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Proving the given two groups are isomorphic

1

$begingroup$

So I am given a group $mathbb R^3$ and a group $H$ = {$(y,0,0)|y in mathbb R$}. I have to prove that that $mathbb R^3/H$ $cong$ $mathbb R^2$. I am not sure how to even begin. My difficulty is coming up with a map between the the two sets. I have already verified that $H unlhd mathbb R^3$. So all I know is $mathbb R^3/H$ is a group. Also, from first isomorphism theorem, I know that the group is isomorphic to the image of the map $f: mathbb R^3 to A$, and I do not know what that $A$ is supposed to be. Today is the first day I learned about isomorphism, and I am very confused what is going on. Can anyone provide some help on this?

abstract-algebra group-isomorphism

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asked 59 mins ago

UfomammutUfomammut

391314

$endgroup$

add a comment | 

1

$begingroup$

So I am given a group $mathbb R^3$ and a group $H$ = {$(y,0,0)|y in mathbb R$}. I have to prove that that $mathbb R^3/H$ $cong$ $mathbb R^2$. I am not sure how to even begin. My difficulty is coming up with a map between the the two sets. I have already verified that $H unlhd mathbb R^3$. So all I know is $mathbb R^3/H$ is a group. Also, from first isomorphism theorem, I know that the group is isomorphic to the image of the map $f: mathbb R^3 to A$, and I do not know what that $A$ is supposed to be. Today is the first day I learned about isomorphism, and I am very confused what is going on. Can anyone provide some help on this?

abstract-algebra group-isomorphism

share|cite|improve this question

asked 59 mins ago

UfomammutUfomammut

391314

$endgroup$

add a comment | 

$begingroup$

So I am given a group $mathbb R^3$ and a group $H$ = {$(y,0,0)|y in mathbb R$}. I have to prove that that $mathbb R^3/H$ $cong$ $mathbb R^2$. I am not sure how to even begin. My difficulty is coming up with a map between the the two sets. I have already verified that $H unlhd mathbb R^3$. So all I know is $mathbb R^3/H$ is a group. Also, from first isomorphism theorem, I know that the group is isomorphic to the image of the map $f: mathbb R^3 to A$, and I do not know what that $A$ is supposed to be. Today is the first day I learned about isomorphism, and I am very confused what is going on. Can anyone provide some help on this?

abstract-algebra group-isomorphism

share|cite|improve this question

asked 59 mins ago

UfomammutUfomammut

391314

$endgroup$

share|cite|improve this question

asked 59 mins ago

UfomammutUfomammut

391314

share|cite|improve this question

asked 59 mins ago

UfomammutUfomammut

391314

asked 59 mins ago

UfomammutUfomammut

391314

asked 59 mins ago

UfomammutUfomammut

391314

2 Answers
2

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2

$begingroup$

The first isomorphism theorem asserts that, if $varphi: Ato B$ is a surjective homomorphism, then $Bcong A/kervarphi$. In your problem, you wish to show $mathbb R^2congmathbb R^3/H $, so a natural guess would be to take $A=mathbb R^3$ and $B=mathbb R^2$. Now it remains to construct the homomorphism so that $kervarphi=H$. I will leave the rest to you.

share|cite|improve this answer

answered 28 mins ago

lEmlEm

3,4521921

$endgroup$

add a comment | 

2

$begingroup$

We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbb{R}^3 longrightarrow mathbb{R}^2$ as follows, $f(x,y,z) = (y,y+z)$. First, we show that this map is a well-defined group homomorphism and next show that $H = {(y,0,0) | y in mathbb{R} }$ is its kernel.
If $(x_1,y_1,z_1) = (x_2,y_2,z_2)$ then $f(x_1,y_1,z_1) = f(x_2,y_2,z_2)$ hence map is well defined.
Next we show this map is homomorphism. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). text{Moreover,} f(0,0,0) = (0,0)$
The kernel of this map is seen to be all $(x,y,z) in mathbb{R}$ such that $y,z$ are $0$ , i.e., $H$.
Hence first isomorphism theorem applies and $ mathbb{R}^3/H equiv mathbb{R}^2.$

share|cite|improve this answer

edited 3 mins ago

answered 22 mins ago

Mayank MishraMayank Mishra

1068

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I came up with something similar, but what about the map $f((x,y,z)) = (y,z)$? I think this one should be fine too, right?
  $endgroup$
  – Ufomammut
  21 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that will also work.
  $endgroup$
  – Mayank Mishra
  17 mins ago
  

  
  
  
  

add a comment | 

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2

$begingroup$

The first isomorphism theorem asserts that, if $varphi: Ato B$ is a surjective homomorphism, then $Bcong A/kervarphi$. In your problem, you wish to show $mathbb R^2congmathbb R^3/H $, so a natural guess would be to take $A=mathbb R^3$ and $B=mathbb R^2$. Now it remains to construct the homomorphism so that $kervarphi=H$. I will leave the rest to you.

