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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 1 hour 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 1 hour 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 1 hour ago

UfomammutUfomammut

391314

$endgroup$

share|cite|improve this question

asked 1 hour ago

UfomammutUfomammut

391314

share|cite|improve this question

asked 1 hour ago

UfomammutUfomammut

391314

asked 1 hour ago

UfomammutUfomammut

391314

asked 1 hour ago

UfomammutUfomammut

391314

2 Answers
2

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3

$begingroup$

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 35 mins ago

lEmlEm

3,4621921

$endgroup$

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 10 mins ago

answered 29 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
  27 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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3

$begingroup$

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 35 mins ago

lEmlEm

3,4621921

$endgroup$

add a comment | 

3

$begingroup$

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 35 mins ago

lEmlEm

3,4621921

$endgroup$

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 35 mins ago

lEmlEm

3,4621921

$endgroup$

share|cite|improve this answer

answered 35 mins ago

lEmlEm

3,4621921

answered 35 mins ago

lEmlEm

3,4621921

answered 35 mins ago

lEmlEm

3,4621921

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 10 mins ago

answered 29 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
  27 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that will also work.
  $endgroup$
  – Mayank Mishra
  24 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 10 mins ago

answered 29 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
  27 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that will also work.
  $endgroup$
  – Mayank Mishra
  24 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 10 mins ago

answered 29 mins ago

Mayank MishraMayank Mishra

1068

$endgroup$

share|cite|improve this answer

edited 10 mins ago

answered 29 mins ago

Mayank MishraMayank Mishra

1068

answered 29 mins ago

Mayank MishraMayank Mishra

1068

answered 29 mins ago

Mayank MishraMayank Mishra

1068