Lagrange's Theorem: Injectivity.












4














Let $U$ be a subgroup of the finite group $G$ and $ain G$.



To show Lagrange's Theorem we defined a map $U rightarrow aU$, $umapsto au$ and said that it was bijective. Now for the injectivity we used the trivial case that for $au_1 = au_2Leftrightarrow u_1=u_2 $ but this uses that we add $a^{-1}$ from the left side in both cases.




Now my question is that of course there exists a $a^{-1}in G$ but don't we need to have $a^{-1}in aU$ to add it from the left? And if so, how can we show $a^{-1}in aU$?




Best,
KingDingeling






abstract-algebra group-theory finite-groups







share|cite|improve this question
























  • What are the domain and the codomain of your function?
    – José Carlos Santos
    8 hours ago










  • Domain is $U$ and codomain is $aU$.
    – KingDingeling
    8 hours ago










  • How can a function whose domain is $U$ be defined as $Umapsto aU$? And what is $a$ then?
    – José Carlos Santos
    8 hours ago










  • $a$ is an element of the Group $G$. It's a basic lefttranslation.
    – KingDingeling
    8 hours ago












  • Let U be trivial. Is $a^{-1}$ in $aU$?
    – Andres Mejia
    7 hours ago
















4














Let $U$ be a subgroup of the finite group $G$ and $ain G$.



To show Lagrange's Theorem we defined a map $U rightarrow aU$, $umapsto au$ and said that it was bijective. Now for the injectivity we used the trivial case that for $au_1 = au_2Leftrightarrow u_1=u_2 $ but this uses that we add $a^{-1}$ from the left side in both cases.




Now my question is that of course there exists a $a^{-1}in G$ but don't we need to have $a^{-1}in aU$ to add it from the left? And if so, how can we show $a^{-1}in aU$?




Best,
KingDingeling






abstract-algebra group-theory finite-groups







share|cite|improve this question
























  • What are the domain and the codomain of your function?
    – José Carlos Santos
    8 hours ago










  • Domain is $U$ and codomain is $aU$.
    – KingDingeling
    8 hours ago










  • How can a function whose domain is $U$ be defined as $Umapsto aU$? And what is $a$ then?
    – José Carlos Santos
    8 hours ago










  • $a$ is an element of the Group $G$. It's a basic lefttranslation.
    – KingDingeling
    8 hours ago












  • Let U be trivial. Is $a^{-1}$ in $aU$?
    – Andres Mejia
    7 hours ago














4












4








4







Let $U$ be a subgroup of the finite group $G$ and $ain G$.



To show Lagrange's Theorem we defined a map $U rightarrow aU$, $umapsto au$ and said that it was bijective. Now for the injectivity we used the trivial case that for $au_1 = au_2Leftrightarrow u_1=u_2 $ but this uses that we add $a^{-1}$ from the left side in both cases.




Now my question is that of course there exists a $a^{-1}in G$ but don't we need to have $a^{-1}in aU$ to add it from the left? And if so, how can we show $a^{-1}in aU$?




Best,
KingDingeling






abstract-algebra group-theory finite-groups







share|cite|improve this question















Let $U$ be a subgroup of the finite group $G$ and $ain G$.



To show Lagrange's Theorem we defined a map $U rightarrow aU$, $umapsto au$ and said that it was bijective. Now for the injectivity we used the trivial case that for $au_1 = au_2Leftrightarrow u_1=u_2 $ but this uses that we add $a^{-1}$ from the left side in both cases.




Now my question is that of course there exists a $a^{-1}in G$ but don't we need to have $a^{-1}in aU$ to add it from the left? And if so, how can we show $a^{-1}in aU$?




Best,
KingDingeling






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 7 hours ago









janmarqz

6,17241630




6,17241630










asked 8 hours ago









KingDingeling

435




435












  • What are the domain and the codomain of your function?
    – José Carlos Santos
    8 hours ago










  • Domain is $U$ and codomain is $aU$.
    – KingDingeling
    8 hours ago










  • How can a function whose domain is $U$ be defined as $Umapsto aU$? And what is $a$ then?
    – José Carlos Santos
    8 hours ago










  • $a$ is an element of the Group $G$. It's a basic lefttranslation.
    – KingDingeling
    8 hours ago












  • Let U be trivial. Is $a^{-1}$ in $aU$?
    – Andres Mejia
    7 hours ago


















  • What are the domain and the codomain of your function?
    – José Carlos Santos
    8 hours ago










  • Domain is $U$ and codomain is $aU$.
    – KingDingeling
    8 hours ago










  • How can a function whose domain is $U$ be defined as $Umapsto aU$? And what is $a$ then?
    – José Carlos Santos
    8 hours ago










  • $a$ is an element of the Group $G$. It's a basic lefttranslation.
    – KingDingeling
    8 hours ago