share|cite|improve this answer

answered 28 mins ago

lEmlEm

3,4521921

$endgroup$

add a comment | 

2

$begingroup$

The first isomorphism theorem asserts that, if $varphi: Ato B$ is a surjective homomorphism, then $Bcong A/kervarphi$. In your problem, you wish to show $mathbb R^2congmathbb R^3/H $, so a natural guess would be to take $A=mathbb R^3$ and $B=mathbb R^2$. Now it remains to construct the homomorphism so that $kervarphi=H$. I will leave the rest to you.

share|cite|improve this answer

answered 28 mins ago

lEmlEm

3,4521921

$endgroup$

add a comment | 

$begingroup$

The first isomorphism theorem asserts that, if $varphi: Ato B$ is a surjective homomorphism, then $Bcong A/kervarphi$. In your problem, you wish to show $mathbb R^2congmathbb R^3/H $, so a natural guess would be to take $A=mathbb R^3$ and $B=mathbb R^2$. Now it remains to construct the homomorphism so that $kervarphi=H$. I will leave the rest to you.

share|cite|improve this answer

answered 28 mins ago

lEmlEm

3,4521921

$endgroup$

share|cite|improve this answer

answered 28 mins ago

lEmlEm

3,4521921

answered 28 mins ago

lEmlEm

3,4521921

answered 28 mins ago

lEmlEm

3,4521921

2

$begingroup$

We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbb{R}^3 longrightarrow mathbb{R}^2$ as follows, $f(x,y,z) = (y,y+z)$. First, we show that this map is a well-defined group homomorphism and next show that $H = {(y,0,0) | y in mathbb{R} }$ is its kernel.
If $(x_1,y_1,z_1) = (x_2,y_2,z_2)$ then $f(x_1,y_1,z_1) = f(x_2,y_2,z_2)$ hence map is well defined.
Next we show this map is homomorphism. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). text{Moreover,} f(0,0,0) = (0,0)$
The kernel of this map is seen to be all $(x,y,z) in mathbb{R}$ such that $y,z$ are $0$ , i.e., $H$.
Hence first isomorphism theorem applies and $ mathbb{R}^3/H equiv mathbb{R}^2.$

share|cite|improve this answer

edited 3 mins ago

answered 22 mins ago

Mayank MishraMayank Mishra

1068

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I came up with something similar, but what about the map $f((x,y,z)) = (y,z)$? I think this one should be fine too, right?
  $endgroup$
  – Ufomammut
  21 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that will also work.
  $endgroup$
  – Mayank Mishra
  17 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbb{R}^3 longrightarrow mathbb{R}^2$ as follows, $f(x,y,z) = (y,y+z)$. First, we show that this map is a well-defined group homomorphism and next show that $H = {(y,0,0) | y in mathbb{R} }$ is its kernel.
If $(x_1,y_1,z_1) = (x_2,y_2,z_2)$ then $f(x_1,y_1,z_1) = f(x_2,y_2,z_2)$ hence map is well defined.
Next we show this map is homomorphism. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). text{Moreover,} f(0,0,0) = (0,0)$
The kernel of this map is seen to be all $(x,y,z) in mathbb{R}$ such that $y,z$ are $0$ , i.e., $H$.
Hence first isomorphism theorem applies and $ mathbb{R}^3/H equiv mathbb{R}^2.$

share|cite|improve this answer

edited 3 mins ago

answered 22 mins ago

Mayank MishraMayank Mishra

1068

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I came up with something similar, but what about the map $f((x,y,z)) = (y,z)$? I think this one should be fine too, right?
  $endgroup$
  – Ufomammut
  21 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that will also work.
  $endgroup$
  – Mayank Mishra
  17 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbb{R}^3 longrightarrow mathbb{R}^2$ as follows, $f(x,y,z) = (y,y+z)$. First, we show that this map is a well-defined group homomorphism and next show that $H = {(y,0,0) | y in mathbb{R} }$ is its kernel.
If $(x_1,y_1,z_1) = (x_2,y_2,z_2)$ then $f(x_1,y_1,z_1) = f(x_2,y_2,z_2)$ hence map is well defined.
Next we show this map is homomorphism. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). text{Moreover,} f(0,0,0) = (0,0)$
The kernel of this map is seen to be all $(x,y,z) in mathbb{R}$ such that $y,z$ are $0$ , i.e., $H$.
Hence first isomorphism theorem applies and $ mathbb{R}^3/H equiv mathbb{R}^2.$

share|cite|improve this answer

edited 3 mins ago

answered 22 mins ago

Mayank MishraMayank Mishra

1068

$endgroup$

share|cite|improve this answer

edited 3 mins ago

answered 22 mins ago

Mayank MishraMayank Mishra

1068

answered 22 mins ago

Mayank MishraMayank Mishra

1068

answered 22 mins ago

Mayank MishraMayank Mishra

1068