Let $left{ H_i | i in Iright}$ be a family of subgroups of G. State and prove a condition which makes $cup_{i...
$begingroup$
Let $G$ be a group and $left{ H_i | i in Iright}$ be a family of subgroups of G. State and prove a condition which makes $cup_{i in I} H_{i}$ a subgroup of $G$, that is that $cup _{i in I} H_{i} = left< cup_{i in I} H_{i} right>$.
To show that $cup _{i in I} H_{i} subseteq left< cup _{i in I} H_{i} right>$, it seems that I have to show that $cup _{i in I} H_{i}$ is a subgroup of G. Since $left< cup _{i in I} H_{i} right>$ is a collection of subgroups of G that contain $cup _{i in I} H_{i}$ as a subset, I would have the first inclusion proven and quite possibly get the second inclusion for free.
However, I am having a hard time on how to prove $cup _{i in I} H_{i}$ is a subgroup of $G$. The only way this could happen is the collection of subgroups ${ H_{i} | i in I}$ is a chain.
Is there another way to show that $cup _{i in I} H_{i}$ is a subgroup of $G$?
Am I thinking about this the right way?
abstract-algebra group-theory alternative-proof cyclic-groups
edited Feb 12 at 14:41
rschwieb
108k12105253
$endgroup$
|
show 6 more comments
$begingroup$
Let $G$ be a group and $left{ H_i | i in Iright}$ be a family of subgroups of G. State and prove a condition which makes $cup_{i in I} H_{i}$ a subgroup of $G$, that is that $cup _{i in I} H_{i} = left< cup_{i in I} H_{i} right>$.
To show that $cup _{i in I} H_{i} subseteq left< cup _{i in I} H_{i} right>$, it seems that I have to show that $cup _{i in I} H_{i}$ is a subgroup of G. Since $left< cup _{i in I} H_{i} right>$ is a collection of subgroups of G that contain $cup _{i in I} H_{i}$ as a subset, I would have the first inclusion proven and quite possibly get the second inclusion for free.
However, I am having a hard time on how to prove $cup _{i in I} H_{i}$ is a subgroup of $G$. The only way this could happen is the collection of subgroups ${ H_{i} | i in I}$ is a chain.
Is there another way to show that $cup _{i in I} H_{i}$ is a subgroup of $G$?
Am I thinking about this the right way?
abstract-algebra group-theory alternative-proof cyclic-groups
edited Feb 12 at 14:41
rschwieb
108k12105253
$endgroup$
$begingroup$
Yes, you are right that chaining is the only way this can happen. Try first just thinking about two subgroups $H_1$ and $H_2$. The way you wrote your reasoning though is very convoluted and confusing. I'll just say that $cup _{i in I} H_{i} subseteq left< cup _{i in I} H_{i} right>$ is the trivial direction. You want to show that anything in the generated subgroup is already in the union, i.e. the reverse direction.
$endgroup$
– zoidberg
Dec 24 '18 at 5:21
$begingroup$
@norfair although chaining surely gives a positive result, we only need the $H_i$ to be 'directed' in some sense. For example, if for all $i,j$ we have $k$ with $H_i,H_j subset H_k$, this is enough to prove that the union is closed with respect to multiplication, and weaker than having a partial order with respect to inclusion.
$endgroup$
– Guido A.
Dec 24 '18 at 5:30
$begingroup$
Yes, my mistake. It doesn't have to be a total ordering. I'm confused why you say it's weaker than partial order though. Isn't that exactly the characterization?
$endgroup$
– zoidberg
Dec 24 '18 at 5:35
$begingroup$
@norfair My bad. I meant to say that 'directedness' suffices but we get anti-symmetry for free as the order in defined by inclusion, which is a partial order on $mathcal{P}G$ and it is weaker than having a total order (i.e. a chain).
$endgroup$
– Guido A.
Dec 24 '18 at 5:38
$begingroup$
@norfair note that however it is not enough for the groups to be partially ordered. Even if we take two groups $H_1, H_2$ , it may very well be that $H_1 cup H_2$ is not a subgroup.
$endgroup$
– Guido A.
