attempt : I thinks option (A) is correct because inverse properties will not hold
Any hints/solution will be appreciated.
Thank you.
abstract-algebra group-theory
abstract-algebra group-theory
edited Jan 3 at 9:39
Shaun
8,951113682
asked Jan 3 at 7:21
jasminejasmine
1,718417
edited Jan 3 at 9:39
Shaun
8,951113682
asked Jan 3 at 7:21
jasminejasmine
1,718417
edited Jan 3 at 9:39
Shaun
8,951113682
edited Jan 3 at 9:39
Shaun
8,951113682
edited Jan 3 at 9:39
Shaun
8,951113682
8,951113682
asked Jan 3 at 7:21
jasminejasmine
1,718417
asked Jan 3 at 7:21
jasminejasmine
1,718417
asked Jan 3 at 7:21
jasminejasmine
1,718417
1,718417
closed as off-topic by Saad, Paul Frost, user91500, ancientmathematician, mrtaurho Jan 3 at 16:55
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Paul Frost, user91500, ancientmathematician, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Paul Frost, user91500, ancientmathematician, mrtaurho Jan 3 at 16:55
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Paul Frost, user91500, ancientmathematician, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
@Arthur ya inverse will not exist
$endgroup$
– jasmine
Jan 3 at 7:24
1
$begingroup$
Given $(x,y)in G$, is $(x^{-1},y^{-1})$ in $G$? What is $(x,y)*(x^{-1},y^{-1})$? Are you still claiming that we don't have inverses?
$endgroup$
– Arthur
Jan 3 at 7:27
1
$begingroup$
ya $frac{1}{0 }$ is not exist @Arthur
$endgroup$
– jasmine
Jan 3 at 7:29
2
$begingroup$
You can't invert $0$. Sure. Is that an issue? Is $0$ in $G$?
$endgroup$
– Arthur
Jan 3 at 7:30
2
$begingroup$
okss @Arthur..$0$ is not in G ..
$endgroup$
– jasmine
Jan 3 at 7:33
|
show 2 more comments
1
$begingroup$
@Arthur ya inverse will not exist
$endgroup$
– jasmine
Jan 3 at 7:24
1
$begingroup$
Given $(x,y)in G$, is $(x^{-1},y^{-1})$ in $G$? What is $(x,y)*(x^{-1},y^{-1})$? Are you still claiming that we don't have inverses?
$endgroup$
– Arthur
Jan 3 at 7:27
1
$begingroup$
ya $frac{1}{0 }$ is not exist @Arthur
$endgroup$
– jasmine
Jan 3 at 7:29
2
$begingroup$
You can't invert $0$. Sure. Is that an issue? Is $0$ in $G$?
$endgroup$
– Arthur
Jan 3 at 7:30
2
$begingroup$
okss @Arthur..$0$ is not in G ..
$endgroup$
– jasmine
Jan 3 at 7:33
1
1
$begingroup$
@Arthur ya inverse will not exist
$endgroup$
– jasmine
Jan 3 at 7:24
$begingroup$
@Arthur ya inverse will not exist
$endgroup$
– jasmine
Jan 3 at 7:24
1
1
$begingroup$
Given $(x,y)in G$, is $(x^{-1},y^{-1})$ in $G$? What is $(x,y)*(x^{-1},y^{-1})$? Are you still claiming that we don't have inverses?
$endgroup$
– Arthur
Jan 3 at 7:27
$begingroup$
Given $(x,y)in G$, is $(x^{-1},y^{-1})$ in $G$? What is $(x,y)*(x^{-1},y^{-1})$? Are you still claiming that we don't have inverses?
$endgroup$
– Arthur
Jan 3 at 7:27
1
1
$begingroup$
ya $frac{1}{0 }$ is not exist @Arthur
$endgroup$
– jasmine
Jan 3 at 7:29
$begingroup$
ya $frac{1}{0 }$ is not exist @Arthur
$endgroup$
– jasmine
Jan 3 at 7:29
2
2
$begingroup$
You can't invert $0$. Sure. Is that an issue? Is $0$ in $G$?
