Does $G times H = K$ iff $G$ and $H$ are normal in $K$? [on hold]
$begingroup$
Let $G times H = K$. By applying The Isomorphism Theorem to the homomorphism $(g, h) rightarrow g : K rightarrow G$, I get $K/H cong G$, so $H$ is normal in $K$. Similarly, $G$ is normal in $K$.
Is this line of reasoning correct? Is the converse true, that if $H$ and $G$ are normal in $K$ then $G times H = K$? If not what can be said about $H$ and $G$?
group-theory normal-subgroups
asked Dec 13 '18 at 23:46
awsomeguyawsomeguy
163
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put on hold as off-topic by Saad, Eevee Trainer, mrtaurho, Cesareo, Adrian Keister Jan 9 at 14:05
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, Eevee Trainer, mrtaurho, Cesareo, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $G times H = K$. By applying The Isomorphism Theorem to the homomorphism $(g, h) rightarrow g : K rightarrow G$, I get $K/H cong G$, so $H$ is normal in $K$. Similarly, $G$ is normal in $K$.
Is this line of reasoning correct? Is the converse true, that if $H$ and $G$ are normal in $K$ then $G times H = K$? If not what can be said about $H$ and $G$?
group-theory normal-subgroups
asked Dec 13 '18 at 23:46
awsomeguyawsomeguy
163
$endgroup$
put on hold as off-topic by Saad, Eevee Trainer, mrtaurho, Cesareo, Adrian Keister Jan 9 at 14:05
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, Eevee Trainer, mrtaurho, Cesareo, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
@Ted Shifrin edited
$endgroup$
– awsomeguy
Dec 14 '18 at 0:00
$begingroup$
For your converse, you need more hypotheses for sure. Think about what you know if you in fact have $K = Gtimes H$.
$endgroup$
– Ted Shifrin
Dec 14 '18 at 0:03
$begingroup$
If $G,H$ are two subgroups of $K$ such that $K subset {gh, g in G, h in H}$ and $gh = hg$ and $G cap H= {1}$ then $K cong G times H$. The opposite direction is easy.
$endgroup$
– reuns
Dec 14 '18 at 0:14
add a comment |
$begingroup$
Let $G times H = K$. By applying The Isomorphism Theorem to the homomorphism $(g, h) rightarrow g : K rightarrow G$, I get $K/H cong G$, so $H$ is normal in $K$. Similarly, $G$ is normal in $K$.
Is this line of reasoning correct? Is the converse true, that if $H$ and $G$ are normal in $K$ then $G times H = K$? If not what can be said about $H$ and $G$?
group-theory normal-subgroups
asked Dec 13 '18 at 23:46
awsomeguyawsomeguy
163
$endgroup$
Let $G times H = K$. By applying The Isomorphism Theorem to the homomorphism $(g, h) rightarrow g : K rightarrow G$, I get $K/H cong G$, so $H$ is normal in $K$. Similarly, $G$ is normal in $K$.
Is this line of reasoning correct? Is the converse true, that if $H$ and $G$ are normal in $K$ then $G times H = K$? If not what can be said about $H$ and $G$?
group-theory normal-subgroups
group-theory normal-subgroups
asked Dec 13 '18 at 23:46
awsomeguyawsomeguy
163
asked Dec 13 '18 at 23:46
awsomeguyawsomeguy
163
edited Dec 13 '18 at 23:59
awsomeguy
asked Dec 13 '18 at 23:46
awsomeguyawsomeguy
163
asked Dec 13 '18 at 23:46
awsomeguyawsomeguy
163
asked Dec 13 '18 at 23:46
awsomeguyawsomeguy
163
163
put on hold as off-topic by Saad, Eevee Trainer, mrtaurho, Cesareo, Adrian Keister Jan 9 at 14:05
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, Eevee Trainer, mrtaurho, Cesareo, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Saad, Eevee Trainer, mrtaurho, Cesareo, Adrian Keister Jan 9 at 14:05
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, Eevee Trainer, mrtaurho, Cesareo, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
@Ted Shifrin edited
$endgroup$
– awsomeguy
Dec 14 '18 at 0:00
$begingroup$
For your converse, you need more hypotheses for sure. Think about what you know if you in fact have $K = Gtimes H$.
$endgroup$
– Ted Shifrin
Dec 14 '18 at 0:03
$begingroup$
If $G,H$ are two subgroups of $K$ such that $K subset {gh, g in G, h in H}$ and $gh = hg$ and $G cap H= {1}$ then $K cong G times H$. The opposite direction is easy.