  • Let U be trivial. Is $a^{-1}$ in $aU$?
    – Andres Mejia
    7 hours ago
















What are the domain and the codomain of your function?
– José Carlos Santos
8 hours ago




What are the domain and the codomain of your function?
– José Carlos Santos
8 hours ago












Domain is $U$ and codomain is $aU$.
– KingDingeling
8 hours ago




Domain is $U$ and codomain is $aU$.
– KingDingeling
8 hours ago












How can a function whose domain is $U$ be defined as $Umapsto aU$? And what is $a$ then?
– José Carlos Santos
8 hours ago




How can a function whose domain is $U$ be defined as $Umapsto aU$? And what is $a$ then?
– José Carlos Santos
8 hours ago












$a$ is an element of the Group $G$. It's a basic lefttranslation.
– KingDingeling
8 hours ago






$a$ is an element of the Group $G$. It's a basic lefttranslation.
– KingDingeling
8 hours ago














Let U be trivial. Is $a^{-1}$ in $aU$?
– Andres Mejia
7 hours ago




Let U be trivial. Is $a^{-1}$ in $aU$?
– Andres Mejia
7 hours ago










2 Answers
2






active

oldest

votes


















2














Okay so essentially what you're trying to show is that if $au_1=au_2$ then $u_1=u_2$, and you seem to think that we are restricted completely to working within $aU$ to do this. But this isn't the case at all, in fact since $u_1$ and $u_2$ may not be in $aU$ (they are not in the case that $anotin U$), so it doesn't even make sense that we could show $u_1=u_2$ while working entirely within the set $aU$.



The point is that these equalities we are worried about are occurring within $G$, so it is valid to use any element of $G$ to show this claim.






share|cite|improve this answer





















  • Thank you for taking time and explaining :)
    – KingDingeling
    6 hours ago



















3














You don’t need to have $a^{-1} in aU$ and won’t it by the way in mot cases.



This is by the way not required to prove the injectivity. The argument using $a^{-1}$ is not making that hypothesis.






share|cite|improve this answer





















  • But if I take an element in $aU$ to prove the injecticity, may I just use $a^{-1}$ even if it isn't in $aU$?
    – KingDingeling
    8 hours ago













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%2f3057946%2flagranges-theorem-injectivity%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














Okay so essentially what you're trying to show is that if $au_1=au_2$ then $u_1=u_2$, and you seem to think that we are restricted completely to working within $aU$ to do this. But this isn't the case at all, in fact since $u_1$ and $u_2$ may not be in $aU$ (they are not in the case that $anotin U$), so it doesn't even make sense that we could show $u_1=u_2$ while working entirely within the set $aU$.



The point is that these equalities we are worried about are occurring within $G$, so it is valid to use any element of $G$ to show this claim.






share|cite|improve this answer





















  • Thank you for taking time and explaining :)
    – KingDingeling
    6 hours ago
















2














Okay so essentially what you're trying to show is that if $au_1=au_2$ then $u_1=u_2$, and you seem to think that we are restricted completely to working within $aU$ to do this. But this isn't the case at all, in fact since $u_1$ and $u_2$ may not be in $aU$ (they are not in the case that $anotin U$), so it doesn't even make sense that we could show $u_1=u_2$ while working entirely within the set $aU$.



The point is that these equalities we are worried about are occurring within $G$, so it is valid to use any element of $G$ to show this claim.






share|cite|improve this answer





















  • Thank you for taking time and explaining :)
    – KingDingeling
    6 hours ago














2












2








2






Okay so essentially what you're trying to show is that if $au_1=au_2$ then $u_1=u_2$, and you seem to think that we are restricted completely to working within $aU$ to do this. But this isn't the case at all, in fact since $u_1$ and $u_2$ may not be in $aU$ (they are not in the case that $anotin U$), so it doesn't even make sense that we could show $u_1=u_2$ while working entirely within the set $aU$.



The point is that these equalities we are worried about are occurring within $G$, so it is valid to use any element of $G$ to show this claim.






share|cite|improve this answer












Okay so essentially what you're trying to show is that if $au_1=au_2$ then $u_1=u_2$, and you seem to think that we are restricted completely to working within $aU$ to do this. But this isn't the case at all, in fact since $u_1$ and $u_2$ may not be in $aU$ (they are not in the case that $anotin U$), so it doesn't even make sense that we could show $u_1=u_2$ while working entirely within the set $aU$.



The point is that these equalities we are worried about are occurring within $G$, so it is valid to use any element of $G$ to show this claim.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 6 hours ago









Alex Mathers

10.8k21344




10.8k21344












  • Thank you for taking time and explaining :)
    – KingDingeling
    6 hours ago


















  • Thank you for taking time and explaining :)
    – KingDingeling
    6 hours ago
















Thank you for taking time and explaining :)
– KingDingeling
6 hours ago




Thank you for taking time and explaining :)
– KingDingeling
6 hours ago











3














You don’t need to have $a^{-1} in aU$ and won’t it by the way in mot cases.



