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first isomorphism problem - injective homomorphism
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I am learning algebra and I do not understand the first isomorphism theorem correctly.
I have an injective group homomorphism $phi: G to H$. Moreover I have given that $im(phi)cong L$, with $L$ a group.
By the first isomorphism theorem it holds: $L cong G/ker(phi)$. Since, $phi$ is injective it holds $ker(phi)=e$. Does this mean $G cong L$?
linear-algebra abstract-algebra group-theory
asked Jul 20 at 18:31
0123abc
214
add a comment |Â
up vote
3
down vote
favorite
I am learning algebra and I do not understand the first isomorphism theorem correctly.
I have an injective group homomorphism $phi: G to H$. Moreover I have given that $im(phi)cong L$, with $L$ a group.
By the first isomorphism theorem it holds: $L cong G/ker(phi)$. Since, $phi$ is injective it holds $ker(phi)=e$. Does this mean $G cong L$?
linear-algebra abstract-algebra group-theory
asked Jul 20 at 18:31
0123abc
214
3
Yes, this is correct.
– Gal Porat
Jul 20 at 18:32
Looks like you understand it correctly to me.
– The Count
Jul 20 at 18:35
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I am learning algebra and I do not understand the first isomorphism theorem correctly.
I have an injective group homomorphism $phi: G to H$. Moreover I have given that $im(phi)cong L$, with $L$ a group.
By the first isomorphism theorem it holds: $L cong G/ker(phi)$. Since, $phi$ is injective it holds $ker(phi)=e$. Does this mean $G cong L$?
linear-algebra abstract-algebra group-theory
asked Jul 20 at 18:31
0123abc
214
I am learning algebra and I do not understand the first isomorphism theorem correctly.
I have an injective group homomorphism $phi: G to H$. Moreover I have given that $im(phi)cong L$, with $L$ a group.
By the first isomorphism theorem it holds: $L cong G/ker(phi)$. Since, $phi$ is injective it holds $ker(phi)=e$. Does this mean $G cong L$?
linear-algebra abstract-algebra group-theory
asked Jul 20 at 18:31
0123abc
214
asked Jul 20 at 18:31
0123abc
214
asked Jul 20 at 18:31
0123abc
214
asked Jul 20 at 18:31
0123abc
214
214
3
Yes, this is correct.
– Gal Porat
Jul 20 at 18:32
Looks like you understand it correctly to me.
– The Count
Jul 20 at 18:35
add a comment |Â
3
Yes, this is correct.
– Gal Porat
Jul 20 at 18:32
Looks like you understand it correctly to me.
– The Count
Jul 20 at 18:35
3
3
Yes, this is correct.
– Gal Porat
Jul 20 at 18:32
Yes, this is correct.
– Gal Porat
Jul 20 at 18:32
Looks like you understand it correctly to me.
– The Count
Jul 20 at 18:35
Looks like you understand it correctly to me.
– The Count
Jul 20 at 18:35
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
The assumption $rm im(phi)cong L$ says that $phi colon Grightarrow L$ is surjective, the other that it is injective. Hence it is an isomorphism of groups, i.e., $Gcong L$. The first isomorphism theorem should not come to a different conclusion. So you are right.
answered Jul 20 at 19:16
Dietrich Burde
74.6k64184
add a comment |Â
up vote
2
down vote
Whenever you have a homomorphism $phicolon Gto H$, you can factor it as $phi=betacircalpha$, where $alpha$ is surjective and $beta$ is injective; the trick is to define
$$
alphacolon Gtooperatornameim(phi),
qquad alpha(x)=phi(x)
$$
and take $beta$ as the inclusion map $operatornameim(phi)to H$.
In your particular case, $alpha$ is also injective, so it is an isomorphism. Composing it with an isomorphism $gammacolonoperatornameim(phi)to L$ (existing by assumption), we conclude that $gammacircalphacolon Gto L$ is an isomorphism.
answered Jul 20 at 19:36
egreg
164k1180187
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The assumption $rm im(phi)cong L$ says that $phi colon Grightarrow L$ is surjective, the other that it is injective. Hence it is an isomorphism of groups, i.e., $Gcong L$. The first isomorphism theorem should not come to a different conclusion. So you are right.