Dec 24 '18 at 5:40
|
show 6 more comments
$begingroup$
Let $G$ be a group and $left{ H_i | i in Iright}$ be a family of subgroups of G. State and prove a condition which makes $cup_{i in I} H_{i}$ a subgroup of $G$, that is that $cup _{i in I} H_{i} = left< cup_{i in I} H_{i} right>$.
To show that $cup _{i in I} H_{i} subseteq left< cup _{i in I} H_{i} right>$, it seems that I have to show that $cup _{i in I} H_{i}$ is a subgroup of G. Since $left< cup _{i in I} H_{i} right>$ is a collection of subgroups of G that contain $cup _{i in I} H_{i}$ as a subset, I would have the first inclusion proven and quite possibly get the second inclusion for free.
However, I am having a hard time on how to prove $cup _{i in I} H_{i}$ is a subgroup of $G$. The only way this could happen is the collection of subgroups ${ H_{i} | i in I}$ is a chain.
Is there another way to show that $cup _{i in I} H_{i}$ is a subgroup of $G$?
Am I thinking about this the right way?
abstract-algebra group-theory alternative-proof cyclic-groups
edited Feb 12 at 14:41
rschwieb
108k12105253
$endgroup$
Let $G$ be a group and $left{ H_i | i in Iright}$ be a family of subgroups of G. State and prove a condition which makes $cup_{i in I} H_{i}$ a subgroup of $G$, that is that $cup _{i in I} H_{i} = left< cup_{i in I} H_{i} right>$.
To show that $cup _{i in I} H_{i} subseteq left< cup _{i in I} H_{i} right>$, it seems that I have to show that $cup _{i in I} H_{i}$ is a subgroup of G. Since $left< cup _{i in I} H_{i} right>$ is a collection of subgroups of G that contain $cup _{i in I} H_{i}$ as a subset, I would have the first inclusion proven and quite possibly get the second inclusion for free.
However, I am having a hard time on how to prove $cup _{i in I} H_{i}$ is a subgroup of $G$. The only way this could happen is the collection of subgroups ${ H_{i} | i in I}$ is a chain.
Is there another way to show that $cup _{i in I} H_{i}$ is a subgroup of $G$?
Am I thinking about this the right way?
abstract-algebra group-theory alternative-proof cyclic-groups
abstract-algebra group-theory alternative-proof cyclic-groups
edited Feb 12 at 14:41
rschwieb
108k12105253
edited Feb 12 at 14:41
rschwieb
108k12105253
edited Feb 12 at 14:41
rschwieb
108k12105253
edited Feb 12 at 14:41
rschwieb
108k12105253
edited Feb 12 at 14:41
rschwieb
108k12105253
108k12105253
asked Dec 24 '18 at 4:52
user586464
$begingroup$
Yes, you are right that chaining is the only way this can happen. Try first just thinking about two subgroups $H_1$ and $H_2$. The way you wrote your reasoning though is very convoluted and confusing. I'll just say that $cup _{i in I} H_{i} subseteq left< cup _{i in I} H_{i} right>$ is the trivial direction. You want to show that anything in the generated subgroup is already in the union, i.e. the reverse direction.
$endgroup$
– zoidberg
Dec 24 '18 at 5:21
$begingroup$
@norfair although chaining surely gives a positive result, we only need the $H_i$ to be 'directed' in some sense. For example, if for all $i,j$ we have $k$ with $H_i,H_j subset H_k$, this is enough to prove that the union is closed with respect to multiplication, and weaker than having a partial order with respect to inclusion.
$endgroup$
– Guido A.
Dec 24 '18 at 5:30
$begingroup$
Yes, my mistake. It doesn't have to be a total ordering. I'm confused why you say it's weaker than partial order though. Isn't that exactly the characterization?
$endgroup$
– zoidberg
Dec 24 '18 at 5:35
$begingroup$
@norfair My bad. I meant to say that 'directedness' suffices but we get anti-symmetry for free as the order in defined by inclusion, which is a partial order on $mathcal{P}G$ and it is weaker than having a total order (i.e. a chain).
$endgroup$
– Guido A.
