Lagrange's Theorem: Injectivity.
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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 $a^-1in G$ but don't we need to have $a^-1in aU$ to add it from the left? And if so, how can we show $a^-1in aU$?
Best,
KingDingeling
abstract-algebra group-theory finite-groups
edited Dec 31 '18 at 20:09
janmarqz
6,20241630
asked Dec 31 '18 at 18:42
KingDingelingKingDingeling
1216
$endgroup$
|
show 1 more comment
$begingroup$
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 $a^-1in G$ but don't we need to have $a^-1in aU$ to add it from the left? And if so, how can we show $a^-1in aU$?
Best,
KingDingeling
abstract-algebra group-theory finite-groups
edited Dec 31 '18 at 20:09
janmarqz
6,20241630
asked Dec 31 '18 at 18:42
KingDingelingKingDingeling
1216
$endgroup$
$begingroup$
What are the domain and the codomain of your function?
$endgroup$
– José Carlos Santos
Dec 31 '18 at 18:52
$begingroup$
Domain is $U$ and codomain is $aU$.
$endgroup$
– KingDingeling
Dec 31 '18 at 18:53
$begingroup$
How can a function whose domain is $U$ be defined as $Umapsto aU$? And what is $a$ then?
$endgroup$
– José Carlos Santos
Dec 31 '18 at 18:54
$begingroup$
$a$ is an element of the Group $G$. It's a basic lefttranslation.
$endgroup$
– KingDingeling
Dec 31 '18 at 18:57
$begingroup$
Let U be trivial. Is $a^-1$ in $aU$?
$endgroup$
– Andres Mejia
Dec 31 '18 at 19:22
|
show 1 more comment
$begingroup$
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 $a^-1in G$ but don't we need to have $a^-1in aU$ to add it from the left? And if so, how can we show $a^-1in aU$?
Best,
KingDingeling
abstract-algebra group-theory finite-groups
edited Dec 31 '18 at 20:09
janmarqz
6,20241630
asked Dec 31 '18 at 18:42
KingDingelingKingDingeling
1216
$endgroup$
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 $a^-1in G$ but don't we need to have $a^-1in aU$ to add it from the left? And if so, how can we show $a^-1in aU$?
Best,
KingDingeling
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
edited Dec 31 '18 at 20:09
janmarqz
6,20241630
asked Dec 31 '18 at 18:42
KingDingelingKingDingeling
1216
edited Dec 31 '18 at 20:09
janmarqz
6,20241630
asked Dec 31 '18 at 18:42
KingDingelingKingDingeling
1216
edited Dec 31 '18 at 20:09
janmarqz
6,20241630
edited Dec 31 '18 at 20:09
janmarqz
6,20241630
edited Dec 31 '18 at 20:09
janmarqz
6,20241630
6,20241630
asked Dec 31 '18 at 18:42
KingDingelingKingDingeling
1216
asked Dec 31 '18 at 18:42
KingDingelingKingDingeling
1216
asked Dec 31 '18 at 18:42
KingDingelingKingDingeling
1216
1216
$begingroup$
What are the domain and the codomain of your function?
$endgroup$
– José Carlos Santos
Dec 31 '18 at 18:52
$begingroup$
Domain is $U$ and codomain is $aU$.
$endgroup$
– KingDingeling
Dec 31 '18 at 18:53
$begingroup$
How can a function whose domain is $U$ be defined as $Umapsto aU$? And what is $a$ then?
$endgroup$
– José Carlos Santos
Dec 31 '18 at 18:54
$begingroup$
$a$ is an element of the Group $G$. It's a basic lefttranslation.
$endgroup$
– KingDingeling
Dec 31 '18 at 18:57
$begingroup$
Let U be trivial. Is $a^-1$ in $aU$?
$endgroup$
– Andres Mejia
Dec 31 '18 at 19:22
|
show 1 more comment
$begingroup$
What are the domain and the codomain of your function?
$endgroup$
– José Carlos Santos
Dec 31 '18 at 18:52
$begingroup$
Domain is $U$ and codomain is $aU$.
$endgroup$
– KingDingeling
Dec 31 '18 at 18:53
$begingroup$
How can a function whose domain is $U$ be defined as $Umapsto aU$? And what is $a$ then?
$endgroup$
– José Carlos Santos
Dec 31 '18 at 18:54
$begingroup$
$a$ is an element of the Group $G$. It's a basic lefttranslation.
$endgroup$
– KingDingeling
Dec 31 '18 at 18:57
$begingroup$
Let U be trivial. Is $a^-1$ in $aU$?
$endgroup$
– Andres Mejia
Dec 31 '18 at 19:22
$begingroup$
What are the domain and the codomain of your function?
$endgroup$
– José Carlos Santos
Dec 31 '18 at 18:52
$begingroup$
What are the domain and the codomain of your function?
$endgroup$
– José Carlos Santos
Dec 31 '18 at 18:52
$begingroup$
Domain is $U$ and codomain is $aU$.
$endgroup$
– KingDingeling
Dec 31 '18 at 18:53
$begingroup$
Domain is $U$ and codomain is $aU$.
$endgroup$
– KingDingeling
Dec 31 '18 at 18:53
$begingroup$
How can a function whose domain is $U$ be defined as $Umapsto aU$? And what is $a$ then?
$endgroup$
– José Carlos Santos
Dec 31 '18 at 18:54
$begingroup$
How can a function whose domain is $U$ be defined as $Umapsto aU$? And what is $a$ then?
$endgroup$
– José Carlos Santos
Dec 31 '18 at 18:54
$begingroup$
$a$ is an element of the Group $G$. It's a basic lefttranslation.
$endgroup$
– KingDingeling
Dec 31 '18 at 18:57
$begingroup$
$a$ is an element of the Group $G$. It's a basic lefttranslation.