This is by the way not required to prove the injectivity. The argument using $a^{-1}$ is not making that hypothesis.






share|cite|improve this answer





















  • But if I take an element in $aU$ to prove the injecticity, may I just use $a^{-1}$ even if it isn't in $aU$?
    – KingDingeling
    8 hours ago


















3














You don’t need to have $a^{-1} in aU$ and won’t it by the way in mot cases.



This is by the way not required to prove the injectivity. The argument using $a^{-1}$ is not making that hypothesis.






share|cite|improve this answer





















  • But if I take an element in $aU$ to prove the injecticity, may I just use $a^{-1}$ even if it isn't in $aU$?
    – KingDingeling
    8 hours ago
















3












3








3






You don’t need to have $a^{-1} in aU$ and won’t it by the way in mot cases.



This is by the way not required to prove the injectivity. The argument using $a^{-1}$ is not making that hypothesis.






share|cite|improve this answer












You don’t need to have $a^{-1} in aU$ and won’t it by the way in mot cases.



This is by the way not required to prove the injectivity. The argument using $a^{-1}$ is not making that hypothesis.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 8 hours ago









mathcounterexamples.net

24.4k21753




24.4k21753












  • But if I take an element in $aU$ to prove the injecticity, may I just use $a^{-1}$ even if it isn't in $aU$?
    – KingDingeling
    8 hours ago




















  • But if I take an element in $aU$ to prove the injecticity, may I just use $a^{-1}$ even if it isn't in $aU$?
    – KingDingeling
    8 hours ago


















But if I take an element in $aU$ to prove the injecticity, may I just use $a^{-1}$ even if it isn't in $aU$?
– KingDingeling
8 hours ago






But if I take an element in $aU$ to prove the injecticity, may I just use $a^{-1}$ even if it isn't in $aU$?
– KingDingeling
8 hours ago




















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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


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%2f3057946%2flagranges-theorem-injectivity%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

Eastern Orthodox Church

Image
Not to be confused with Oriental Orthodoxy, Eastern Christianity, or Eastern Catholic Churches. "Eastern Orthodoxy", "Orthodox Church", "Orthodox Catholic Church", and "Orthodox Christian Church" redirect here. For other uses, see Orthodox (disambiguation). This article may be too long to read and navigate comfortably . The readable prose size is 132 kilobytes. Please consider splitting content into sub-articles, condensing it, or adding or removing subheadings. ( January 2018 ) Part of a series on the Eastern Orthodox Church Mosaic of Christ Pantocrator, Hagia Sophia Overview Structure Theology (History of theology) Liturgy Church history Holy Mysteries View of salvation View of Mary View of icons Background Crucifixion / Resurrection / Ascension of Jesus Christianity Christian Church Apostolic succession Four Marks of the Church Orthodoxy Organization Autocephaly Patriarchate ...
Read more

Zagreb

Image
This article is about the Croatian capital city. For other uses, see Zagreb (disambiguation). City in City of Zagreb, Croatia Zagreb City Grad Zagreb City of Zagreb Clockwise, from top: St. Mark's Square, Croatian State Archives, Zagreb Tram, Art Pavilion, Cibona & HOTO towers and Croatian National Theatre. Flag Coat of arms Zagreb Location of Zagreb in Croatia Show map of Croatia Zagreb Zagreb (Europe) Show map of Europe Coordinates: 45°49′N 15°59′E  /  45.817°N 15.983°E  / 45.817; 15.983 Coordinates: 45°49′N 15°59′E  /  45.817°N 15.983°E  / 45.817; 15.983 Country   Croatia County City of Zagreb RC diocese 1094 Free royal city 1242 Unified 1850 Subdivisions 17 city districts 218 local committees (70 settlements) Government  • Type Mayor-Council  • Mayor Milan Bandić (BM 365)  • City Assembly 51 members BM 365, ZL, NS-R (14) SDP, GLAS, HNS, HSS, NH-PS (13) ...
Read more

Understanding the information contained in the Deep Space Network XML data?

Image
up vote 1 down vote favorite NASA's Deep Space Network activity can be viewed conveniently at https://eyes.nasa.gov/dsn/dsn.html but if you would like data in a more raw XML form, it can be accessed as described in this answer Below is a small Python script that reads the XML into a list of dictionaries, and saves it to disk as a JSON. The dictionaries are organized at the higher level by dish, then by spacecraft per dish. Question: What are the most-likely meanings of all the data field contained in the Deep Space Network XML data? This is the example url from the other answer, used in the script below: https://eyes.nasa.gov/dsn/data/1365107113.xml def dictify(r,root=True): # https://stackoverflow.com/questions/2148119/how-to-convert-an-xml-string-to-a-dictionary-in-python/30923963#30923963 if root: ret...
Read more