Permutation group of Cosets
$begingroup$
Let $K$ be a subgroup of a group $G$. Let $T$ denote the set of all distinct right cosets of $K$ in $G$ and $A(T)$ be the permutation group of $T$. Prove the following statements.
(a) For each $ain G$, the function $f_a:Trightarrow T$ given by $f_a(Kb)=K(ba)$ is a bijection.
(b) The function $varphi:Grightarrow A(T)$ given by $varphi(a)=f_{a^{-1}}$ is a group homomorphism whose kernel is contained in $K$.
I have been able to work out part (a) and most of part (b). I am stuck on the last part of (b) where I am asked to verify that $Ker(varphi)subset K$. To do this we need to know the identity element of $A(T)$ which is clearly the identity permutation. In this context that should be the $f_e$ mapping. But this gives me that $Ker(varphi)={e}$ which doesn't seem correct. Also there is a part (c) that says to prove that if $K$ is normal then $Ker(varphi)=K$ so I'm pretty sure that I'm wrong but I can't seem to make anymore progress. Any help is greatly appreciated.
group-theory proof-verification permutations
asked Dec 5 '18 at 22:34
WaltWalt
374114
$endgroup$
add a comment |
$begingroup$
Let $K$ be a subgroup of a group $G$. Let $T$ denote the set of all distinct right cosets of $K$ in $G$ and $A(T)$ be the permutation group of $T$. Prove the following statements.
(a) For each $ain G$, the function $f_a:Trightarrow T$ given by $f_a(Kb)=K(ba)$ is a bijection.
(b) The function $varphi:Grightarrow A(T)$ given by $varphi(a)=f_{a^{-1}}$ is a group homomorphism whose kernel is contained in $K$.
I have been able to work out part (a) and most of part (b). I am stuck on the last part of (b) where I am asked to verify that $Ker(varphi)subset K$. To do this we need to know the identity element of $A(T)$ which is clearly the identity permutation. In this context that should be the $f_e$ mapping. But this gives me that $Ker(varphi)={e}$ which doesn't seem correct. Also there is a part (c) that says to prove that if $K$ is normal then $Ker(varphi)=K$ so I'm pretty sure that I'm wrong but I can't seem to make anymore progress. Any help is greatly appreciated.
group-theory proof-verification permutations
asked Dec 5 '18 at 22:34
WaltWalt
374114
$endgroup$
$begingroup$
Thanks for the adjustment! You can always ask "part 2" after understanding answers to (a, b), if you need to!
$endgroup$
– amWhy
Dec 5 '18 at 22:43
$begingroup$
Suppose that $gin G$ has $f_{g^{-1}}$ the identity map. that means that for every $bin G$, $Kbg^{-1} = Kb$, which requires $bgb^{-1}in K$ for all $bin G$. That does not mean, necessarily, that $g=e$. However, you should be able to deduce that it does imply that $gin K$.
$endgroup$
– Arturo Magidin
Dec 5 '18 at 23:01
$begingroup$
I did write that down at one point, however all I could come up with was that $g=b^{-1}kb$ for some $kin K$. Why is $b^{-1}kbin K$ for all $bin G$? All we really know is that $K$ is a general subgroup.
$endgroup$
– Walt
Dec 5 '18 at 23:09
add a comment |
$begingroup$
Let $K$ be a subgroup of a group $G$. Let $T$ denote the set of all distinct right cosets of $K$ in $G$ and $A(T)$ be the permutation group of $T$. Prove the following statements.
(a) For each $ain G$, the function $f_a:Trightarrow T$ given by $f_a(Kb)=K(ba)$ is a bijection.
(b) The function $varphi:Grightarrow A(T)$ given by $varphi(a)=f_{a^{-1}}$ is a group homomorphism whose kernel is contained in $K$.
