Index of a subgroup is preserved under group isomorphism The Next CEO of Stack Overflowgroup,subgroup and isomorphismProve that there is a subgroup of index 2Normal subgroup, quotient group, isomorphism.Finding a normal subgroup of finite indexNormal subgroup and index problemIf $phi : G rightarrow G'$ is a group isomorphism, and $H$ is a subgroup of $G$, prove that the image set of $H$ is a subgroup of $G'$Surjection from $G$ into direct product $prod_i G/H_i$Existence of free abelian subgroup and isomorphismGroup isomorphism on quotient groups?Group isomorphism induces a homomorphism on the quotient group?
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Index of a subgroup is preserved under group isomorphism
The Next CEO of Stack Overflowgroup,subgroup and isomorphismProve that there is a subgroup of index 2Normal subgroup, quotient group, isomorphism.Finding a normal subgroup of finite indexNormal subgroup and index problemIf $phi : G rightarrow G'$ is a group isomorphism, and $H$ is a subgroup of $G$, prove that the image set of $H$ is a subgroup of $G'$Surjection from $G$ into direct product $prod_i G/H_i$Existence of free abelian subgroup and isomorphismGroup isomorphism on quotient groups?Group isomorphism induces a homomorphism on the quotient group?
$begingroup$
Let $varphi : Gto G'$ be a group isomorphism and let $Nlt G$.
prove: if $[G:N]lt infty$ then $[G:N]=[G':varphi(N)]$
It is known that $varphi(N)$ is indeed a subgroup of $G'$,
but how can I show equality of indices?
Note: I have yet to learn any isomorphism theorems.
Any help would be appreciated
abstract-algebra group-theory group-isomorphism
edited Mar 19 at 20:54
Shaun
9,869113684
asked Mar 19 at 20:51
Galush BalushGalush Balush
895
$endgroup$
|
show 3 more comments
$begingroup$
Let $varphi : Gto G'$ be a group isomorphism and let $Nlt G$.
prove: if $[G:N]lt infty$ then $[G:N]=[G':varphi(N)]$
It is known that $varphi(N)$ is indeed a subgroup of $G'$,
but how can I show equality of indices?
Note: I have yet to learn any isomorphism theorems.
Any help would be appreciated
abstract-algebra group-theory group-isomorphism
edited Mar 19 at 20:54
Shaun
9,869113684
asked Mar 19 at 20:51
Galush BalushGalush Balush
895
$endgroup$
1
$begingroup$
What is the definition of index you are using? Because there are various approaches, and depending on the definition you are using one fits better than the other.
$endgroup$
– b00n heT
Mar 19 at 20:52
$begingroup$
Index is defined as the number of lest or right cosets
$endgroup$
– Galush Balush
Mar 19 at 20:53
3
$begingroup$
Perfect: then take each and every coset of $N$ in $G$ and map it to $G'$ using $varphi$. Doing so will result in cosets of $varphi(N)$.
$endgroup$
– b00n heT
Mar 19 at 20:54
1
$begingroup$
@jgon I know of only one, but I cannot exclude that there can be more than one.
$endgroup$
– b00n heT
Mar 19 at 20:59
2
$begingroup$
@b00nheT Completely reasonable, I was just curious, since I also only knew of one.
$endgroup$
– jgon
Mar 19 at 21:00
|
show 3 more comments
$begingroup$
Let $varphi : Gto G'$ be a group isomorphism and let $Nlt G$.
prove: if $[G:N]lt infty$ then $[G:N]=[G':varphi(N)]$
It is known that $varphi(N)$ is indeed a subgroup of $G'$,
but how can I show equality of indices?
Note: I have yet to learn any isomorphism theorems.
Any help would be appreciated
abstract-algebra group-theory group-isomorphism
edited Mar 19 at 20:54
Shaun
9,869113684
asked Mar 19 at 20:51
Galush BalushGalush Balush
895
$endgroup$
Let $varphi : Gto G'$ be a group isomorphism and let $Nlt G$.
prove: if $[G:N]lt infty$ then $[G:N]=[G':varphi(N)]$
It is known that $varphi(N)$ is indeed a subgroup of $G'$,
but how can I show equality of indices?
Note: I have yet to learn any isomorphism theorems.
Any help would be appreciated
abstract-algebra group-theory group-isomorphism
abstract-algebra group-theory group-isomorphism
edited Mar 19 at 20:54
Shaun
9,869113684
asked Mar 19 at 20:51
Galush BalushGalush Balush
895
edited Mar 19 at 20:54
Shaun
9,869113684
asked Mar 19 at 20:51
Galush BalushGalush Balush
895
edited Mar 19 at 20:54
Shaun
9,869113684
edited Mar 19 at 20:54
Shaun
9,869113684
edited Mar 19 at 20:54
Shaun
9,869113684
9,869113684
asked Mar 19 at 20:51
Galush BalushGalush Balush
895
asked Mar 19 at 20:51
Galush BalushGalush Balush
895
asked Mar 19 at 20:51
Galush BalushGalush Balush
895
895
1
$begingroup$
What is the definition of index you are using? Because there are various approaches, and depending on the definition you are using one fits better than the other.