$endgroup$
– reuns
Dec 14 '18 at 0:14
add a comment |
$begingroup$
@Ted Shifrin edited
$endgroup$
– awsomeguy
Dec 14 '18 at 0:00
$begingroup$
For your converse, you need more hypotheses for sure. Think about what you know if you in fact have $K = Gtimes H$.
$endgroup$
– Ted Shifrin
Dec 14 '18 at 0:03
$begingroup$
If $G,H$ are two subgroups of $K$ such that $K subset {gh, g in G, h in H}$ and $gh = hg$ and $G cap H= {1}$ then $K cong G times H$. The opposite direction is easy.
$endgroup$
– reuns
Dec 14 '18 at 0:14
$begingroup$
@Ted Shifrin edited
$endgroup$
– awsomeguy
Dec 14 '18 at 0:00
$begingroup$
@Ted Shifrin edited
$endgroup$
– awsomeguy
Dec 14 '18 at 0:00
$begingroup$
For your converse, you need more hypotheses for sure. Think about what you know if you in fact have $K = Gtimes H$.
$endgroup$
– Ted Shifrin
Dec 14 '18 at 0:03
$begingroup$
For your converse, you need more hypotheses for sure. Think about what you know if you in fact have $K = Gtimes H$.
$endgroup$
– Ted Shifrin
Dec 14 '18 at 0:03
$begingroup$
If $G,H$ are two subgroups of $K$ such that $K subset {gh, g in G, h in H}$ and $gh = hg$ and $G cap H= {1}$ then $K cong G times H$. The opposite direction is easy.
$endgroup$
– reuns
Dec 14 '18 at 0:14
$begingroup$
If $G,H$ are two subgroups of $K$ such that $K subset {gh, g in G, h in H}$ and $gh = hg$ and $G cap H= {1}$ then $K cong G times H$. The opposite direction is easy.
$endgroup$
– reuns
Dec 14 '18 at 0:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I assume that when you write $H$ in $K$, you mean ${1} times H$ and $G times {1}$. If that is the case, then you are correct. Another way of seeing that is 'manually', i.e. if $(1,h') in {1} times H$, then for any $(g,h) in K$ we have that
$$
(g,h)(1,h')(g^{-1},h^{-1}) = (1,hh'h^{-1}) in {1} times H.
$$
The same works for $G$. As for the other direction, you need more hypotheses: for example, if the group is abelian any pair of subgroups is normal, but their product may not be the whole group. A concrete case: $mathbb{Z} oplus 0 oplus 0$ and $0 oplus mathbb{Z} oplus 0$ are normal in $mathbb{Z}^3$, but their sum (which is the product in this case) is $mathbb{Z}^2 not simeq mathbb{Z}^3$.
What does hold is the following: if $H,G triangleleft K$ and $HG = K$, $G cap H = {1}$, then $K simeq H times G$. Hint: take the function that maps $(g,h)$ to $gh in K$, and prove that it is an isomorphism.
answered Dec 14 '18 at 0:08
Guido A.Guido A.
7,2861730
$endgroup$
$begingroup$
Why is it that ${1} times H$ is correct whereas $H$ alone is incorrect? Are the two not isomorphic?
$endgroup$
– awsomeguy
Dec 14 '18 at 0:14
$begingroup$
They are, but $H$ is not a subgroup of $K$ strictly speaking. For a subgroup to be normal, it has to be subgroup in the first place, and $H$ is not a subset (let alone a subgroup) of $K$.
$endgroup$
– Guido A.
Dec 14 '18 at 0:17
$begingroup$
@awsomeguy by the way, a hypothesis was missing from the converse statement, we need both $GH = K$ and $G cap H = 1$. I have edited it accordingly :)
$endgroup$
– Guido A.
Dec 14 '18 at 0:20
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I assume that when you write $H$ in $K$, you mean ${1} times H$ and $G times {1}$. If that is the case, then you are correct. Another way of seeing that is 'manually', i.e. if $(1,h') in {1} times H$, then for any $(g,h) in K$ we have that
$$
(g,h)(1,h')(g^{-1},h^{-1}) = (1,hh'h^{-1}) in {1} times H.
$$
The same works for $G$. As for the other direction, you need more hypotheses: for example, if the group is abelian any pair of subgroups is normal, but their product may not be the whole group. A concrete case: $mathbb{Z} oplus 0 oplus 0$ and $0 oplus mathbb{Z} oplus 0$ are normal in $mathbb{Z}^3$, but their sum (which is the product in this case) is $mathbb{Z}^2 not simeq mathbb{Z}^3$.
