Given that $G/H={xH:x in G}$ is group under operation $(xH)(yH)=(xyH)$ . Then $H$ is normal subgroup of $G$
$begingroup$
let $G$ is any group and $H$ is its subgroup , such that $G/H={xH:x in G}$ is group under operation $(xH)(yH)=(xyH)$ . Then show that $H$ is normal subgroup of $G$
if $H$ is not normal ,then there exist some $xin G$ such that $xH neq Hx$, clearly $xnotin H$ now how can i get a contradiction . I know then operation on $G/H$ is not well defined. but how to prove that ?
any hint. Thanks in advanced
abstract-algebra group-theory finite-groups normal-subgroups quotient-group
asked Dec 29 '18 at 14:49
EklavyaEklavya
982515
$endgroup$
add a comment |
$begingroup$
let $G$ is any group and $H$ is its subgroup , such that $G/H={xH:x in G}$ is group under operation $(xH)(yH)=(xyH)$ . Then show that $H$ is normal subgroup of $G$
if $H$ is not normal ,then there exist some $xin G$ such that $xH neq Hx$, clearly $xnotin H$ now how can i get a contradiction . I know then operation on $G/H$ is not well defined. but how to prove that ?
any hint. Thanks in advanced
abstract-algebra group-theory finite-groups normal-subgroups quotient-group
asked Dec 29 '18 at 14:49
EklavyaEklavya
982515
$endgroup$
add a comment |
$begingroup$
let $G$ is any group and $H$ is its subgroup , such that $G/H={xH:x in G}$ is group under operation $(xH)(yH)=(xyH)$ . Then show that $H$ is normal subgroup of $G$
if $H$ is not normal ,then there exist some $xin G$ such that $xH neq Hx$, clearly $xnotin H$ now how can i get a contradiction . I know then operation on $G/H$ is not well defined. but how to prove that ?
any hint. Thanks in advanced
abstract-algebra group-theory finite-groups normal-subgroups quotient-group
asked Dec 29 '18 at 14:49
EklavyaEklavya
982515
$endgroup$
let $G$ is any group and $H$ is its subgroup , such that $G/H={xH:x in G}$ is group under operation $(xH)(yH)=(xyH)$ . Then show that $H$ is normal subgroup of $G$
if $H$ is not normal ,then there exist some $xin G$ such that $xH neq Hx$, clearly $xnotin H$ now how can i get a contradiction . I know then operation on $G/H$ is not well defined. but how to prove that ?
any hint. Thanks in advanced
abstract-algebra group-theory finite-groups normal-subgroups quotient-group
abstract-algebra group-theory finite-groups normal-subgroups quotient-group
asked Dec 29 '18 at 14:49
EklavyaEklavya
982515
asked Dec 29 '18 at 14:49
EklavyaEklavya
982515
edited Dec 29 '18 at 14:58
Eklavya
asked Dec 29 '18 at 14:49
EklavyaEklavya
982515
asked Dec 29 '18 at 14:49
EklavyaEklavya
982515
asked Dec 29 '18 at 14:49
EklavyaEklavya
982515
982515
add a comment |
add a comment |
2 Answers
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$begingroup$
We clearly have $xhH=xH$ for $xin G, hin H$. Then, by being well defined, we arrive to
$$(xhx^{-1})H=xhHcdot x^{-1}H=xHcdot x^{-1}H=(xx^{-1})H=eH=H$$
Thus, $xhx^{-1}in H$.
answered Dec 29 '18 at 15:03
BerciBerci
61.6k23674
$endgroup$
add a comment |
$begingroup$
The set $G/H$ is well-defined even if $H$ is not normal, because we can just define $G/H$ as the set of all left cosets on $H$. The problem is with the fact that, if $H$ is not normal, this set does not satisfy the group axioms under the operation $(xH)(yH)=(xy)H$. Now, there is probably a way to do this through a proof by contradiction using the group axioms, like you are trying to do. However, I feel like it might be easier to do a more indirect proof using homomorphisms.
Let's assume $G/H$ is a group under the operation $(xH)(yH)=(xy)H$. Then, let us define the homomorphism $phi : G rightarrow G/H$ by $phi(g)=gH$. This is a homomorphism because:
$$phi(gH)phi(g_2H)=(gH)(g_2)H=(gg_2)H=phi(gg_2)$$
Now, what is the kernel of this homomorphism? Well, the identity in $G/H$ is $H$, so we need to solve $phi(g)=gH=H$. This happens if and only if $g in H$, so the kernel of $phi$ is $H$. However, the kernel of any homomorphism is a normal subgroup of the domain, so $H$ is a normal subgroup of $G$.
answered Dec 29 '18 at 15:02
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
We clearly have $xhH=xH$ for $xin G, hin H$. Then, by being well defined, we arrive to
$$(xhx^{-1})H=xhHcdot x^{-1}H=xHcdot x^{-1}H=(xx^{-1})H=eH=H$$
Thus, $xhx^{-1}in H$.
answered Dec 29 '18 at 15:03
BerciBerci
61.6k23674
$endgroup$
add a comment |
$begingroup$
We clearly have $xhH=xH$ for $xin G, hin H$. Then, by being well defined, we arrive to
$$(xhx^{-1})H=xhHcdot x^{-1}H=xHcdot x^{-1}H=(xx^{-1})H=eH=H$$
Thus, $xhx^{-1}in H$.
answered Dec 29 '18 at 15:03
BerciBerci
61.6k23674
$endgroup$
add a comment |
$begingroup$
We clearly have $xhH=xH$ for $xin G, hin H$. Then, by being well defined, we arrive to
$$(xhx^{-1})H=xhHcdot x^{-1}H=xHcdot x^{-1}H=(xx^{-1})H=eH=H$$
Thus, $xhx^{-1}in H$.