$endgroup$
– b00n heT
Mar 19 at 20:52
$begingroup$
Index is defined as the number of lest or right cosets
$endgroup$
– Galush Balush
Mar 19 at 20:53
3
$begingroup$
Perfect: then take each and every coset of $N$ in $G$ and map it to $G'$ using $varphi$. Doing so will result in cosets of $varphi(N)$.
$endgroup$
– b00n heT
Mar 19 at 20:54
1
$begingroup$
@jgon I know of only one, but I cannot exclude that there can be more than one.
$endgroup$
– b00n heT
Mar 19 at 20:59
2
$begingroup$
@b00nheT Completely reasonable, I was just curious, since I also only knew of one.
$endgroup$
– jgon
Mar 19 at 21:00
|
show 3 more comments
1
$begingroup$
What is the definition of index you are using? Because there are various approaches, and depending on the definition you are using one fits better than the other.
$endgroup$
– b00n heT
Mar 19 at 20:52
$begingroup$
Index is defined as the number of lest or right cosets
$endgroup$
– Galush Balush
Mar 19 at 20:53
3
$begingroup$
Perfect: then take each and every coset of $N$ in $G$ and map it to $G'$ using $varphi$. Doing so will result in cosets of $varphi(N)$.
$endgroup$
– b00n heT
Mar 19 at 20:54
1
$begingroup$
@jgon I know of only one, but I cannot exclude that there can be more than one.
$endgroup$
– b00n heT
Mar 19 at 20:59
2
$begingroup$
@b00nheT Completely reasonable, I was just curious, since I also only knew of one.
$endgroup$
– jgon
Mar 19 at 21:00
1
1
$begingroup$
What is the definition of index you are using? Because there are various approaches, and depending on the definition you are using one fits better than the other.
$endgroup$
– b00n heT
Mar 19 at 20:52
$begingroup$
What is the definition of index you are using? Because there are various approaches, and depending on the definition you are using one fits better than the other.
$endgroup$
– b00n heT
Mar 19 at 20:52
$begingroup$
Index is defined as the number of lest or right cosets
$endgroup$
– Galush Balush
Mar 19 at 20:53
$begingroup$
Index is defined as the number of lest or right cosets
$endgroup$
– Galush Balush
Mar 19 at 20:53
3
3
$begingroup$
Perfect: then take each and every coset of $N$ in $G$ and map it to $G'$ using $varphi$. Doing so will result in cosets of $varphi(N)$.
$endgroup$
– b00n heT
Mar 19 at 20:54
$begingroup$
Perfect: then take each and every coset of $N$ in $G$ and map it to $G'$ using $varphi$. Doing so will result in cosets of $varphi(N)$.
$endgroup$
– b00n heT
Mar 19 at 20:54
1
1
$begingroup$
@jgon I know of only one, but I cannot exclude that there can be more than one.
$endgroup$
– b00n heT
Mar 19 at 20:59
$begingroup$
@jgon I know of only one, but I cannot exclude that there can be more than one.
$endgroup$
– b00n heT
Mar 19 at 20:59
2
2
$begingroup$
@b00nheT Completely reasonable, I was just curious, since I also only knew of one.
$endgroup$
– jgon
Mar 19 at 21:00
$begingroup$
@b00nheT Completely reasonable, I was just curious, since I also only knew of one.
$endgroup$
– jgon
Mar 19 at 21:00
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Let $gN$ be a left coset for $gin G$. Then, since $varphi$ is a group homomorphism, if you take $nin N$, then $varphi(gn)=varphi(g)varphi(n)$, so clearly $varphi(gN)subseteqvarphi(g)varphi(N)$. Since $varphi$ is surjective, one has that every left coset of $G'$ is of the form $varphi(g)varphi(N)$.
Injectivity gives us that no two distinct cosets $g'N$ and $gN$ map to the same $varphi(g)varphi(N)$, since if $varphi(g')varphi(N)=varphi(g)varphi(N)$, then for some $n,n'in N$
$varphi(g')varphi(n')=varphi(g)varphi(n).$
But this is the same as saying $g'n'=gn$, so $g'N=gN$.
