Are all groups of order 175 abelian?












4












$begingroup$


Question. Are all groups of order 175 abelian?



I can show that there exists only one Sylow 5-subgroup of order 25, call it $H$, and one Sylow 7-subgroup of order 7, denote $K$.



I know that $K$ is cyclic, and thus abelian. I know that $|H| = p^2$, where $p=5$ is prime, and so $H$ is abelian too. I also know the $|G| = |H| cdot |K|$.



Further, I know that $G$ happens to be the direct product of these two groups as they intersect trivially, and this completes the proof.




Could somebody please explain:




  1. Why the group is the direct product, is this always so if the groups intersect trivially, and the product of the orders of subgroups matches the group order?


  2. Why the direct product is abelian. Is this always the case if the subgroups $H$ and $K$ intersect trivially, or is it because they are both abelian too?


  3. Anything else I should know?



Thanks






abstract-algebra group-theory finite-groups







share|cite|improve this question











$endgroup$












  • $begingroup$
    Ad 2) The direct product of abelian groups is abelian. As the product is direct by (1), $G cong H times K$ is abelian.
    $endgroup$
    – martini
    May 21 '12 at 11:07










  • $begingroup$
    Only because they intersect trivially?
    $endgroup$
    – rk101
    May 21 '12 at 11:10












  • $begingroup$
    For (1) , you need normality too. As in, you need $H$ and $K$ to be both normal (i.e. there is precisely one Sylow $p$-group and one Sylow $q$ groups where $|G|=p^aq^b$). This works, as for $G$ to be a direct product you need to find two normal subgroups $H$ and $K$ which intersect trivially (which you have) and such that $G=HK$ (which you also have, but proving this is slightly more subtle - but I am sure this is not outwith your abilities. For (3), there are many things you should know. See, for example, bbc.co.uk/news and tearfund.org)
    $endgroup$
    – user1729
    May 21 '12 at 11:12








  • 3




    $begingroup$
    (Also, for (1), let $G$ be a group where $G=HK$ and $Hcap K=1$. If both of $H$ and $K$ are normal $G$ is called a direct product of $H$ and $K$, $G=Htimes K$, if one one is normal ($H$ say) then $G$ is called a semidirect direct product of $H$ and $K$, $Hrtimes K$, while if neither is normal then $G$ is called a Zappa-Szep product of $H$ and $K$, $G=Hbowtie K$ (see my answer here math.stackexchange.com/questions/107781/…)
    $endgroup$
    – user1729
    May 21 '12 at 11:16








  • 1




    $begingroup$
    @rk101, see also math.stackexchange.com/questions/67407/…
    $endgroup$
    – Nicky Hekster
    May 21 '12 at 22:28
















4












$begingroup$


Question. Are all groups of order 175 abelian?



I can show that there exists only one Sylow 5-subgroup of order 25, call it $H$, and one Sylow 7-subgroup of order 7, denote $K$.



I know that $K$ is cyclic, and thus abelian. I know that $|H| = p^2$, where $p=5$ is prime, and so $H$ is abelian too. I also know the $|G| = |H| cdot |K|$.



Further, I know that $G$ happens to be the direct product of these two groups as they intersect trivially, and this completes the proof.




Could somebody please explain:




  1. Why the group is the direct product, is this always so if the groups intersect trivially, and the product of the orders of subgroups matches the group order?


  2. Why the direct product is abelian. Is this always the case if the subgroups $H$ and $K$ intersect trivially, or is it because they are both abelian too?


  3. Anything else I should know?



Thanks






abstract-algebra group-theory finite-groups







share|cite|improve this question











$endgroup$












  • $begingroup$
    Ad 2) The direct product of abelian groups is abelian. As the product is direct by (1), $G cong H times K$ is abelian.
    $endgroup$
    – martini
    May 21 '12 at 11:07










  • $begingroup$
    Only because they intersect trivially?
    $endgroup$
    – rk101
    May 21 '12 at 11:10












  • $begingroup$
    For (1) , you need normality too. As in, you need $H$ and $K$ to be both normal (i.e. there is precisely one Sylow $p$-group and one Sylow $q$ groups where $|G|=p^aq^b$). This works, as for $G$ to be a direct product you need to find two normal subgroups $H$ and $K$ which intersect trivially (which you have) and such that $G=HK$ (which you also have, but proving this is slightly more subtle - but I am sure this is not outwith your abilities. For (3), there are many things you should know. See, for example, bbc.co.uk/news and tearfund.org)
    $endgroup$
    – user1729
    May 21 '12 at 11:12








