the group homomorphism $f:Grightarrow H$ is exactly then an injection when $ker(f)=eG$
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I have to show that a group homomorphism $f:Grightarrow H$ is a injective function when $ker(f):=[uin G:f(u)=eH]=eG$,$:e$ is the neutral element.
Attempt:
If we take $ker(f)={a}$
$f(a#b)=f(a)*f(b)$
$f(a#b)=eH*f(b)=eH$
$implies ker(f)=[a,(aneq b)]$ then it wouldnt be an injection
but if $ker(f)=eG$ then it works
$f(eG#b)=f(eG)*f(b)$
$f(eG) =eH*f(b)=eH$
is that correct?
linear-algebra group-homomorphism
edited Nov 16 at 9:02
Yadati Kiran
1,049317
asked Nov 16 at 8:51
mimon67
11
add a comment |
up vote
0
down vote
favorite
I have to show that a group homomorphism $f:Grightarrow H$ is a injective function when $ker(f):=[uin G:f(u)=eH]=eG$,$:e$ is the neutral element.
Attempt:
If we take $ker(f)={a}$
$f(a#b)=f(a)*f(b)$
$f(a#b)=eH*f(b)=eH$
$implies ker(f)=[a,(aneq b)]$ then it wouldnt be an injection
but if $ker(f)=eG$ then it works
$f(eG#b)=f(eG)*f(b)$
$f(eG) =eH*f(b)=eH$
is that correct?
linear-algebra group-homomorphism
edited Nov 16 at 9:02
Yadati Kiran
1,049317
asked Nov 16 at 8:51
mimon67
11
Welcome to MSE! Please use mathjax for typing.
– Wuestenfux
Nov 16 at 9:03
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have to show that a group homomorphism $f:Grightarrow H$ is a injective function when $ker(f):=[uin G:f(u)=eH]=eG$,$:e$ is the neutral element.
Attempt:
If we take $ker(f)={a}$
$f(a#b)=f(a)*f(b)$
$f(a#b)=eH*f(b)=eH$
$implies ker(f)=[a,(aneq b)]$ then it wouldnt be an injection
but if $ker(f)=eG$ then it works
$f(eG#b)=f(eG)*f(b)$
$f(eG) =eH*f(b)=eH$
is that correct?
linear-algebra group-homomorphism
edited Nov 16 at 9:02
Yadati Kiran
1,049317
asked Nov 16 at 8:51
mimon67
11
I have to show that a group homomorphism $f:Grightarrow H$ is a injective function when $ker(f):=[uin G:f(u)=eH]=eG$,$:e$ is the neutral element.
Attempt:
If we take $ker(f)={a}$
$f(a#b)=f(a)*f(b)$
$f(a#b)=eH*f(b)=eH$
$implies ker(f)=[a,(aneq b)]$ then it wouldnt be an injection
but if $ker(f)=eG$ then it works
$f(eG#b)=f(eG)*f(b)$
$f(eG) =eH*f(b)=eH$
is that correct?
linear-algebra group-homomorphism
linear-algebra group-homomorphism
edited Nov 16 at 9:02
Yadati Kiran
1,049317
asked Nov 16 at 8:51
mimon67
11
edited Nov 16 at 9:02
Yadati Kiran
1,049317
asked Nov 16 at 8:51
mimon67
11
edited Nov 16 at 9:02
Yadati Kiran
1,049317
edited Nov 16 at 9:02
Yadati Kiran
1,049317
edited Nov 16 at 9:02
Yadati Kiran
1,049317
1,049317
asked Nov 16 at 8:51
mimon67
11
asked Nov 16 at 8:51
mimon67
11
asked Nov 16 at 8:51
mimon67
11
11
Welcome to MSE! Please use mathjax for typing.
– Wuestenfux
Nov 16 at 9:03
add a comment |
Welcome to MSE! Please use mathjax for typing.
– Wuestenfux
Nov 16 at 9:03
Welcome to MSE! Please use mathjax for typing.
– Wuestenfux
Nov 16 at 9:03
Welcome to MSE! Please use mathjax for typing.
– Wuestenfux
Nov 16 at 9:03
add a comment |
1 Answer
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$ker(f)={xin G : f(x)=e^*}$ where $e^*$ is the identity element of $H$. Its given $ker(f)={e}$. $f(x)=f(g)implies f(xg^{-1})=f(g)f(g^{-1})implies f(xg^{-1})=f(e)=e^*implies x=g $.
answered Nov 16 at 9:05
Yadati Kiran
1,049317
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
$ker(f)={xin G : f(x)=e^*}$ where $e^*$ is the identity element of $H$. Its given $ker(f)={e}$. $f(x)=f(g)implies f(xg^{-1})=f(g)f(g^{-1})implies f(xg^{-1})=f(e)=e^*implies x=g $.
answered Nov 16 at 9:05
Yadati Kiran
1,049317
add a comment |
up vote
0
down vote
$ker(f)={xin G : f(x)=e^*}$ where $e^*$ is the identity element of $H$. Its given $ker(f)={e}$. $f(x)=f(g)implies f(xg^{-1})=f(g)f(g^{-1})implies f(xg^{-1})=f(e)=e^*implies x=g $.
answered Nov 16 at 9:05
Yadati Kiran
1,049317
add a comment |
up vote
0
down vote
up vote
0
down vote
$ker(f)={xin G : f(x)=e^*}$ where $e^*$ is the identity element of $H$. Its given $ker(f)={e}$. $f(x)=f(g)implies f(xg^{-1})=f(g)f(g^{-1})implies f(xg^{-1})=f(e)=e^*implies x=g $.
answered Nov 16 at 9:05
Yadati Kiran
1,049317
$ker(f)={xin G : f(x)=e^*}$ where $e^*$ is the identity element of $H$. Its given $ker(f)={e}$. $f(x)=f(g)implies f(xg^{-1})=f(g)f(g^{-1})implies f(xg^{-1})=f(e)=e^*implies x=g $.
answered Nov 16 at 9:05
Yadati Kiran
1,049317
answered Nov 16 at 9:05
Yadati Kiran
1,049317
answered Nov 16 at 9:05
Yadati Kiran
1,049317
answered Nov 16 at 9:05
Yadati Kiran
1,049317
1,049317
add a comment |
add a comment |
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