What does hold is the following: if $H,G triangleleft K$ and $HG = K$, $G cap H = {1}$, then $K simeq H times G$. Hint: take the function that maps $(g,h)$ to $gh in K$, and prove that it is an isomorphism.
answered Dec 14 '18 at 0:08
Guido A.Guido A.
7,2861730
$endgroup$
$begingroup$
Why is it that ${1} times H$ is correct whereas $H$ alone is incorrect? Are the two not isomorphic?
$endgroup$
– awsomeguy
Dec 14 '18 at 0:14
$begingroup$
They are, but $H$ is not a subgroup of $K$ strictly speaking. For a subgroup to be normal, it has to be subgroup in the first place, and $H$ is not a subset (let alone a subgroup) of $K$.
$endgroup$
– Guido A.
Dec 14 '18 at 0:17
$begingroup$
@awsomeguy by the way, a hypothesis was missing from the converse statement, we need both $GH = K$ and $G cap H = 1$. I have edited it accordingly :)
$endgroup$
– Guido A.
Dec 14 '18 at 0:20
add a comment |
$begingroup$
I assume that when you write $H$ in $K$, you mean ${1} times H$ and $G times {1}$. If that is the case, then you are correct. Another way of seeing that is 'manually', i.e. if $(1,h') in {1} times H$, then for any $(g,h) in K$ we have that
$$
(g,h)(1,h')(g^{-1},h^{-1}) = (1,hh'h^{-1}) in {1} times H.
$$
The same works for $G$. As for the other direction, you need more hypotheses: for example, if the group is abelian any pair of subgroups is normal, but their product may not be the whole group. A concrete case: $mathbb{Z} oplus 0 oplus 0$ and $0 oplus mathbb{Z} oplus 0$ are normal in $mathbb{Z}^3$, but their sum (which is the product in this case) is $mathbb{Z}^2 not simeq mathbb{Z}^3$.
What does hold is the following: if $H,G triangleleft K$ and $HG = K$, $G cap H = {1}$, then $K simeq H times G$. Hint: take the function that maps $(g,h)$ to $gh in K$, and prove that it is an isomorphism.
answered Dec 14 '18 at 0:08
Guido A.Guido A.
7,2861730
$endgroup$
$begingroup$
Why is it that ${1} times H$ is correct whereas $H$ alone is incorrect? Are the two not isomorphic?
$endgroup$
– awsomeguy
Dec 14 '18 at 0:14
$begingroup$
They are, but $H$ is not a subgroup of $K$ strictly speaking. For a subgroup to be normal, it has to be subgroup in the first place, and $H$ is not a subset (let alone a subgroup) of $K$.
$endgroup$
– Guido A.
Dec 14 '18 at 0:17
$begingroup$
@awsomeguy by the way, a hypothesis was missing from the converse statement, we need both $GH = K$ and $G cap H = 1$. I have edited it accordingly :)
$endgroup$
– Guido A.
Dec 14 '18 at 0:20
add a comment |
$begingroup$
I assume that when you write $H$ in $K$, you mean ${1} times H$ and $G times {1}$. If that is the case, then you are correct. Another way of seeing that is 'manually', i.e. if $(1,h') in {1} times H$, then for any $(g,h) in K$ we have that
$$
(g,h)(1,h')(g^{-1},h^{-1}) = (1,hh'h^{-1}) in {1} times H.
$$
The same works for $G$. As for the other direction, you need more hypotheses: for example, if the group is abelian any pair of subgroups is normal, but their product may not be the whole group. A concrete case: $mathbb{Z} oplus 0 oplus 0$ and $0 oplus mathbb{Z} oplus 0$ are normal in $mathbb{Z}^3$, but their sum (which is the product in this case) is $mathbb{Z}^2 not simeq mathbb{Z}^3$.
What does hold is the following: if $H,G triangleleft K$ and $HG = K$, $G cap H = {1}$, then $K simeq H times G$. Hint: take the function that maps $(g,h)$ to $gh in K$, and prove that it is an isomorphism.
answered Dec 14 '18 at 0:08
Guido A.Guido A.
7,2861730
$endgroup$
I assume that when you write $H$ in $K$, you mean ${1} times H$ and $G times {1}$. If that is the case, then you are correct. Another way of seeing that is 'manually', i.e. if $(1,h') in {1} times H$, then for any $(g,h) in K$ we have that
$$
(g,h)(1,h')(g^{-1},h^{-1}) = (1,hh'h^{-1}) in {1} times H.
$$
The same works for $G$. As for the other direction, you need more hypotheses: for example, if the group is abelian any pair of subgroups is normal, but their product may not be the whole group. A concrete case: $mathbb{Z} oplus 0 oplus 0$ and $0 oplus mathbb{Z} oplus 0$ are normal in $mathbb{Z}^3$, but their sum (which is the product in this case) is $mathbb{Z}^2 not simeq mathbb{Z}^3$.