Puting all together one concludes that the number of left cosets are the same.
answered Mar 19 at 21:07
JaviJavi
3,1242832
$endgroup$
add a comment |
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$begingroup$
Let $gN$ be a left coset for $gin G$. Then, since $varphi$ is a group homomorphism, if you take $nin N$, then $varphi(gn)=varphi(g)varphi(n)$, so clearly $varphi(gN)subseteqvarphi(g)varphi(N)$. Since $varphi$ is surjective, one has that every left coset of $G'$ is of the form $varphi(g)varphi(N)$.
Injectivity gives us that no two distinct cosets $g'N$ and $gN$ map to the same $varphi(g)varphi(N)$, since if $varphi(g')varphi(N)=varphi(g)varphi(N)$, then for some $n,n'in N$
$varphi(g')varphi(n')=varphi(g)varphi(n).$
But this is the same as saying $g'n'=gn$, so $g'N=gN$.
Puting all together one concludes that the number of left cosets are the same.
answered Mar 19 at 21:07
JaviJavi
3,1242832
$endgroup$
add a comment |
$begingroup$
Let $gN$ be a left coset for $gin G$. Then, since $varphi$ is a group homomorphism, if you take $nin N$, then $varphi(gn)=varphi(g)varphi(n)$, so clearly $varphi(gN)subseteqvarphi(g)varphi(N)$. Since $varphi$ is surjective, one has that every left coset of $G'$ is of the form $varphi(g)varphi(N)$.
Injectivity gives us that no two distinct cosets $g'N$ and $gN$ map to the same $varphi(g)varphi(N)$, since if $varphi(g')varphi(N)=varphi(g)varphi(N)$, then for some $n,n'in N$
$varphi(g')varphi(n')=varphi(g)varphi(n).$
But this is the same as saying $g'n'=gn$, so $g'N=gN$.
Puting all together one concludes that the number of left cosets are the same.
answered Mar 19 at 21:07
JaviJavi
3,1242832
$endgroup$
add a comment |
$begingroup$
Let $gN$ be a left coset for $gin G$. Then, since $varphi$ is a group homomorphism, if you take $nin N$, then $varphi(gn)=varphi(g)varphi(n)$, so clearly $varphi(gN)subseteqvarphi(g)varphi(N)$. Since $varphi$ is surjective, one has that every left coset of $G'$ is of the form $varphi(g)varphi(N)$.
Injectivity gives us that no two distinct cosets $g'N$ and $gN$ map to the same $varphi(g)varphi(N)$, since if $varphi(g')varphi(N)=varphi(g)varphi(N)$, then for some $n,n'in N$
$varphi(g')varphi(n')=varphi(g)varphi(n).$
But this is the same as saying $g'n'=gn$, so $g'N=gN$.
Puting all together one concludes that the number of left cosets are the same.
answered Mar 19 at 21:07
JaviJavi
3,1242832
$endgroup$
Let $gN$ be a left coset for $gin G$. Then, since $varphi$ is a group homomorphism, if you take $nin N$, then $varphi(gn)=varphi(g)varphi(n)$, so clearly $varphi(gN)subseteqvarphi(g)varphi(N)$. Since $varphi$ is surjective, one has that every left coset of $G'$ is of the form $varphi(g)varphi(N)$.
Injectivity gives us that no two distinct cosets $g'N$ and $gN$ map to the same $varphi(g)varphi(N)$, since if $varphi(g')varphi(N)=varphi(g)varphi(N)$, then for some $n,n'in N$
$varphi(g')varphi(n')=varphi(g)varphi(n).$
But this is the same as saying $g'n'=gn$, so $g'N=gN$.
Puting all together one concludes that the number of left cosets are the same.
answered Mar 19 at 21:07
JaviJavi
3,1242832
answered Mar 19 at 21:07
JaviJavi
3,1242832
answered Mar 19 at 21:07
JaviJavi
3,1242832
answered Mar 19 at 21:07
JaviJavi
3,1242832
3,1242832
add a comment |
add a comment |
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1
$begingroup$
What is the definition of index you are using? Because there are various approaches, and depending on the definition you are using one fits better than the other.
$endgroup$
– b00n heT
Mar 19 at 20:52
$begingroup$
Index is defined as the number of lest or right cosets
$endgroup$
– Galush Balush
Mar 19 at 20:53
3
$begingroup$
Perfect: then take each and every coset of $N$ in $G$ and map it to $G'$ using $varphi$. Doing so will result in cosets of $varphi(N)$.
$endgroup$
– b00n heT
Mar 19 at 20:54
1
$begingroup$
@jgon I know of only one, but I cannot exclude that there can be more than one.
$endgroup$
– b00n heT
Mar 19 at 20:59
2
$begingroup$
@b00nheT Completely reasonable, I was just curious, since I also only knew of one.
$endgroup$
– jgon
Mar 19 at 21:00