Proving the given two groups are isomorphic The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Are $(mathbbR,+)$ and $(mathbbC,+)$ isomorphic as additive groups?How do I show that these two presentations are isomorphic?Determine whether or not the two given groups are isomorphic.Surjective Homomorphisms of Isomorphic Abelian GroupsGroup isomorphism between two groups .How to use the first isomorphism theorem to show that two groups are isomorphic?Showing that these two groups are isomorphic?Showing that $2$ of the following groups are not isomorphicShow that the Two Given Groups are IsomorphicAre given groups isomorphic
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Proving the given two groups are isomorphic
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Are $(mathbbR,+)$ and $(mathbbC,+)$ isomorphic as additive groups?How do I show that these two presentations are isomorphic?Determine whether or not the two given groups are isomorphic.Surjective Homomorphisms of Isomorphic Abelian GroupsGroup isomorphism between two groups .How to use the first isomorphism theorem to show that two groups are isomorphic?Showing that these two groups are isomorphic?Showing that $2$ of the following groups are not isomorphicShow that the Two Given Groups are IsomorphicAre given groups isomorphic
$begingroup$
So I am given a group $mathbb R^3$ and a group $H$ = 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
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 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
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 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
asked 59 mins ago
UfomammutUfomammut
391314
$endgroup$
So I am given a group $mathbb R^3$ and a group $H$ = 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
abstract-algebra group-isomorphism
asked 59 mins ago
UfomammutUfomammut
391314
asked 59 mins ago
UfomammutUfomammut
391314
asked 59 mins ago
UfomammutUfomammut
391314
asked 59 mins ago
UfomammutUfomammut
391314
asked 59 mins ago
UfomammutUfomammut
391314
391314
add a comment |
add a comment |
2 Answers
2
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$begingroup$
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.
answered 28 mins ago
lEmlEm
3,4521921
$endgroup$
add a comment |
$begingroup$
We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbbR^3 longrightarrow mathbbR^2$ 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 in mathbbR $ 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. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). textMoreover, f(0,0,0) = (0,0)$
The kernel of this map is seen to be all $(x,y,z) in mathbbR$ such that $y,z$ are $0$ , i.e., $H$.
Hence first isomorphism theorem applies and $ mathbbR^3/H equiv mathbbR^2.$
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
20 mins ago
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
17 mins ago
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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.
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.
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.
answered 28 mins ago
lEmlEm
3,4521921
$endgroup$
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.
answered 28 mins ago
lEmlEm
3,4521921
answered 28 mins ago
lEmlEm
3,4521921
answered 28 mins ago
lEmlEm
3,4521921
answered 28 mins ago
lEmlEm
3,4521921
3,4521921
add a comment |
add a comment |
$begingroup$
We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbbR^3 longrightarrow mathbbR^2$ 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 in mathbbR $ 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. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). textMoreover, f(0,0,0) = (0,0)$
The kernel of this map is seen to be all $(x,y,z) in mathbbR$ such that $y,z$ are $0$ , i.e., $H$.
Hence first isomorphism theorem applies and $ mathbbR^3/H equiv mathbbR^2.$
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
20 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 $f : mathbbR^3 longrightarrow mathbbR^2$ 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 in mathbbR $ 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. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). textMoreover, f(0,0,0) = (0,0)$
The kernel of this map is seen to be all $(x,y,z) in mathbbR$ such that $y,z$ are $0$ , i.e., $H$.
Hence first isomorphism theorem applies and $ mathbbR^3/H equiv mathbbR^2.$
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
20 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 $f : mathbbR^3 longrightarrow mathbbR^2$ 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 in mathbbR $ 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. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). textMoreover, f(0,0,0) = (0,0)$
The kernel of this map is seen to be all $(x,y,z) in mathbbR$ such that $y,z$ are $0$ , i.e., $H$.
Hence first isomorphism theorem applies and $ mathbbR^3/H equiv mathbbR^2.$
answered 22 mins ago
Mayank MishraMayank Mishra
1068
$endgroup$
We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbbR^3 longrightarrow mathbbR^2$ 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 in mathbbR $ 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. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). textMoreover, f(0,0,0) = (0,0)$
The kernel of this map is seen to be all $(x,y,z) in mathbbR$ such that $y,z$ are $0$ , i.e., $H$.
Hence first isomorphism theorem applies and $ mathbbR^3/H equiv mathbbR^2.$
answered 22 mins ago
Mayank MishraMayank Mishra
1068
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
1068
$begingroup$
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
20 mins ago
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
17 mins ago
add a comment |
$begingroup$
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
20 mins ago
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
17 mins ago
$begingroup$
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
20 mins ago
$begingroup$
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
20 mins ago
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
17 mins ago
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
17 mins ago
add a comment |
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