  • 3




    $begingroup$
    (Also, for (1), let $G$ be a group where $G=HK$ and $Hcap K=1$. If both of $H$ and $K$ are normal $G$ is called a direct product of $H$ and $K$, $G=Htimes K$, if one one is normal ($H$ say) then $G$ is called a semidirect direct product of $H$ and $K$, $Hrtimes K$, while if neither is normal then $G$ is called a Zappa-Szep product of $H$ and $K$, $G=Hbowtie K$ (see my answer here math.stackexchange.com/questions/107781/…)
    $endgroup$
    – user1729
    May 21 '12 at 11:16








  • 1




    $begingroup$
    @rk101, see also math.stackexchange.com/questions/67407/…
    $endgroup$
    – Nicky Hekster
    May 21 '12 at 22:28














4












4








4


1



$begingroup$


Question. Are all groups of order 175 abelian?



I can show that there exists only one Sylow 5-subgroup of order 25, call it $H$, and one Sylow 7-subgroup of order 7, denote $K$.



I know that $K$ is cyclic, and thus abelian. I know that $|H| = p^2$, where $p=5$ is prime, and so $H$ is abelian too. I also know the $|G| = |H| cdot |K|$.



Further, I know that $G$ happens to be the direct product of these two groups as they intersect trivially, and this completes the proof.




Could somebody please explain:




  1. Why the group is the direct product, is this always so if the groups intersect trivially, and the product of the orders of subgroups matches the group order?


  2. Why the direct product is abelian. Is this always the case if the subgroups $H$ and $K$ intersect trivially, or is it because they are both abelian too?


  3. Anything else I should know?



Thanks






abstract-algebra group-theory finite-groups







share|cite|improve this question











$endgroup$




Question. Are all groups of order 175 abelian?



I can show that there exists only one Sylow 5-subgroup of order 25, call it $H$, and one Sylow 7-subgroup of order 7, denote $K$.



I know that $K$ is cyclic, and thus abelian. I know that $|H| = p^2$, where $p=5$ is prime, and so $H$ is abelian too. I also know the $|G| = |H| cdot |K|$.



Further, I know that $G$ happens to be the direct product of these two groups as they intersect trivially, and this completes the proof.




Could somebody please explain:




  1. Why the group is the direct product, is this always so if the groups intersect trivially, and the product of the orders of subgroups matches the group order?


  2. Why the direct product is abelian. Is this always the case if the subgroups $H$ and $K$ intersect trivially, or is it because they are both abelian too?


  3. Anything else I should know?



Thanks






abstract-algebra group-theory finite-groups




abstract-algebra group-theory finite-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 23:37









the_fox

2,58711533




2,58711533










asked May 21 '12 at 11:03









rk101rk101

12829




12829












  • $begingroup$
    Ad 2) The direct product of abelian groups is abelian. As the product is direct by (1), $G cong H times K$ is abelian.
    $endgroup$
    – martini
    May 21 '12 at 11:07










  • $begingroup$
    Only because they intersect trivially?
    $endgroup$
    – rk101
    May 21 '12 at 11:10












  • $begingroup$
    For (1) , you need normality too. As in, you need $H$ and $K$ to be both normal (i.e. there is precisely one Sylow $p$-group and one Sylow $q$ groups where $|G|=p^aq^b$). This works, as for $G$ to be a direct product you need to find two normal subgroups $H$ and $K$ which intersect trivially (which you have) and such that $G=HK$ (which you also have, but proving this is slightly more subtle - but I am sure this is not outwith your abilities. For (3), there are many things you should know. See, for example, bbc.co.uk/news and tearfund.org)
    $endgroup$
    – user1729
    May 21 '12 at 11:12








  • 3




    $begingroup$
    (Also, for (1), let $G$ be a group where $G=HK$ and $Hcap K=1$. If both of $H$ and $K$ are normal $G$ is called a direct product of $H$ and $K$, $G=Htimes K$, if one one is normal ($H$ say) then $G$ is called a semidirect direct product of $H$ and $K$, $Hrtimes K$, while if neither is normal then $G$ is called a Zappa-Szep product of $H$ and $K$, $G=Hbowtie K$ (see my answer here math.stackexchange.com/questions/107781/…)
    $endgroup$
    – user1729
    May 21 '12 at 11:16








