Let $left{ H_i | i in Iright}$ be a family of subgroups of G. State and prove a condition which makes $cup_{i...












2












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















2












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















2












2








2


1



$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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




















  • $begingroup$
    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












1 Answer
1






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oldest

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1












$begingroup$

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 $gg' in H_k subset H square$.







share|cite|improve this answer









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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 $gg' in H_k subset H square$.







    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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 $gg' in H_k subset H square$.







      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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 $gg' in H_k subset H square$.







        share|cite|improve this answer









        $endgroup$



        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 $gg' in H_k subset H square$.








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 24 '18 at 5:20









        Guido A.Guido A.

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            Musa

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            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
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