answered Dec 29 '18 at 15:03
BerciBerci
61.6k23674
$endgroup$
We clearly have $xhH=xH$ for $xin G, hin H$. Then, by being well defined, we arrive to
$$(xhx^{-1})H=xhHcdot x^{-1}H=xHcdot x^{-1}H=(xx^{-1})H=eH=H$$
Thus, $xhx^{-1}in H$.
answered Dec 29 '18 at 15:03
BerciBerci
61.6k23674
answered Dec 29 '18 at 15:03
BerciBerci
61.6k23674
answered Dec 29 '18 at 15:03
BerciBerci
61.6k23674
answered Dec 29 '18 at 15:03
BerciBerci
61.6k23674
61.6k23674
add a comment |
add a comment |
$begingroup$
The set $G/H$ is well-defined even if $H$ is not normal, because we can just define $G/H$ as the set of all left cosets on $H$. The problem is with the fact that, if $H$ is not normal, this set does not satisfy the group axioms under the operation $(xH)(yH)=(xy)H$. Now, there is probably a way to do this through a proof by contradiction using the group axioms, like you are trying to do. However, I feel like it might be easier to do a more indirect proof using homomorphisms.
Let's assume $G/H$ is a group under the operation $(xH)(yH)=(xy)H$. Then, let us define the homomorphism $phi : G rightarrow G/H$ by $phi(g)=gH$. This is a homomorphism because:
$$phi(gH)phi(g_2H)=(gH)(g_2)H=(gg_2)H=phi(gg_2)$$
Now, what is the kernel of this homomorphism? Well, the identity in $G/H$ is $H$, so we need to solve $phi(g)=gH=H$. This happens if and only if $g in H$, so the kernel of $phi$ is $H$. However, the kernel of any homomorphism is a normal subgroup of the domain, so $H$ is a normal subgroup of $G$.
answered Dec 29 '18 at 15:02
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
The set $G/H$ is well-defined even if $H$ is not normal, because we can just define $G/H$ as the set of all left cosets on $H$. The problem is with the fact that, if $H$ is not normal, this set does not satisfy the group axioms under the operation $(xH)(yH)=(xy)H$. Now, there is probably a way to do this through a proof by contradiction using the group axioms, like you are trying to do. However, I feel like it might be easier to do a more indirect proof using homomorphisms.
Let's assume $G/H$ is a group under the operation $(xH)(yH)=(xy)H$. Then, let us define the homomorphism $phi : G rightarrow G/H$ by $phi(g)=gH$. This is a homomorphism because:
$$phi(gH)phi(g_2H)=(gH)(g_2)H=(gg_2)H=phi(gg_2)$$
Now, what is the kernel of this homomorphism? Well, the identity in $G/H$ is $H$, so we need to solve $phi(g)=gH=H$. This happens if and only if $g in H$, so the kernel of $phi$ is $H$. However, the kernel of any homomorphism is a normal subgroup of the domain, so $H$ is a normal subgroup of $G$.
answered Dec 29 '18 at 15:02
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
The set $G/H$ is well-defined even if $H$ is not normal, because we can just define $G/H$ as the set of all left cosets on $H$. The problem is with the fact that, if $H$ is not normal, this set does not satisfy the group axioms under the operation $(xH)(yH)=(xy)H$. Now, there is probably a way to do this through a proof by contradiction using the group axioms, like you are trying to do. However, I feel like it might be easier to do a more indirect proof using homomorphisms.
Let's assume $G/H$ is a group under the operation $(xH)(yH)=(xy)H$. Then, let us define the homomorphism $phi : G rightarrow G/H$ by $phi(g)=gH$. This is a homomorphism because:
$$phi(gH)phi(g_2H)=(gH)(g_2)H=(gg_2)H=phi(gg_2)$$
Now, what is the kernel of this homomorphism? Well, the identity in $G/H$ is $H$, so we need to solve $phi(g)=gH=H$. This happens if and only if $g in H$, so the kernel of $phi$ is $H$. However, the kernel of any homomorphism is a normal subgroup of the domain, so $H$ is a normal subgroup of $G$.
answered Dec 29 '18 at 15:02
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
The set $G/H$ is well-defined even if $H$ is not normal, because we can just define $G/H$ as the set of all left cosets on $H$. The problem is with the fact that, if $H$ is not normal, this set does not satisfy the group axioms under the operation $(xH)(yH)=(xy)H$. Now, there is probably a way to do this through a proof by contradiction using the group axioms, like you are trying to do. However, I feel like it might be easier to do a more indirect proof using homomorphisms.
Let's assume $G/H$ is a group under the operation $(xH)(yH)=(xy)H$. Then, let us define the homomorphism $phi : G rightarrow G/H$ by $phi(g)=gH$. This is a homomorphism because:
$$phi(gH)phi(g_2H)=(gH)(g_2)H=(gg_2)H=phi(gg_2)$$
Now, what is the kernel of this homomorphism? Well, the identity in $G/H$ is $H$, so we need to solve $phi(g)=gH=H$. This happens if and only if $g in H$, so the kernel of $phi$ is $H$. However, the kernel of any homomorphism is a normal subgroup of the domain, so $H$ is a normal subgroup of $G$.
answered Dec 29 '18 at 15:02
Noble MushtakNoble Mushtak
15.3k1835
answered Dec 29 '18 at 15:02
Noble MushtakNoble Mushtak
15.3k1835
answered Dec 29 '18 at 15:02
Noble MushtakNoble Mushtak
15.3k1835
answered Dec 29 '18 at 15:02
Noble MushtakNoble Mushtak
15.3k1835
15.3k1835
add a comment |
add a comment |
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