Dec 24 '18 at 5:38
$begingroup$
@norfair note that however it is not enough for the groups to be partially ordered. Even if we take two groups $H_1, H_2$ , it may very well be that $H_1 cup H_2$ is not a subgroup.
$endgroup$
– Guido A.
Dec 24 '18 at 5:40
|
show 6 more comments
$begingroup$
Yes, you are right that chaining is the only way this can happen. Try first just thinking about two subgroups $H_1$ and $H_2$. The way you wrote your reasoning though is very convoluted and confusing. I'll just say that $cup _{i in I} H_{i} subseteq left< cup _{i in I} H_{i} right>$ is the trivial direction. You want to show that anything in the generated subgroup is already in the union, i.e. the reverse direction.
$endgroup$
– zoidberg
Dec 24 '18 at 5:21
$begingroup$
@norfair although chaining surely gives a positive result, we only need the $H_i$ to be 'directed' in some sense. For example, if for all $i,j$ we have $k$ with $H_i,H_j subset H_k$, this is enough to prove that the union is closed with respect to multiplication, and weaker than having a partial order with respect to inclusion.
$endgroup$
– Guido A.
Dec 24 '18 at 5:30
$begingroup$
Yes, my mistake. It doesn't have to be a total ordering. I'm confused why you say it's weaker than partial order though. Isn't that exactly the characterization?
$endgroup$
– zoidberg
Dec 24 '18 at 5:35
$begingroup$
@norfair My bad. I meant to say that 'directedness' suffices but we get anti-symmetry for free as the order in defined by inclusion, which is a partial order on $mathcal{P}G$ and it is weaker than having a total order (i.e. a chain).
$endgroup$
– Guido A.
Dec 24 '18 at 5:38
$begingroup$
@norfair note that however it is not enough for the groups to be partially ordered. Even if we take two groups $H_1, H_2$ , it may very well be that $H_1 cup H_2$ is not a subgroup.
$endgroup$
– Guido A.
Dec 24 '18 at 5:40
$begingroup$
Yes, you are right that chaining is the only way this can happen. Try first just thinking about two subgroups $H_1$ and $H_2$. The way you wrote your reasoning though is very convoluted and confusing. I'll just say that $cup _{i in I} H_{i} subseteq left< cup _{i in I} H_{i} right>$ is the trivial direction. You want to show that anything in the generated subgroup is already in the union, i.e. the reverse direction.
$endgroup$
– zoidberg
Dec 24 '18 at 5:21
$begingroup$
Yes, you are right that chaining is the only way this can happen. Try first just thinking about two subgroups $H_1$ and $H_2$. The way you wrote your reasoning though is very convoluted and confusing. I'll just say that $cup _{i in I} H_{i} subseteq left< cup _{i in I} H_{i} right>$ is the trivial direction. You want to show that anything in the generated subgroup is already in the union, i.e. the reverse direction.
$endgroup$
– zoidberg
Dec 24 '18 at 5:21
$begingroup$
@norfair although chaining surely gives a positive result, we only need the $H_i$ to be 'directed' in some sense. For example, if for all $i,j$ we have $k$ with $H_i,H_j subset H_k$, this is enough to prove that the union is closed with respect to multiplication, and weaker than having a partial order with respect to inclusion.
$endgroup$
– Guido A.
Dec 24 '18 at 5:30
$begingroup$
@norfair although chaining surely gives a positive result, we only need the $H_i$ to be 'directed' in some sense. For example, if for all $i,j$ we have $k$ with $H_i,H_j subset H_k$, this is enough to prove that the union is closed with respect to multiplication, and weaker than having a partial order with respect to inclusion.
$endgroup$
– Guido A.
Dec 24 '18 at 5:30
$begingroup$
Yes, my mistake. It doesn't have to be a total ordering. I'm confused why you say it's weaker than partial order though. Isn't that exactly the characterization?
$endgroup$
– zoidberg
Dec 24 '18 at 5:35
$begingroup$
Yes, my mistake. It doesn't have to be a total ordering. I'm confused why you say it's weaker than partial order though. Isn't that exactly the characterization?
