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
    3 hours ago










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










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










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












  • Let U be trivial. Is $a^{-1}$ in $aU$?
    – Andres Mejia
    2 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
    3 hours ago










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










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










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












  • Let U be trivial. Is $a^{-1}$ in $aU$?
    – Andres Mejia
    2 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 1 hour ago









janmarqz

6,17241630




6,17241630










asked 3 hours ago









KingDingeling

435




435












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










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










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










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












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


















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










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










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










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












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
















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




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












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




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












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




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












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






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














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




Let U be trivial. Is $a^{-1}$ in $aU$?
– Andres Mejia
2 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
    1 hour 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
    3 hours ago













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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
    1 hour 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
    1 hour 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 1 hour ago









Alex Mathers

10.8k21344




10.8k21344












  • Thank you for taking time and explaining :)
    – KingDingeling
    1 hour ago


















  • Thank you for taking time and explaining :)
    – KingDingeling
    1 hour ago
















Thank you for taking time and explaining :)
– KingDingeling
1 hour ago




Thank you for taking time and explaining :)
– KingDingeling
1 hour 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
    3 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
    3 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 3 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
    3 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
    3 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
3 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
3 hours ago




















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