Direct product, homomorphism Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Direct Product Of Abelian GroupHomomorphism between direct products and quotients.D_4 cannot be written as direct product of groups A x B.prove homomorphic image has order 4Homomorphism from Q8 to VDirect product of abelian group of $4$Equivalence of Internal & External Direct ProductShowing that such a homomorphism existsIsomorphism from inner direct product to external.How to find homomorphism of direct product?
Classification of bundles, Postnikov towers, obstruction theory, local coefficients
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Direct product, homomorphism
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Direct Product Of Abelian GroupHomomorphism between direct products and quotients.D_4 cannot be written as direct product of groups A x B.prove homomorphic image has order 4Homomorphism from Q8 to VDirect product of abelian group of $4$Equivalence of Internal & External Direct ProductShowing that such a homomorphism existsIsomorphism from inner direct product to external.How to find homomorphism of direct product?
$begingroup$
Prove (G1 x G2)/(K1 x K2) is isomorphic to G1/K1 x G2/K2.
Attempt:
First, we need to show a homomorphism exists, such that f : G1 x G2 → G1/K1 x G2/K2.
Second, we need to verify Imf = G1/K1 x G2/K2.
Last, we need to verify Kerf = K1 x K2.
Then, I have define a homomorphism such that f[(g1, g2)] = (g1k1, g2k2), where g1,g2 are elements in G1 x G2, and k1,k2 are elements in K1 x K2.
Then we need to show f[(g1,g2)(g1',g2')] = f[(g1,g2)]f[(g1',g2')].
So, f[(g1,g2)(g1',g2')] = [(g1k1, g2k2)][(g1'k1, g2'k2)] = [(g1g1'k1, g2g2'k2)] = g1g1'k1 x g2g2'k2 = G/K1 x G2/K2.
Can anyone please help me verify this is a good defined homomorphism.
I am confused because of the direct product part. Thank you!
abstract-algebra group-theory
asked May 5 '14 at 1:41
user1523user1523
283
$endgroup$
add a comment |
$begingroup$
Prove (G1 x G2)/(K1 x K2) is isomorphic to G1/K1 x G2/K2.
Attempt:
First, we need to show a homomorphism exists, such that f : G1 x G2 → G1/K1 x G2/K2.
Second, we need to verify Imf = G1/K1 x G2/K2.
Last, we need to verify Kerf = K1 x K2.
Then, I have define a homomorphism such that f[(g1, g2)] = (g1k1, g2k2), where g1,g2 are elements in G1 x G2, and k1,k2 are elements in K1 x K2.
Then we need to show f[(g1,g2)(g1',g2')] = f[(g1,g2)]f[(g1',g2')].
So, f[(g1,g2)(g1',g2')] = [(g1k1, g2k2)][(g1'k1, g2'k2)] = [(g1g1'k1, g2g2'k2)] = g1g1'k1 x g2g2'k2 = G/K1 x G2/K2.
Can anyone please help me verify this is a good defined homomorphism.
I am confused because of the direct product part. Thank you!
abstract-algebra group-theory
asked May 5 '14 at 1:41
user1523user1523
283
$endgroup$
add a comment |
$begingroup$
Prove (G1 x G2)/(K1 x K2) is isomorphic to G1/K1 x G2/K2.
Attempt:
First, we need to show a homomorphism exists, such that f : G1 x G2 → G1/K1 x G2/K2.
Second, we need to verify Imf = G1/K1 x G2/K2.
Last, we need to verify Kerf = K1 x K2.
Then, I have define a homomorphism such that f[(g1, g2)] = (g1k1, g2k2), where g1,g2 are elements in G1 x G2, and k1,k2 are elements in K1 x K2.
Then we need to show f[(g1,g2)(g1',g2')] = f[(g1,g2)]f[(g1',g2')].
So, f[(g1,g2)(g1',g2')] = [(g1k1, g2k2)][(g1'k1, g2'k2)] = [(g1g1'k1, g2g2'k2)] = g1g1'k1 x g2g2'k2 = G/K1 x G2/K2.