$endgroup$
– zoidberg
Dec 24 '18 at 5:35
$begingroup$
@norfair My bad. I meant to say that 'directedness' suffices but we get anti-symmetry for free as the order in defined by inclusion, which is a partial order on $mathcal{P}G$ and it is weaker than having a total order (i.e. a chain).
$endgroup$
– Guido A.
Dec 24 '18 at 5:38
$begingroup$
@norfair My bad. I meant to say that 'directedness' suffices but we get anti-symmetry for free as the order in defined by inclusion, which is a partial order on $mathcal{P}G$ and it is weaker than having a total order (i.e. a chain).
$endgroup$
– Guido A.
Dec 24 '18 at 5:38
$begingroup$
@norfair note that however it is not enough for the groups to be partially ordered. Even if we take two groups $H_1, H_2$ , it may very well be that $H_1 cup H_2$ is not a subgroup.
$endgroup$
– Guido A.
Dec 24 '18 at 5:40
$begingroup$
@norfair note that however it is not enough for the groups to be partially ordered. Even if we take two groups $H_1, H_2$ , it may very well be that $H_1 cup H_2$ is not a subgroup.
$endgroup$
– Guido A.
Dec 24 '18 at 5:40
|
show 6 more comments
1 Answer
1
active
oldest
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$begingroup$
As stated, it seems like the condition is of your choosing. You are on the right track when thinking about a collection of 'chained' groups. I would advise you to try and prove this for a countable collection $G_1 subset G_2 dots$. This could lead you to stronger results which only use the essentials of the previous proof.
A sketch of a possible generalization is below,
Suppose that $(H_i)_{i in I}$ is such that for each $g in H_i, g' in H_j$ there exists $k in I$ with $g,g' in H_k$. Then, the set $H = bigcup_{i in I} H_i$ is a subgroup of $G$: clearly we have that $1 in G$; and if $g in H_i subset H$, then $g^{-1} in H_i subset H$. So far we have used no assumptions. Now, take $g,g' in H$. Hence there exist $i,j in I$ with $g in H_i$ and $g in H_j$. By hypothesis we have $k in I$ with $g,g' in H_k$ and so $gg' in H_k subset H square$.
answered Dec 24 '18 at 5:20
Guido A.Guido A.
8,3351730
$endgroup$
add a comment |
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$begingroup$
As stated, it seems like the condition is of your choosing. You are on the right track when thinking about a collection of 'chained' groups. I would advise you to try and prove this for a countable collection $G_1 subset G_2 dots$. This could lead you to stronger results which only use the essentials of the previous proof.
A sketch of a possible generalization is below,
Suppose that $(H_i)_{i in I}$ is such that for each $g in H_i, g' in H_j$ there exists $k in I$ with $g,g' in H_k$. Then, the set $H = bigcup_{i in I} H_i$ is a subgroup of $G$: clearly we have that $1 in G$; and if $g in H_i subset H$, then $g^{-1} in H_i subset H$. So far we have used no assumptions. Now, take $g,g' in H$. Hence there exist $i,j in I$ with $g in H_i$ and $g in H_j$. By hypothesis we have $k in I$ with $g,g' in H_k$ and so $gg' in H_k subset H square$.
answered Dec 24 '18 at 5:20
Guido A.Guido A.
8,3351730
$endgroup$
add a comment |
$begingroup$
As stated, it seems like the condition is of your choosing. You are on the right track when thinking about a collection of 'chained' groups. I would advise you to try and prove this for a countable collection $G_1 subset G_2 dots$. This could lead you to stronger results which only use the essentials of the previous proof.
A sketch of a possible generalization is below,
Suppose that $(H_i)_{i in I}$ is such that for each $g in H_i, g' in H_j$ there exists $k in I$ with $g,g' in H_k$. Then, the set $H = bigcup_{i in I} H_i$ is a subgroup of $G$: clearly we have that $1 in G$; and if $g in H_i subset H$, then $g^{-1} in H_i subset H$. So far we have used no assumptions. Now, take $g,g' in H$. Hence there exist $i,j in I$ with $g in H_i$ and $g in H_j$. By hypothesis we have $k in I$ with $g,g' in H_k$ and so $gg' in H_k subset H square$.
answered Dec 24 '18 at 5:20
Guido A.Guido A.