I have been able to work out part (a) and most of part (b). I am stuck on the last part of (b) where I am asked to verify that $Ker(varphi)subset K$. To do this we need to know the identity element of $A(T)$ which is clearly the identity permutation. In this context that should be the $f_e$ mapping. But this gives me that $Ker(varphi)={e}$ which doesn't seem correct. Also there is a part (c) that says to prove that if $K$ is normal then $Ker(varphi)=K$ so I'm pretty sure that I'm wrong but I can't seem to make anymore progress. Any help is greatly appreciated.
group-theory proof-verification permutations
asked Dec 5 '18 at 22:34
WaltWalt
374114
$endgroup$
Let $K$ be a subgroup of a group $G$. Let $T$ denote the set of all distinct right cosets of $K$ in $G$ and $A(T)$ be the permutation group of $T$. Prove the following statements.
(a) For each $ain G$, the function $f_a:Trightarrow T$ given by $f_a(Kb)=K(ba)$ is a bijection.
(b) The function $varphi:Grightarrow A(T)$ given by $varphi(a)=f_{a^{-1}}$ is a group homomorphism whose kernel is contained in $K$.
I have been able to work out part (a) and most of part (b). I am stuck on the last part of (b) where I am asked to verify that $Ker(varphi)subset K$. To do this we need to know the identity element of $A(T)$ which is clearly the identity permutation. In this context that should be the $f_e$ mapping. But this gives me that $Ker(varphi)={e}$ which doesn't seem correct. Also there is a part (c) that says to prove that if $K$ is normal then $Ker(varphi)=K$ so I'm pretty sure that I'm wrong but I can't seem to make anymore progress. Any help is greatly appreciated.
group-theory proof-verification permutations
group-theory proof-verification permutations
asked Dec 5 '18 at 22:34
WaltWalt
374114
asked Dec 5 '18 at 22:34
WaltWalt
374114
edited Dec 5 '18 at 22:42
Walt
asked Dec 5 '18 at 22:34
WaltWalt
374114
asked Dec 5 '18 at 22:34
WaltWalt
374114
asked Dec 5 '18 at 22:34
WaltWalt
374114
374114
$begingroup$
Thanks for the adjustment! You can always ask "part 2" after understanding answers to (a, b), if you need to!
$endgroup$
– amWhy
Dec 5 '18 at 22:43
$begingroup$
Suppose that $gin G$ has $f_{g^{-1}}$ the identity map. that means that for every $bin G$, $Kbg^{-1} = Kb$, which requires $bgb^{-1}in K$ for all $bin G$. That does not mean, necessarily, that $g=e$. However, you should be able to deduce that it does imply that $gin K$.
$endgroup$
– Arturo Magidin
Dec 5 '18 at 23:01
$begingroup$
I did write that down at one point, however all I could come up with was that $g=b^{-1}kb$ for some $kin K$. Why is $b^{-1}kbin K$ for all $bin G$? All we really know is that $K$ is a general subgroup.
$endgroup$
– Walt
Dec 5 '18 at 23:09
add a comment |
$begingroup$
Thanks for the adjustment! You can always ask "part 2" after understanding answers to (a, b), if you need to!
$endgroup$
– amWhy
Dec 5 '18 at 22:43
$begingroup$
Suppose that $gin G$ has $f_{g^{-1}}$ the identity map. that means that for every $bin G$, $Kbg^{-1} = Kb$, which requires $bgb^{-1}in K$ for all $bin G$. That does not mean, necessarily, that $g=e$. However, you should be able to deduce that it does imply that $gin K$.
$endgroup$
– Arturo Magidin
Dec 5 '18 at 23:01
$begingroup$
I did write that down at one point, however all I could come up with was that $g=b^{-1}kb$ for some $kin K$. Why is $b^{-1}kbin K$ for all $bin G$? All we really know is that $K$ is a general subgroup.
$endgroup$
– Walt
Dec 5 '18 at 23:09
$begingroup$
Thanks for the adjustment! You can always ask "part 2" after understanding answers to (a, b), if you need to!
$endgroup$
– amWhy
Dec 5 '18 at 22:43
$begingroup$
Thanks for the adjustment! You can always ask "part 2" after understanding answers to (a, b), if you need to!
$endgroup$
– amWhy
Dec 5 '18 at 22:43
$begingroup$
Suppose that $gin G$ has $f_{g^{-1}}$ the identity map. that means that for every $bin G$, $Kbg^{-1} = Kb$, which requires $bgb^{-1}in K$ for all $bin G$. That does not mean, necessarily, that $g=e$. However, you should be able to deduce that it does imply that $gin K$.