Can anyone please help me verify this is a good defined homomorphism.
I am confused because of the direct product part. Thank you!
abstract-algebra group-theory
asked May 5 '14 at 1:41
user1523user1523
283
$endgroup$
Prove (G1 x G2)/(K1 x K2) is isomorphic to G1/K1 x G2/K2.
Attempt:
First, we need to show a homomorphism exists, such that f : G1 x G2 → G1/K1 x G2/K2.
Second, we need to verify Imf = G1/K1 x G2/K2.
Last, we need to verify Kerf = K1 x K2.
Then, I have define a homomorphism such that f[(g1, g2)] = (g1k1, g2k2), where g1,g2 are elements in G1 x G2, and k1,k2 are elements in K1 x K2.
Then we need to show f[(g1,g2)(g1',g2')] = f[(g1,g2)]f[(g1',g2')].
So, f[(g1,g2)(g1',g2')] = [(g1k1, g2k2)][(g1'k1, g2'k2)] = [(g1g1'k1, g2g2'k2)] = g1g1'k1 x g2g2'k2 = G/K1 x G2/K2.
Can anyone please help me verify this is a good defined homomorphism.
I am confused because of the direct product part. Thank you!
abstract-algebra group-theory
abstract-algebra group-theory
asked May 5 '14 at 1:41
user1523user1523
283
asked May 5 '14 at 1:41
user1523user1523
283
edited May 5 '14 at 1:59
user1523
asked May 5 '14 at 1:41
user1523user1523
283
asked May 5 '14 at 1:41
user1523user1523
283
asked May 5 '14 at 1:41
user1523user1523
283
283
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your homomorphism (as it is written now) does not even appear to have the appropriate domain and codomain. You'd like to map an element of $(G_1 times G_2) / (K_1 times K_2)$ to an element of $G_1 / K_1 times G_2 / K_2$. What do elements of these two groups look like?
They should look like cosets. Elements of the quotient group $(G_1 times G_2) / (K_1 times K_1)$ are of the form $(g_1,g_2)(K_1times K_2)$; they should not be written merely using their coset representatives from the group $G_1 times G_2$. It is important that you have some grasp of the objects you are mapping between before you try to write out this mapping.
EDIT: The domain and codomain of a function $f$ are the sets that function maps from and into, respectively. If your goal is to write a homomorphism (which is a special kind of function) between two sets (which in your case happen to be groups), then you will need to make sure that your function takes in elements of the former set and yields elements of the latter. None of your attempts (either in the question or in comments) has satisfied this. You have not even defined a function between the correct sets, much less a well-defined homomorphism!
answered May 5 '14 at 2:07
matthugsmatthugs
313
$endgroup$
$begingroup$
A correspondence between things that look like $(g_1, g_2)(K_1times K_2)$ and things that looks like $(h_1 K_1, h_2 K_2)$, where $g_1,h_1in G_1$, and $g_2,h_2in G_2$.
$endgroup$
– matthugs
May 5 '14 at 2:18
$begingroup$
Notice that I am using capital letters to denote groups; your use of lower-case $k_1$ and $k_2$ seems to me to be evidence of misunderstanding.
$endgroup$
– matthugs
May 5 '14 at 2:20
$begingroup$
So for example: f[(g1,g2)] = (g1,g2)(K1 x K2). Then f[(g1,g2)]*f[(g1', g2')] = (g1,g2)(K1x K2)*(g1', g2')(K1 xK2) = (g1g1', g2g2')(K1 x K2)? Is this correct?
$endgroup$
– user1523
May 5 '14 at 2:24
$begingroup$
$f$ is meant to be your homomorphism, right? What are the domain and codomain of the $f$ that you've just written?