  • 1




    $begingroup$
    @rk101, see also math.stackexchange.com/questions/67407/…
    $endgroup$
    – Nicky Hekster
    May 21 '12 at 22:28


















  • $begingroup$
    Ad 2) The direct product of abelian groups is abelian. As the product is direct by (1), $G cong H times K$ is abelian.
    $endgroup$
    – martini
    May 21 '12 at 11:07










  • $begingroup$
    Only because they intersect trivially?
    $endgroup$
    – rk101
    May 21 '12 at 11:10












  • $begingroup$
    For (1) , you need normality too. As in, you need $H$ and $K$ to be both normal (i.e. there is precisely one Sylow $p$-group and one Sylow $q$ groups where $|G|=p^aq^b$). This works, as for $G$ to be a direct product you need to find two normal subgroups $H$ and $K$ which intersect trivially (which you have) and such that $G=HK$ (which you also have, but proving this is slightly more subtle - but I am sure this is not outwith your abilities. For (3), there are many things you should know. See, for example, bbc.co.uk/news and tearfund.org)
    $endgroup$
    – user1729
    May 21 '12 at 11:12








  • 3




    $begingroup$
    (Also, for (1), let $G$ be a group where $G=HK$ and $Hcap K=1$. If both of $H$ and $K$ are normal $G$ is called a direct product of $H$ and $K$, $G=Htimes K$, if one one is normal ($H$ say) then $G$ is called a semidirect direct product of $H$ and $K$, $Hrtimes K$, while if neither is normal then $G$ is called a Zappa-Szep product of $H$ and $K$, $G=Hbowtie K$ (see my answer here math.stackexchange.com/questions/107781/…)
    $endgroup$
    – user1729
    May 21 '12 at 11:16








  • 1




    $begingroup$
    @rk101, see also math.stackexchange.com/questions/67407/…
    $endgroup$
    – Nicky Hekster
    May 21 '12 at 22:28
















$begingroup$
Ad 2) The direct product of abelian groups is abelian. As the product is direct by (1), $G cong H times K$ is abelian.
$endgroup$
– martini
May 21 '12 at 11:07




$begingroup$
Ad 2) The direct product of abelian groups is abelian. As the product is direct by (1), $G cong H times K$ is abelian.
$endgroup$
– martini
May 21 '12 at 11:07












$begingroup$
Only because they intersect trivially?
$endgroup$
– rk101
May 21 '12 at 11:10






$begingroup$
Only because they intersect trivially?
$endgroup$
– rk101
May 21 '12 at 11:10














$begingroup$
For (1) , you need normality too. As in, you need $H$ and $K$ to be both normal (i.e. there is precisely one Sylow $p$-group and one Sylow $q$ groups where $|G|=p^aq^b$). This works, as for $G$ to be a direct product you need to find two normal subgroups $H$ and $K$ which intersect trivially (which you have) and such that $G=HK$ (which you also have, but proving this is slightly more subtle - but I am sure this is not outwith your abilities. For (3), there are many things you should know. See, for example, bbc.co.uk/news and tearfund.org)
$endgroup$
– user1729
May 21 '12 at 11:12






$begingroup$
For (1) , you need normality too. As in, you need $H$ and $K$ to be both normal (i.e. there is precisely one Sylow $p$-group and one Sylow $q$ groups where $|G|=p^aq^b$). This works, as for $G$ to be a direct product you need to find two normal subgroups $H$ and $K$ which intersect trivially (which you have) and such that $G=HK$ (which you also have, but proving this is slightly more subtle - but I am sure this is not outwith your abilities. For (3), there are many things you should know. See, for example, bbc.co.uk/news and tearfund.org)
$endgroup$
– user1729
May 21 '12 at 11:12






3




3




$begingroup$
(Also, for (1), let $G$ be a group where $G=HK$ and $Hcap K=1$. If both of $H$ and $K$ are normal $G$ is called a direct product of $H$ and $K$, $G=Htimes K$, if one one is normal ($H$ say) then $G$ is called a semidirect direct product of $H$ and $K$, $Hrtimes K$, while if neither is normal then $G$ is called a Zappa-Szep product of $H$ and $K$, $G=Hbowtie K$ (see my answer here math.stackexchange.com/questions/107781/…)
$endgroup$
– user1729
May 21 '12 at 11:16