$endgroup$
– Arturo Magidin
Dec 5 '18 at 23:01
$begingroup$
Suppose that $gin G$ has $f_{g^{-1}}$ the identity map. that means that for every $bin G$, $Kbg^{-1} = Kb$, which requires $bgb^{-1}in K$ for all $bin G$. That does not mean, necessarily, that $g=e$. However, you should be able to deduce that it does imply that $gin K$.
$endgroup$
– Arturo Magidin
Dec 5 '18 at 23:01
$begingroup$
I did write that down at one point, however all I could come up with was that $g=b^{-1}kb$ for some $kin K$. Why is $b^{-1}kbin K$ for all $bin G$? All we really know is that $K$ is a general subgroup.
$endgroup$
– Walt
Dec 5 '18 at 23:09
$begingroup$
I did write that down at one point, however all I could come up with was that $g=b^{-1}kb$ for some $kin K$. Why is $b^{-1}kbin K$ for all $bin G$? All we really know is that $K$ is a general subgroup.
$endgroup$
– Walt
Dec 5 '18 at 23:09
add a comment |
1 Answer
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$begingroup$
Let $hin Ker(varphi)$. Note that if your $f_{h^{-1}}$ is the identity permutation, then in particular $f_{h^{-1}}(K)=Kh^{-1}=K$ and this implies that $h^{-1}in K$ and also its inverse does since $K$ is a subgroup.
Conversely, suppose $K$ is normal and $h in K$. Then $f_{g^{-1}}(Kb)=Kbg^{-1}$, $f_{g^{-1}h^{-1}}(Kb)=Kbg^{-1}h^{-1}$ and $f_{h^{-1}g^{-1}}(Kb)=Kbh^{-1}g^{-1}$ for all $gin G$. But $Kbg^{-1}h^{-1}=bg^{-1}Kh^{-1}=bg^{-1}K=Kbg^{-1}$ and $Kbh^{-1}g^{-1}=Kbg^{-1}$ by normality of $K$, so you have that $varphi(hg)=varphi(gh)=varphi(g)$, which implies that $varphi(h)$ is the identity. So $K$ is included in the kernel and using the first inclusion we conclude that $K=Ker(varphi)$.
answered Dec 5 '18 at 23:15
Niki Di GianoNiki Di Giano
941211
$endgroup$
add a comment |
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$begingroup$
Let $hin Ker(varphi)$. Note that if your $f_{h^{-1}}$ is the identity permutation, then in particular $f_{h^{-1}}(K)=Kh^{-1}=K$ and this implies that $h^{-1}in K$ and also its inverse does since $K$ is a subgroup.
Conversely, suppose $K$ is normal and $h in K$. Then $f_{g^{-1}}(Kb)=Kbg^{-1}$, $f_{g^{-1}h^{-1}}(Kb)=Kbg^{-1}h^{-1}$ and $f_{h^{-1}g^{-1}}(Kb)=Kbh^{-1}g^{-1}$ for all $gin G$. But $Kbg^{-1}h^{-1}=bg^{-1}Kh^{-1}=bg^{-1}K=Kbg^{-1}$ and $Kbh^{-1}g^{-1}=Kbg^{-1}$ by normality of $K$, so you have that $varphi(hg)=varphi(gh)=varphi(g)$, which implies that $varphi(h)$ is the identity. So $K$ is included in the kernel and using the first inclusion we conclude that $K=Ker(varphi)$.
answered Dec 5 '18 at 23:15
Niki Di GianoNiki Di Giano
941211
$endgroup$
add a comment |
$begingroup$
Let $hin Ker(varphi)$. Note that if your $f_{h^{-1}}$ is the identity permutation, then in particular $f_{h^{-1}}(K)=Kh^{-1}=K$ and this implies that $h^{-1}in K$ and also its inverse does since $K$ is a subgroup.