answered Jul 20 at 19:16
Dietrich Burde
74.6k64184
add a comment |Â
up vote
3
down vote
accepted
The assumption $rm im(phi)cong L$ says that $phi colon Grightarrow L$ is surjective, the other that it is injective. Hence it is an isomorphism of groups, i.e., $Gcong L$. The first isomorphism theorem should not come to a different conclusion. So you are right.
answered Jul 20 at 19:16
Dietrich Burde
74.6k64184
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The assumption $rm im(phi)cong L$ says that $phi colon Grightarrow L$ is surjective, the other that it is injective. Hence it is an isomorphism of groups, i.e., $Gcong L$. The first isomorphism theorem should not come to a different conclusion. So you are right.
answered Jul 20 at 19:16
Dietrich Burde
74.6k64184
The assumption $rm im(phi)cong L$ says that $phi colon Grightarrow L$ is surjective, the other that it is injective. Hence it is an isomorphism of groups, i.e., $Gcong L$. The first isomorphism theorem should not come to a different conclusion. So you are right.
answered Jul 20 at 19:16
Dietrich Burde
74.6k64184
answered Jul 20 at 19:16
Dietrich Burde
74.6k64184
answered Jul 20 at 19:16
Dietrich Burde
74.6k64184
answered Jul 20 at 19:16
Dietrich Burde
74.6k64184
74.6k64184
add a comment |Â
add a comment |Â
up vote
2
down vote
Whenever you have a homomorphism $phicolon Gto H$, you can factor it as $phi=betacircalpha$, where $alpha$ is surjective and $beta$ is injective; the trick is to define
$$
alphacolon Gtooperatornameim(phi),
qquad alpha(x)=phi(x)
$$
and take $beta$ as the inclusion map $operatornameim(phi)to H$.
In your particular case, $alpha$ is also injective, so it is an isomorphism. Composing it with an isomorphism $gammacolonoperatornameim(phi)to L$ (existing by assumption), we conclude that $gammacircalphacolon Gto L$ is an isomorphism.
answered Jul 20 at 19:36
egreg
164k1180187
add a comment |Â
up vote
2
down vote
Whenever you have a homomorphism $phicolon Gto H$, you can factor it as $phi=betacircalpha$, where $alpha$ is surjective and $beta$ is injective; the trick is to define
$$
alphacolon Gtooperatornameim(phi),
qquad alpha(x)=phi(x)
$$
and take $beta$ as the inclusion map $operatornameim(phi)to H$.
In your particular case, $alpha$ is also injective, so it is an isomorphism. Composing it with an isomorphism $gammacolonoperatornameim(phi)to L$ (existing by assumption), we conclude that $gammacircalphacolon Gto L$ is an isomorphism.
answered Jul 20 at 19:36
egreg
164k1180187
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Whenever you have a homomorphism $phicolon Gto H$, you can factor it as $phi=betacircalpha$, where $alpha$ is surjective and $beta$ is injective; the trick is to define
$$
alphacolon Gtooperatornameim(phi),
qquad alpha(x)=phi(x)
$$
and take $beta$ as the inclusion map $operatornameim(phi)to H$.
In your particular case, $alpha$ is also injective, so it is an isomorphism. Composing it with an isomorphism $gammacolonoperatornameim(phi)to L$ (existing by assumption), we conclude that $gammacircalphacolon Gto L$ is an isomorphism.
answered Jul 20 at 19:36
egreg
164k1180187
Whenever you have a homomorphism $phicolon Gto H$, you can factor it as $phi=betacircalpha$, where $alpha$ is surjective and $beta$ is injective; the trick is to define
$$
alphacolon Gtooperatornameim(phi),
qquad alpha(x)=phi(x)
$$
and take $beta$ as the inclusion map $operatornameim(phi)to H$.
In your particular case, $alpha$ is also injective, so it is an isomorphism. Composing it with an isomorphism $gammacolonoperatornameim(phi)to L$ (existing by assumption), we conclude that $gammacircalphacolon Gto L$ is an isomorphism.
answered Jul 20 at 19:36
egreg
164k1180187
answered Jul 20 at 19:36
egreg
164k1180187
answered Jul 20 at 19:36
egreg
164k1180187
answered Jul 20 at 19:36
egreg
164k1180187
164k1180187
add a comment |Â
add a comment |Â
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3
Yes, this is correct.
– Gal Porat
Jul 20 at 18:32
Looks like you understand it correctly to me.
– The Count
Jul 20 at 18:35