Lagrange's Theorem: Injectivity.
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
edited Dec 31 '18 at 20:09
janmarqz
6,18241630
asked Dec 31 '18 at 18:42
KingDingeling
505
|
show 1 more comment
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
edited Dec 31 '18 at 20:09
janmarqz
6,18241630
asked Dec 31 '18 at 18:42
KingDingeling
505
What are the domain and the codomain of your function?
– José Carlos Santos
Dec 31 '18 at 18:52
Domain is $U$ and codomain is $aU$.
– KingDingeling
Dec 31 '18 at 18:53
How can a function whose domain is $U$ be defined as $Umapsto aU$? And what is $a$ then?
– José Carlos Santos
Dec 31 '18 at 18:54
$a$ is an element of the Group $G$. It's a basic lefttranslation.
– KingDingeling
Dec 31 '18 at 18:57
Let U be trivial. Is $a^{-1}$ in $aU$?
– Andres Mejia
Dec 31 '18 at 19:22
|
show 1 more comment
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
edited Dec 31 '18 at 20:09
janmarqz
6,18241630
asked Dec 31 '18 at 18:42
KingDingeling
505
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
edited Dec 31 '18 at 20:09
janmarqz
6,18241630
asked Dec 31 '18 at 18:42
KingDingeling
505
edited Dec 31 '18 at 20:09
janmarqz
6,18241630
asked Dec 31 '18 at 18:42
KingDingeling
505
edited Dec 31 '18 at 20:09
janmarqz
6,18241630
edited Dec 31 '18 at 20:09
janmarqz
6,18241630
edited Dec 31 '18 at 20:09
janmarqz
6,18241630
6,18241630
asked Dec 31 '18 at 18:42
KingDingeling
505
asked Dec 31 '18 at 18:42
KingDingeling
505
asked Dec 31 '18 at 18:42
KingDingeling
505
505
What are the domain and the codomain of your function?
– José Carlos Santos
Dec 31 '18 at 18:52
Domain is $U$ and codomain is $aU$.
– KingDingeling
Dec 31 '18 at 18:53
How can a function whose domain is $U$ be defined as $Umapsto aU$? And what is $a$ then?
– José Carlos Santos
Dec 31 '18 at 18:54
$a$ is an element of the Group $G$. It's a basic lefttranslation.
– KingDingeling
Dec 31 '18 at 18:57
Let U be trivial. Is $a^{-1}$ in $aU$?
– Andres Mejia
Dec 31 '18 at 19:22
|
show 1 more comment
What are the domain and the codomain of your function?
– José Carlos Santos
Dec 31 '18 at 18:52
Domain is $U$ and codomain is $aU$.
– KingDingeling
Dec 31 '18 at 18:53
How can a function whose domain is $U$ be defined as $Umapsto aU$? And what is $a$ then?
– José Carlos Santos
Dec 31 '18 at 18:54
$a$ is an element of the Group $G$. It's a basic lefttranslation.
– KingDingeling
Dec 31 '18 at 18:57
Let U be trivial. Is $a^{-1}$ in $aU$?
– Andres Mejia
Dec 31 '18 at 19:22
What are the domain and the codomain of your function?
– José Carlos Santos
Dec 31 '18 at 18:52
What are the domain and the codomain of your function?
– José Carlos Santos
Dec 31 '18 at 18:52
Domain is $U$ and codomain is $aU$.
– KingDingeling
Dec 31 '18 at 18:53
Domain is $U$ and codomain is $aU$.
– KingDingeling
Dec 31 '18 at 18:53
How can a function whose domain is $U$ be defined as $Umapsto aU$? And what is $a$ then?
– José Carlos Santos
Dec 31 '18 at 18:54
How can a function whose domain is $U$ be defined as $Umapsto aU$? And what is $a$ then?
– José Carlos Santos
Dec 31 '18 at 18:54
$a$ is an element of the Group $G$. It's a basic lefttranslation.
– KingDingeling
Dec 31 '18 at 18:57
$a$ is an element of the Group $G$. It's a basic lefttranslation.
– KingDingeling
Dec 31 '18 at 18:57
Let U be trivial. Is $a^{-1}$ in $aU$?
– Andres Mejia
Dec 31 '18 at 19:22
Let U be trivial. Is $a^{-1}$ in $aU$?
– Andres Mejia
Dec 31 '18 at 19:22
|
show 1 more comment
2 Answers
2
active
oldest
votes
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 Mathers
10.8k21344
Thank you for taking time and explaining :)
– KingDingeling
Dec 31 '18 at 20:19
add a comment |
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.net
24.7k21753
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
Dec 31 '18 at 18:59
add a comment |
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2 Answers
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2 Answers
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oldest
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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 Mathers
10.8k21344
Thank you for taking time and explaining :)
– KingDingeling
Dec 31 '18 at 20:19
add a comment |
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 Mathers
10.8k21344
Thank you for taking time and explaining :)
– KingDingeling
Dec 31 '18 at 20:19
add a comment |
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 Mathers
10.8k21344
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 Mathers
10.8k21344
answered Dec 31 '18 at 20:16
Alex Mathers
10.8k21344
answered Dec 31 '18 at 20:16
Alex Mathers
10.8k21344
answered Dec 31 '18 at 20:16
Alex Mathers
10.8k21344
10.8k21344
Thank you for taking time and explaining :)
– KingDingeling
Dec 31 '18 at 20:19
add a comment |
Thank you for taking time and explaining :)
– KingDingeling
Dec 31 '18 at 20:19
Thank you for taking time and explaining :)
– KingDingeling
Dec 31 '18 at 20:19
Thank you for taking time and explaining :)
– KingDingeling
Dec 31 '18 at 20:19
add a comment |
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.net
24.7k21753
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
Dec 31 '18 at 18:59
add a comment |
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.net
24.7k21753
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
Dec 31 '18 at 18:59
add a comment |
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.net
24.7k21753
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.net
24.7k21753
answered Dec 31 '18 at 18:51
mathcounterexamples.net
24.7k21753
answered Dec 31 '18 at 18:51
mathcounterexamples.net
24.7k21753
answered Dec 31 '18 at 18:51
mathcounterexamples.net
24.7k21753
24.7k21753
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
Dec 31 '18 at 18:59
add a comment |
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
Dec 31 '18 at 18:59
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
Dec 31 '18 at 18:59
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
Dec 31 '18 at 18:59
add a comment |
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What are the domain and the codomain of your function?
– José Carlos Santos
Dec 31 '18 at 18:52
Domain is $U$ and codomain is $aU$.
– KingDingeling
Dec 31 '18 at 18:53
How can a function whose domain is $U$ be defined as $Umapsto aU$? And what is $a$ then?
– José Carlos Santos
Dec 31 '18 at 18:54
$a$ is an element of the Group $G$. It's a basic lefttranslation.
– KingDingeling
Dec 31 '18 at 18:57
Let U be trivial. Is $a^{-1}$ in $aU$?
– Andres Mejia
Dec 31 '18 at 19:22