Conversely, suppose $K$ is normal and $h in K$. Then $f_{g^{-1}}(Kb)=Kbg^{-1}$, $f_{g^{-1}h^{-1}}(Kb)=Kbg^{-1}h^{-1}$ and $f_{h^{-1}g^{-1}}(Kb)=Kbh^{-1}g^{-1}$ for all $gin G$. But $Kbg^{-1}h^{-1}=bg^{-1}Kh^{-1}=bg^{-1}K=Kbg^{-1}$ and $Kbh^{-1}g^{-1}=Kbg^{-1}$ by normality of $K$, so you have that $varphi(hg)=varphi(gh)=varphi(g)$, which implies that $varphi(h)$ is the identity. So $K$ is included in the kernel and using the first inclusion we conclude that $K=Ker(varphi)$.
answered Dec 5 '18 at 23:15
Niki Di GianoNiki Di Giano
941211
$endgroup$
add a comment |
$begingroup$
Let $hin Ker(varphi)$. Note that if your $f_{h^{-1}}$ is the identity permutation, then in particular $f_{h^{-1}}(K)=Kh^{-1}=K$ and this implies that $h^{-1}in K$ and also its inverse does since $K$ is a subgroup.
Conversely, suppose $K$ is normal and $h in K$. Then $f_{g^{-1}}(Kb)=Kbg^{-1}$, $f_{g^{-1}h^{-1}}(Kb)=Kbg^{-1}h^{-1}$ and $f_{h^{-1}g^{-1}}(Kb)=Kbh^{-1}g^{-1}$ for all $gin G$. But $Kbg^{-1}h^{-1}=bg^{-1}Kh^{-1}=bg^{-1}K=Kbg^{-1}$ and $Kbh^{-1}g^{-1}=Kbg^{-1}$ by normality of $K$, so you have that $varphi(hg)=varphi(gh)=varphi(g)$, which implies that $varphi(h)$ is the identity. So $K$ is included in the kernel and using the first inclusion we conclude that $K=Ker(varphi)$.
answered Dec 5 '18 at 23:15
Niki Di GianoNiki Di Giano
941211
$endgroup$
Let $hin Ker(varphi)$. Note that if your $f_{h^{-1}}$ is the identity permutation, then in particular $f_{h^{-1}}(K)=Kh^{-1}=K$ and this implies that $h^{-1}in K$ and also its inverse does since $K$ is a subgroup.
Conversely, suppose $K$ is normal and $h in K$. Then $f_{g^{-1}}(Kb)=Kbg^{-1}$, $f_{g^{-1}h^{-1}}(Kb)=Kbg^{-1}h^{-1}$ and $f_{h^{-1}g^{-1}}(Kb)=Kbh^{-1}g^{-1}$ for all $gin G$. But $Kbg^{-1}h^{-1}=bg^{-1}Kh^{-1}=bg^{-1}K=Kbg^{-1}$ and $Kbh^{-1}g^{-1}=Kbg^{-1}$ by normality of $K$, so you have that $varphi(hg)=varphi(gh)=varphi(g)$, which implies that $varphi(h)$ is the identity. So $K$ is included in the kernel and using the first inclusion we conclude that $K=Ker(varphi)$.
answered Dec 5 '18 at 23:15
Niki Di GianoNiki Di Giano
941211
edited Dec 6 '18 at 11:36
answered Dec 5 '18 at 23:15
Niki Di GianoNiki Di Giano
941211
answered Dec 5 '18 at 23:15
Niki Di GianoNiki Di Giano
941211
answered Dec 5 '18 at 23:15
Niki Di GianoNiki Di Giano
941211
941211
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$begingroup$
Thanks for the adjustment! You can always ask "part 2" after understanding answers to (a, b), if you need to!
$endgroup$
– amWhy
Dec 5 '18 at 22:43
$begingroup$
Suppose that $gin G$ has $f_{g^{-1}}$ the identity map. that means that for every $bin G$, $Kbg^{-1} = Kb$, which requires $bgb^{-1}in K$ for all $bin G$. That does not mean, necessarily, that $g=e$. However, you should be able to deduce that it does imply that $gin K$.
$endgroup$
– Arturo Magidin
Dec 5 '18 at 23:01
$begingroup$
I did write that down at one point, however all I could come up with was that $g=b^{-1}kb$ for some $kin K$. Why is $b^{-1}kbin K$ for all $bin G$? All we really know is that $K$ is a general subgroup.
$endgroup$
– Walt
Dec 5 '18 at 23:09