8,3351730
$endgroup$
add a comment |
$begingroup$
As stated, it seems like the condition is of your choosing. You are on the right track when thinking about a collection of 'chained' groups. I would advise you to try and prove this for a countable collection $G_1 subset G_2 dots$. This could lead you to stronger results which only use the essentials of the previous proof.
A sketch of a possible generalization is below,
Suppose that $(H_i)_{i in I}$ is such that for each $g in H_i, g' in H_j$ there exists $k in I$ with $g,g' in H_k$. Then, the set $H = bigcup_{i in I} H_i$ is a subgroup of $G$: clearly we have that $1 in G$; and if $g in H_i subset H$, then $g^{-1} in H_i subset H$. So far we have used no assumptions. Now, take $g,g' in H$. Hence there exist $i,j in I$ with $g in H_i$ and $g in H_j$. By hypothesis we have $k in I$ with $g,g' in H_k$ and so $gg' in H_k subset H square$.
answered Dec 24 '18 at 5:20
Guido A.Guido A.
8,3351730
$endgroup$
As stated, it seems like the condition is of your choosing. You are on the right track when thinking about a collection of 'chained' groups. I would advise you to try and prove this for a countable collection $G_1 subset G_2 dots$. This could lead you to stronger results which only use the essentials of the previous proof.
A sketch of a possible generalization is below,
Suppose that $(H_i)_{i in I}$ is such that for each $g in H_i, g' in H_j$ there exists $k in I$ with $g,g' in H_k$. Then, the set $H = bigcup_{i in I} H_i$ is a subgroup of $G$: clearly we have that $1 in G$; and if $g in H_i subset H$, then $g^{-1} in H_i subset H$. So far we have used no assumptions. Now, take $g,g' in H$. Hence there exist $i,j in I$ with $g in H_i$ and $g in H_j$. By hypothesis we have $k in I$ with $g,g' in H_k$ and so $gg' in H_k subset H square$.
answered Dec 24 '18 at 5:20
Guido A.Guido A.
8,3351730
answered Dec 24 '18 at 5:20
Guido A.Guido A.
8,3351730
answered Dec 24 '18 at 5:20
Guido A.Guido A.
8,3351730
answered Dec 24 '18 at 5:20
Guido A.Guido A.
8,3351730
8,3351730
add a comment |
add a comment |
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$begingroup$
Yes, you are right that chaining is the only way this can happen. Try first just thinking about two subgroups $H_1$ and $H_2$. The way you wrote your reasoning though is very convoluted and confusing. I'll just say that $cup _{i in I} H_{i} subseteq left< cup _{i in I} H_{i} right>$ is the trivial direction. You want to show that anything in the generated subgroup is already in the union, i.e. the reverse direction.
$endgroup$
– zoidberg
Dec 24 '18 at 5:21
$begingroup$
@norfair although chaining surely gives a positive result, we only need the $H_i$ to be 'directed' in some sense. For example, if for all $i,j$ we have $k$ with $H_i,H_j subset H_k$, this is enough to prove that the union is closed with respect to multiplication, and weaker than having a partial order with respect to inclusion.
$endgroup$
– Guido A.
Dec 24 '18 at 5:30
$begingroup$
Yes, my mistake. It doesn't have to be a total ordering. I'm confused why you say it's weaker than partial order though. Isn't that exactly the characterization?
$endgroup$
– zoidberg
Dec 24 '18 at 5:35
$begingroup$
@norfair My bad. I meant to say that 'directedness' suffices but we get anti-symmetry for free as the order in defined by inclusion, which is a partial order on $mathcal{P}G$ and it is weaker than having a total order (i.e. a chain).
$endgroup$
– Guido A.
Dec 24 '18 at 5:38
$begingroup$
@norfair note that however it is not enough for the groups to be partially ordered. Even if we take two groups $H_1, H_2$ , it may very well be that $H_1 cup H_2$ is not a subgroup.
$endgroup$
– Guido A.
Dec 24 '18 at 5:40