$begingroup$
(Also, for (1), let $G$ be a group where $G=HK$ and $Hcap K=1$. If both of $H$ and $K$ are normal $G$ is called a direct product of $H$ and $K$, $G=Htimes K$, if one one is normal ($H$ say) then $G$ is called a semidirect direct product of $H$ and $K$, $Hrtimes K$, while if neither is normal then $G$ is called a Zappa-Szep product of $H$ and $K$, $G=Hbowtie K$ (see my answer here math.stackexchange.com/questions/107781/…)
$endgroup$
– user1729
May 21 '12 at 11:16






1




1




$begingroup$
@rk101, see also math.stackexchange.com/questions/67407/…
$endgroup$
– Nicky Hekster
May 21 '12 at 22:28




$begingroup$
@rk101, see also math.stackexchange.com/questions/67407/…
$endgroup$
– Nicky Hekster
May 21 '12 at 22:28










1 Answer
1






active

oldest

votes


















3












$begingroup$

Your questions:



1) This is not enough: it must be also that both sbgps. are normal in G



2) Because they're abelian, too.



3) Lots more, as anyone else...but not for this particular question, imo.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why do we need to show they're normal?
    $endgroup$
    – TheLast Cipher
    Sep 26 '18 at 11:21










  • $begingroup$
    Is it because we want to show that $G$ is a direct product?
    $endgroup$
    – TheLast Cipher
    Sep 26 '18 at 11:54











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

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%2f147701%2fare-all-groups-of-order-175-abelian%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









3












$begingroup$

Your questions:



1) This is not enough: it must be also that both sbgps. are normal in G



2) Because they're abelian, too.



3) Lots more, as anyone else...but not for this particular question, imo.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why do we need to show they're normal?
    $endgroup$
    – TheLast Cipher
    Sep 26 '18 at 11:21










  • $begingroup$
    Is it because we want to show that $G$ is a direct product?
    $endgroup$
    – TheLast Cipher
    Sep 26 '18 at 11:54
















3












$begingroup$

Your questions:



1) This is not enough: it must be also that both sbgps. are normal in G



2) Because they're abelian, too.



3) Lots more, as anyone else...but not for this particular question, imo.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why do we need to show they're normal?
    $endgroup$
    – TheLast Cipher
    Sep 26 '18 at 11:21










  • $begingroup$
    Is it because we want to show that $G$ is a direct product?
    $endgroup$
    – TheLast Cipher
    Sep 26 '18 at 11:54














3












3








3





$begingroup$

Your questions:



1) This is not enough: it must be also that both sbgps. are normal in G



2) Because they're abelian, too.



3) Lots more, as anyone else...but not for this particular question, imo.






share|cite|improve this answer









$endgroup$



Your questions:



1) This is not enough: it must be also that both sbgps. are normal in G



2) Because they're abelian, too.



3) Lots more, as anyone else...but not for this particular question, imo.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered May 21 '12 at 11:12









DonAntonioDonAntonio

177k1492227




177k1492227












  • $begingroup$
    Why do we need to show they're normal?
    $endgroup$
    – TheLast Cipher
    Sep 26 '18 at 11:21










  • $begingroup$
    Is it because we want to show that $G$ is a direct product?
    $endgroup$
    – TheLast Cipher
    Sep 26 '18 at 11:54


















  • $begingroup$
    Why do we need to show they're normal?
    $endgroup$
    – TheLast Cipher
    Sep 26 '18 at 11:21










  • $begingroup$
    Is it because we want to show that $G$ is a direct product?
    $endgroup$
    – TheLast Cipher
    Sep 26 '18 at 11:54
















$begingroup$
Why do we need to show they're normal?
$endgroup$
– TheLast Cipher
Sep 26 '18 at 11:21




$begingroup$
Why do we need to show they're normal?
$endgroup$
– TheLast Cipher
Sep 26 '18 at 11:21












$begingroup$
Is it because we want to show that $G$ is a direct product?
$endgroup$
– TheLast Cipher
Sep 26 '18 at 11:54




$begingroup$
Is it because we want to show that $G$ is a direct product?
$endgroup$
– TheLast Cipher
Sep 26 '18 at 11:54


















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%2f147701%2fare-all-groups-of-order-175-abelian%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