What does hold is the following: if $H,G triangleleft K$ and $HG = K$, $G cap H = {1}$, then $K simeq H times G$. Hint: take the function that maps $(g,h)$ to $gh in K$, and prove that it is an isomorphism.
answered Dec 14 '18 at 0:08
Guido A.Guido A.
7,2861730
edited Dec 14 '18 at 0:19
answered Dec 14 '18 at 0:08
Guido A.Guido A.
7,2861730
answered Dec 14 '18 at 0:08
Guido A.Guido A.
7,2861730
answered Dec 14 '18 at 0:08
Guido A.Guido A.
7,2861730
7,2861730
$begingroup$
Why is it that ${1} times H$ is correct whereas $H$ alone is incorrect? Are the two not isomorphic?
$endgroup$
– awsomeguy
Dec 14 '18 at 0:14
$begingroup$
They are, but $H$ is not a subgroup of $K$ strictly speaking. For a subgroup to be normal, it has to be subgroup in the first place, and $H$ is not a subset (let alone a subgroup) of $K$.
$endgroup$
– Guido A.
Dec 14 '18 at 0:17
$begingroup$
@awsomeguy by the way, a hypothesis was missing from the converse statement, we need both $GH = K$ and $G cap H = 1$. I have edited it accordingly :)
$endgroup$
– Guido A.
Dec 14 '18 at 0:20
add a comment |
$begingroup$
Why is it that ${1} times H$ is correct whereas $H$ alone is incorrect? Are the two not isomorphic?
$endgroup$
– awsomeguy
Dec 14 '18 at 0:14
$begingroup$
They are, but $H$ is not a subgroup of $K$ strictly speaking. For a subgroup to be normal, it has to be subgroup in the first place, and $H$ is not a subset (let alone a subgroup) of $K$.
$endgroup$
– Guido A.
Dec 14 '18 at 0:17
$begingroup$
@awsomeguy by the way, a hypothesis was missing from the converse statement, we need both $GH = K$ and $G cap H = 1$. I have edited it accordingly :)
$endgroup$
– Guido A.
Dec 14 '18 at 0:20
$begingroup$
Why is it that ${1} times H$ is correct whereas $H$ alone is incorrect? Are the two not isomorphic?
$endgroup$
– awsomeguy
Dec 14 '18 at 0:14
$begingroup$
Why is it that ${1} times H$ is correct whereas $H$ alone is incorrect? Are the two not isomorphic?
$endgroup$
– awsomeguy
Dec 14 '18 at 0:14
$begingroup$
They are, but $H$ is not a subgroup of $K$ strictly speaking. For a subgroup to be normal, it has to be subgroup in the first place, and $H$ is not a subset (let alone a subgroup) of $K$.
$endgroup$
– Guido A.
Dec 14 '18 at 0:17
$begingroup$
They are, but $H$ is not a subgroup of $K$ strictly speaking. For a subgroup to be normal, it has to be subgroup in the first place, and $H$ is not a subset (let alone a subgroup) of $K$.
$endgroup$
– Guido A.
Dec 14 '18 at 0:17
$begingroup$
@awsomeguy by the way, a hypothesis was missing from the converse statement, we need both $GH = K$ and $G cap H = 1$. I have edited it accordingly :)
$endgroup$
– Guido A.
Dec 14 '18 at 0:20
$begingroup$
@awsomeguy by the way, a hypothesis was missing from the converse statement, we need both $GH = K$ and $G cap H = 1$. I have edited it accordingly :)
$endgroup$
– Guido A.
Dec 14 '18 at 0:20
add a comment |
$begingroup$
@Ted Shifrin edited
$endgroup$
– awsomeguy
Dec 14 '18 at 0:00
$begingroup$
For your converse, you need more hypotheses for sure. Think about what you know if you in fact have $K = Gtimes H$.
$endgroup$
– Ted Shifrin
Dec 14 '18 at 0:03
$begingroup$
If $G,H$ are two subgroups of $K$ such that $K subset {gh, g in G, h in H}$ and $gh = hg$ and $G cap H= {1}$ then $K cong G times H$. The opposite direction is easy.
$endgroup$
– reuns
Dec 14 '18 at 0:14