$endgroup$
– matthugs
May 5 '14 at 2:25
$begingroup$
Yes, there is no restriction. I don't know, I would say it could be any elements
$endgroup$
– user1523
May 5 '14 at 2:27
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Your homomorphism (as it is written now) does not even appear to have the appropriate domain and codomain. You'd like to map an element of $(G_1 times G_2) / (K_1 times K_2)$ to an element of $G_1 / K_1 times G_2 / K_2$. What do elements of these two groups look like?
They should look like cosets. Elements of the quotient group $(G_1 times G_2) / (K_1 times K_1)$ are of the form $(g_1,g_2)(K_1times K_2)$; they should not be written merely using their coset representatives from the group $G_1 times G_2$. It is important that you have some grasp of the objects you are mapping between before you try to write out this mapping.
EDIT: The domain and codomain of a function $f$ are the sets that function maps from and into, respectively. If your goal is to write a homomorphism (which is a special kind of function) between two sets (which in your case happen to be groups), then you will need to make sure that your function takes in elements of the former set and yields elements of the latter. None of your attempts (either in the question or in comments) has satisfied this. You have not even defined a function between the correct sets, much less a well-defined homomorphism!
answered May 5 '14 at 2:07
matthugsmatthugs
313
$endgroup$
$begingroup$
A correspondence between things that look like $(g_1, g_2)(K_1times K_2)$ and things that looks like $(h_1 K_1, h_2 K_2)$, where $g_1,h_1in G_1$, and $g_2,h_2in G_2$.
$endgroup$
– matthugs
May 5 '14 at 2:18
$begingroup$
Notice that I am using capital letters to denote groups; your use of lower-case $k_1$ and $k_2$ seems to me to be evidence of misunderstanding.
$endgroup$
– matthugs
May 5 '14 at 2:20
$begingroup$
So for example: f[(g1,g2)] = (g1,g2)(K1 x K2). Then f[(g1,g2)]*f[(g1', g2')] = (g1,g2)(K1x K2)*(g1', g2')(K1 xK2) = (g1g1', g2g2')(K1 x K2)? Is this correct?
$endgroup$
– user1523
May 5 '14 at 2:24
$begingroup$
$f$ is meant to be your homomorphism, right? What are the domain and codomain of the $f$ that you've just written?
$endgroup$
– matthugs
May 5 '14 at 2:25
$begingroup$
Yes, there is no restriction. I don't know, I would say it could be any elements
$endgroup$
– user1523
May 5 '14 at 2:27
add a comment |
$begingroup$
Your homomorphism (as it is written now) does not even appear to have the appropriate domain and codomain. You'd like to map an element of $(G_1 times G_2) / (K_1 times K_2)$ to an element of $G_1 / K_1 times G_2 / K_2$. What do elements of these two groups look like?
They should look like cosets. Elements of the quotient group $(G_1 times G_2) / (K_1 times K_1)$ are of the form $(g_1,g_2)(K_1times K_2)$; they should not be written merely using their coset representatives from the group $G_1 times G_2$. It is important that you have some grasp of the objects you are mapping between before you try to write out this mapping.
EDIT: The domain and codomain of a function $f$ are the sets that function maps from and into, respectively. If your goal is to write a homomorphism (which is a special kind of function) between two sets (which in your case happen to be groups), then you will need to make sure that your function takes in elements of the former set and yields elements of the latter. None of your attempts (either in the question or in comments) has satisfied this. You have not even defined a function between the correct sets, much less a well-defined homomorphism!
answered May 5 '14 at 2:07
matthugsmatthugs
313
$endgroup$
$begingroup$
A correspondence between things that look like $(g_1, g_2)(K_1times K_2)$ and things that looks like $(h_1 K_1, h_2 K_2)$, where $g_1,h_1in G_1$, and $g_2,h_2in G_2$.
$endgroup$
– matthugs
May 5 '14 at 2:18
$begingroup$
Notice that I am using capital letters to denote groups; your use of lower-case $k_1$ and $k_2$ seems to me to be evidence of misunderstanding.