$endgroup$
– KingDingeling
Dec 31 '18 at 18:57
$begingroup$
Let U be trivial. Is $a^-1$ in $aU$?
$endgroup$
– Andres Mejia
Dec 31 '18 at 19:22
$begingroup$
Let U be trivial. Is $a^-1$ in $aU$?
$endgroup$
– Andres Mejia
Dec 31 '18 at 19:22
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Dec 31 '18 at 20:16
Alex MathersAlex Mathers
10.8k21344
$endgroup$
$begingroup$
Thank you for taking time and explaining :)
$endgroup$
– KingDingeling
Dec 31 '18 at 20:19
add a comment |
$begingroup$
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.
answered Dec 31 '18 at 18:51
mathcounterexamples.netmathcounterexamples.net
25.5k21953
$endgroup$
$begingroup$
But if I take an element in $aU$ to prove the injecticity, may I just use $a^-1$ even if it isn't in $aU$?
$endgroup$
– KingDingeling
Dec 31 '18 at 18:59
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Dec 31 '18 at 20:16
Alex MathersAlex Mathers
10.8k21344
$endgroup$
$begingroup$
Thank you for taking time and explaining :)
$endgroup$
– KingDingeling
Dec 31 '18 at 20:19
add a comment |
$begingroup$
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.
answered Dec 31 '18 at 20:16
Alex MathersAlex Mathers
10.8k21344
$endgroup$
$begingroup$
Thank you for taking time and explaining :)
$endgroup$
– KingDingeling
Dec 31 '18 at 20:19
add a comment |
$begingroup$
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.
answered Dec 31 '18 at 20:16
Alex MathersAlex Mathers
10.8k21344
$endgroup$
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.
answered Dec 31 '18 at 20:16
Alex MathersAlex Mathers
10.8k21344
answered Dec 31 '18 at 20:16
Alex MathersAlex Mathers
10.8k21344
answered Dec 31 '18 at 20:16
Alex MathersAlex Mathers
10.8k21344
answered Dec 31 '18 at 20:16
Alex MathersAlex Mathers
10.8k21344
10.8k21344
$begingroup$
Thank you for taking time and explaining :)
$endgroup$
– KingDingeling
Dec 31 '18 at 20:19
add a comment |
$begingroup$
Thank you for taking time and explaining :)
$endgroup$
– KingDingeling
Dec 31 '18 at 20:19
$begingroup$
Thank you for taking time and explaining :)
$endgroup$
– KingDingeling
Dec 31 '18 at 20:19
$begingroup$
Thank you for taking time and explaining :)
$endgroup$
– KingDingeling
Dec 31 '18 at 20:19
add a comment |
$begingroup$
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.
answered Dec 31 '18 at 18:51
mathcounterexamples.netmathcounterexamples.net
25.5k21953
$endgroup$
$begingroup$
But if I take an element in $aU$ to prove the injecticity, may I just use $a^-1$ even if it isn't in $aU$?
$endgroup$
– KingDingeling
Dec 31 '18 at 18:59
add a comment |
$begingroup$
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.
answered Dec 31 '18 at 18:51
mathcounterexamples.netmathcounterexamples.net
25.5k21953
$endgroup$
$begingroup$
But if I take an element in $aU$ to prove the injecticity, may I just use $a^-1$ even if it isn't in $aU$?
$endgroup$
– KingDingeling
Dec 31 '18 at 18:59
add a comment |
$begingroup$
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.
answered Dec 31 '18 at 18:51
mathcounterexamples.netmathcounterexamples.net
25.5k21953
$endgroup$
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.
answered Dec 31 '18 at 18:51
mathcounterexamples.netmathcounterexamples.net
25.5k21953
answered Dec 31 '18 at 18:51
mathcounterexamples.netmathcounterexamples.net
25.5k21953
answered Dec 31 '18 at 18:51
mathcounterexamples.netmathcounterexamples.net
25.5k21953
answered Dec 31 '18 at 18:51
mathcounterexamples.netmathcounterexamples.net
25.5k21953
25.5k21953
$begingroup$
But if I take an element in $aU$ to prove the injecticity, may I just use $a^-1$ even if it isn't in $aU$?
$endgroup$
– KingDingeling
Dec 31 '18 at 18:59
add a comment |
$begingroup$
But if I take an element in $aU$ to prove the injecticity, may I just use $a^-1$ even if it isn't in $aU$?
$endgroup$
– KingDingeling
Dec 31 '18 at 18:59
$begingroup$
But if I take an element in $aU$ to prove the injecticity, may I just use $a^-1$ even if it isn't in $aU$?
$endgroup$
– KingDingeling
Dec 31 '18 at 18:59
$begingroup$
But if I take an element in $aU$ to prove the injecticity, may I just use $a^-1$ even if it isn't in $aU$?
$endgroup$
– KingDingeling
Dec 31 '18 at 18:59
add a comment |
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$begingroup$
What are the domain and the codomain of your function?
$endgroup$
– José Carlos Santos
Dec 31 '18 at 18:52
$begingroup$
Domain is $U$ and codomain is $aU$.
$endgroup$
– KingDingeling
Dec 31 '18 at 18:53
$begingroup$
How can a function whose domain is $U$ be defined as $Umapsto aU$? And what is $a$ then?
$endgroup$
– José Carlos Santos
Dec 31 '18 at 18:54
$begingroup$
$a$ is an element of the Group $G$. It's a basic lefttranslation.
$endgroup$
– KingDingeling
Dec 31 '18 at 18:57
$begingroup$
Let U be trivial. Is $a^-1$ in $aU$?
$endgroup$
– Andres Mejia
Dec 31 '18 at 19:22