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






active

oldest

votes


















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









$endgroup$














    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050948%2flet-left-h-i-i-in-i-right-be-a-family-of-subgroups-of-g-state-and-pro%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown
























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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












      $begingroup$

      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.

        8,3351730




        8,3351730






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050948%2flet-left-h-i-i-in-i-right-be-a-family-of-subgroups-of-g-state-and-pro%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
            Read more

            Puebla de Zaragoza

            Image
            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
            Read more

            Musa

            Image
            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...
            Read more