$endgroup$
– matthugs
May 5 '14 at 2:20
$begingroup$
So for example: f[(g1,g2)] = (g1,g2)(K1 x K2). Then f[(g1,g2)]*f[(g1', g2')] = (g1,g2)(K1x K2)*(g1', g2')(K1 xK2) = (g1g1', g2g2')(K1 x K2)? Is this correct?
$endgroup$
– user1523
May 5 '14 at 2:24
$begingroup$
$f$ is meant to be your homomorphism, right? What are the domain and codomain of the $f$ that you've just written?
$endgroup$
– matthugs
May 5 '14 at 2:25
$begingroup$
Yes, there is no restriction. I don't know, I would say it could be any elements
$endgroup$
– user1523
May 5 '14 at 2:27
add a comment |
$begingroup$
Your homomorphism (as it is written now) does not even appear to have the appropriate domain and codomain. You'd like to map an element of $(G_1 times G_2) / (K_1 times K_2)$ to an element of $G_1 / K_1 times G_2 / K_2$. What do elements of these two groups look like?
They should look like cosets. Elements of the quotient group $(G_1 times G_2) / (K_1 times K_1)$ are of the form $(g_1,g_2)(K_1times K_2)$; they should not be written merely using their coset representatives from the group $G_1 times G_2$. It is important that you have some grasp of the objects you are mapping between before you try to write out this mapping.
EDIT: The domain and codomain of a function $f$ are the sets that function maps from and into, respectively. If your goal is to write a homomorphism (which is a special kind of function) between two sets (which in your case happen to be groups), then you will need to make sure that your function takes in elements of the former set and yields elements of the latter. None of your attempts (either in the question or in comments) has satisfied this. You have not even defined a function between the correct sets, much less a well-defined homomorphism!
answered May 5 '14 at 2:07
matthugsmatthugs
313
$endgroup$
Your homomorphism (as it is written now) does not even appear to have the appropriate domain and codomain. You'd like to map an element of $(G_1 times G_2) / (K_1 times K_2)$ to an element of $G_1 / K_1 times G_2 / K_2$. What do elements of these two groups look like?
They should look like cosets. Elements of the quotient group $(G_1 times G_2) / (K_1 times K_1)$ are of the form $(g_1,g_2)(K_1times K_2)$; they should not be written merely using their coset representatives from the group $G_1 times G_2$. It is important that you have some grasp of the objects you are mapping between before you try to write out this mapping.
EDIT: The domain and codomain of a function $f$ are the sets that function maps from and into, respectively. If your goal is to write a homomorphism (which is a special kind of function) between two sets (which in your case happen to be groups), then you will need to make sure that your function takes in elements of the former set and yields elements of the latter. None of your attempts (either in the question or in comments) has satisfied this. You have not even defined a function between the correct sets, much less a well-defined homomorphism!
answered May 5 '14 at 2:07
matthugsmatthugs
313
edited May 5 '14 at 2:36
answered May 5 '14 at 2:07
matthugsmatthugs
313
answered May 5 '14 at 2:07
matthugsmatthugs
313
answered May 5 '14 at 2:07
matthugsmatthugs
313
313
$begingroup$
A correspondence between things that look like $(g_1, g_2)(K_1times K_2)$ and things that looks like $(h_1 K_1, h_2 K_2)$, where $g_1,h_1in G_1$, and $g_2,h_2in G_2$.
$endgroup$
– matthugs
May 5 '14 at 2:18
$begingroup$
Notice that I am using capital letters to denote groups; your use of lower-case $k_1$ and $k_2$ seems to me to be evidence of misunderstanding.
$endgroup$
– matthugs
May 5 '14 at 2:20
$begingroup$
So for example: f[(g1,g2)] = (g1,g2)(K1 x K2). Then f[(g1,g2)]*f[(g1', g2')] = (g1,g2)(K1x K2)*(g1', g2')(K1 xK2) = (g1g1', g2g2')(K1 x K2)? Is this correct?
$endgroup$
– user1523
May 5 '14 at 2:24
$begingroup$
$f$ is meant to be your homomorphism, right? What are the domain and codomain of the $f$ that you've just written?