$endgroup$
– Arthur
Jan 3 at 7:30
$begingroup$
You can't invert $0$. Sure. Is that an issue? Is $0$ in $G$?
$endgroup$
– Arthur
Jan 3 at 7:30
2
2
$begingroup$
okss @Arthur..$0$ is not in G ..
$endgroup$
– jasmine
Jan 3 at 7:33
$begingroup$
okss @Arthur..$0$ is not in G ..
$endgroup$
– jasmine
Jan 3 at 7:33
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Hint: We have $i^4=1$ and $omega^3=1$.
Hint 2: Every element of $G$ is a product of $(1, omega)$ and $(i, 1)$ with respect to $ast$. Do these particular elements commute? Can you write one as a power of the other?
Hover over or click the box below for the solution.
The answer is (B) as $(G, ast)$ is the direct product of $G_1$ and $G_2$.
answered Jan 3 at 9:11
ShaunShaun
8,951113682
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: We have $i^4=1$ and $omega^3=1$.
Hint 2: Every element of $G$ is a product of $(1, omega)$ and $(i, 1)$ with respect to $ast$. Do these particular elements commute? Can you write one as a power of the other?
Hover over or click the box below for the solution.
The answer is (B) as $(G, ast)$ is the direct product of $G_1$ and $G_2$.
answered Jan 3 at 9:11
ShaunShaun
8,951113682
$endgroup$
add a comment |
$begingroup$
Hint: We have $i^4=1$ and $omega^3=1$.
Hint 2: Every element of $G$ is a product of $(1, omega)$ and $(i, 1)$ with respect to $ast$. Do these particular elements commute? Can you write one as a power of the other?
Hover over or click the box below for the solution.
The answer is (B) as $(G, ast)$ is the direct product of $G_1$ and $G_2$.
answered Jan 3 at 9:11
ShaunShaun
8,951113682
$endgroup$
add a comment |
$begingroup$
Hint: We have $i^4=1$ and $omega^3=1$.
Hint 2: Every element of $G$ is a product of $(1, omega)$ and $(i, 1)$ with respect to $ast$. Do these particular elements commute? Can you write one as a power of the other?
Hover over or click the box below for the solution.
The answer is (B) as $(G, ast)$ is the direct product of $G_1$ and $G_2$.
answered Jan 3 at 9:11
ShaunShaun
8,951113682
$endgroup$
Hint: We have $i^4=1$ and $omega^3=1$.
Hint 2: Every element of $G$ is a product of $(1, omega)$ and $(i, 1)$ with respect to $ast$. Do these particular elements commute? Can you write one as a power of the other?
Hover over or click the box below for the solution.
The answer is (B) as $(G, ast)$ is the direct product of $G_1$ and $G_2$.
answered Jan 3 at 9:11
ShaunShaun
8,951113682
edited Jan 3 at 9:41
answered Jan 3 at 9:11
ShaunShaun
8,951113682
answered Jan 3 at 9:11
ShaunShaun
8,951113682
answered Jan 3 at 9:11
ShaunShaun
8,951113682
8,951113682
add a comment |
add a comment |
1
$begingroup$
@Arthur ya inverse will not exist
$endgroup$
– jasmine
Jan 3 at 7:24
1
$begingroup$
Given $(x,y)in G$, is $(x^{-1},y^{-1})$ in $G$? What is $(x,y)*(x^{-1},y^{-1})$? Are you still claiming that we don't have inverses?
$endgroup$
– Arthur
Jan 3 at 7:27
1
$begingroup$
ya $frac{1}{0 }$ is not exist @Arthur
$endgroup$
– jasmine
Jan 3 at 7:29
2
$begingroup$
You can't invert $0$. Sure. Is that an issue? Is $0$ in $G$?
$endgroup$
– Arthur
Jan 3 at 7:30
2
$begingroup$
okss @Arthur..$0$ is not in G ..
$endgroup$
– jasmine
Jan 3 at 7:33