$endgroup$
– matthugs
May 5 '14 at 2:25
$begingroup$
Yes, there is no restriction. I don't know, I would say it could be any elements
$endgroup$
– user1523
May 5 '14 at 2:27
add a comment |
$begingroup$
A correspondence between things that look like $(g_1, g_2)(K_1times K_2)$ and things that looks like $(h_1 K_1, h_2 K_2)$, where $g_1,h_1in G_1$, and $g_2,h_2in G_2$.
$endgroup$
– matthugs
May 5 '14 at 2:18
$begingroup$
Notice that I am using capital letters to denote groups; your use of lower-case $k_1$ and $k_2$ seems to me to be evidence of misunderstanding.
$endgroup$
– matthugs
May 5 '14 at 2:20
$begingroup$
So for example: f[(g1,g2)] = (g1,g2)(K1 x K2). Then f[(g1,g2)]*f[(g1', g2')] = (g1,g2)(K1x K2)*(g1', g2')(K1 xK2) = (g1g1', g2g2')(K1 x K2)? Is this correct?
$endgroup$
– user1523
May 5 '14 at 2:24
$begingroup$
$f$ is meant to be your homomorphism, right? What are the domain and codomain of the $f$ that you've just written?
$endgroup$
– matthugs
May 5 '14 at 2:25
$begingroup$
Yes, there is no restriction. I don't know, I would say it could be any elements
$endgroup$
– user1523
May 5 '14 at 2:27
$begingroup$
A correspondence between things that look like $(g_1, g_2)(K_1times K_2)$ and things that looks like $(h_1 K_1, h_2 K_2)$, where $g_1,h_1in G_1$, and $g_2,h_2in G_2$.
$endgroup$
– matthugs
May 5 '14 at 2:18
$begingroup$
A correspondence between things that look like $(g_1, g_2)(K_1times K_2)$ and things that looks like $(h_1 K_1, h_2 K_2)$, where $g_1,h_1in G_1$, and $g_2,h_2in G_2$.
$endgroup$
– matthugs
May 5 '14 at 2:18
$begingroup$
Notice that I am using capital letters to denote groups; your use of lower-case $k_1$ and $k_2$ seems to me to be evidence of misunderstanding.
$endgroup$
– matthugs
May 5 '14 at 2:20
$begingroup$
Notice that I am using capital letters to denote groups; your use of lower-case $k_1$ and $k_2$ seems to me to be evidence of misunderstanding.
$endgroup$
– matthugs
May 5 '14 at 2:20
$begingroup$
So for example: f[(g1,g2)] = (g1,g2)(K1 x K2). Then f[(g1,g2)]*f[(g1', g2')] = (g1,g2)(K1x K2)*(g1', g2')(K1 xK2) = (g1g1', g2g2')(K1 x K2)? Is this correct?
$endgroup$
– user1523
May 5 '14 at 2:24
$begingroup$
So for example: f[(g1,g2)] = (g1,g2)(K1 x K2). Then f[(g1,g2)]*f[(g1', g2')] = (g1,g2)(K1x K2)*(g1', g2')(K1 xK2) = (g1g1', g2g2')(K1 x K2)? Is this correct?
$endgroup$
– user1523
May 5 '14 at 2:24
$begingroup$
$f$ is meant to be your homomorphism, right? What are the domain and codomain of the $f$ that you've just written?
$endgroup$
– matthugs
May 5 '14 at 2:25
$begingroup$
$f$ is meant to be your homomorphism, right? What are the domain and codomain of the $f$ that you've just written?
$endgroup$
– matthugs
May 5 '14 at 2:25
$begingroup$
Yes, there is no restriction. I don't know, I would say it could be any elements
$endgroup$
– user1523
May 5 '14 at 2:27
$begingroup$
Yes, there is no restriction. I don't know, I would say it could be any elements
$endgroup$
– user1523
May 5 '14 at 2:27
add a comment |
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