Mixing
Show that $f^{-1}{f(a)}=atext{Ker}f$
$begingroup$
Let $G,G'$ be groups and assume $f: space G longrightarrow G'$ is a
homomorphism. Show that $forall f(a) in operatorname{Im}f$: $$
f^{-1}{f(a)}=a operatorname{Ker}f $$
To begin with, the kernel of the homomorphism is a normal subgroup of $G$, $ operatorname{Ker}f triangleleft G.$ This implies that the left and right cosets coincide. Thus, the set of cosets of $ operatorname{Ker} f$ is:
$$
G/ operatorname{ker}f:=left{ a operatorname{Ker}f, space forall a in G right}
$$
Now, since $ operatorname{ker}f triangleleft G$, $G/ operatorname{Ker}f$ is a group. Moreover, the fundamental theorem of homomorphisms gives:
$$
G/ operatorname{Ker}f cong operatorname{Im}f
$$
and so there exists a bijection $varphi: space G/ operatorname{Ker}f longrightarrow operatorname{Im}f;$ s.t.
$$
varphi(a operatorname{Ker}f)= operatorname{Im}f, space forall a in G
$$
Given that, we must show:
$$
f^{-1}[varphi(a operatorname{Ker}f)]=a operatorname{Ker}f
$$
Question:
Which argument about the relation of $f$ and $varphi$ should one use to prove the initial statement?
abstract-algebra group-theory group-isomorphism group-homomorphism
edited Feb 1 at 17:19
Shaun
10.5k113687
asked Feb 1 at 15:44
JevautJevaut
1,168312
$endgroup$
add a comment |
$begingroup$
Let $G,G'$ be groups and assume $f: space G longrightarrow G'$ is a
homomorphism. Show that $forall f(a) in operatorname{Im}f$: $$
f^{-1}{f(a)}=a operatorname{Ker}f $$
To begin with, the kernel of the homomorphism is a normal subgroup of $G$, $ operatorname{Ker}f triangleleft G.$ This implies that the left and right cosets coincide. Thus, the set of cosets of $ operatorname{Ker} f$ is:
$$
G/ operatorname{ker}f:=left{ a operatorname{Ker}f, space forall a in G right}
$$
Now, since $ operatorname{ker}f triangleleft G$, $G/ operatorname{Ker}f$ is a group. Moreover, the fundamental theorem of homomorphisms gives:
$$
G/ operatorname{Ker}f cong operatorname{Im}f
$$
and so there exists a bijection $varphi: space G/ operatorname{Ker}f longrightarrow operatorname{Im}f;$ s.t.
$$
varphi(a operatorname{Ker}f)= operatorname{Im}f, space forall a in G
$$
Given that, we must show:
$$
f^{-1}[varphi(a operatorname{Ker}f)]=a operatorname{Ker}f
$$
Question:
Which argument about the relation of $f$ and $varphi$ should one use to prove the initial statement?
abstract-algebra group-theory group-isomorphism group-homomorphism
edited Feb 1 at 17:19
Shaun
10.5k113687
asked Feb 1 at 15:44
JevautJevaut
1,168312
$endgroup$
5
$begingroup$
Are you imposed to use all this armada of theorems? I mean a direct proof of $f^{-1}{f(a)}=atext{Ker}f$ is quite simple.
$endgroup$
– mathcounterexamples.net
Feb 1 at 15:52
add a comment |
$begingroup$
Let $G,G'$ be groups and assume $f: space G longrightarrow G'$ is a
homomorphism. Show that $forall f(a) in operatorname{Im}f$: $$
f^{-1}{f(a)}=a operatorname{Ker}f $$
To begin with, the kernel of the homomorphism is a normal subgroup of $G$, $ operatorname{Ker}f triangleleft G.$ This implies that the left and right cosets coincide. Thus, the set of cosets of $ operatorname{Ker} f$ is:
$$
G/ operatorname{ker}f:=left{ a operatorname{Ker}f, space forall a in G right}
$$
Now, since $ operatorname{ker}f triangleleft G$, $G/ operatorname{Ker}f$ is a group. Moreover, the fundamental theorem of homomorphisms gives:
$$
G/ operatorname{Ker}f cong operatorname{Im}f
$$
and so there exists a bijection $varphi: space G/ operatorname{Ker}f longrightarrow operatorname{Im}f;$ s.t.
$$
varphi(a operatorname{Ker}f)= operatorname{Im}f, space forall a in G
$$
Given that, we must show:
$$
f^{-1}[varphi(a operatorname{Ker}f)]=a operatorname{Ker}f
$$
Question:
Which argument about the relation of $f$ and $varphi$ should one use to prove the initial statement?
abstract-algebra group-theory group-isomorphism group-homomorphism
edited Feb 1 at 17:19
Shaun
10.5k113687
asked Feb 1 at 15:44
JevautJevaut
1,168312
$endgroup$
Let $G,G'$ be groups and assume $f: space G longrightarrow G'$ is a
homomorphism. Show that $forall f(a) in operatorname{Im}f$: $$
f^{-1}{f(a)}=a operatorname{Ker}f $$
To begin with, the kernel of the homomorphism is a normal subgroup of $G$, $ operatorname{Ker}f triangleleft G.$ This implies that the left and right cosets coincide. Thus, the set of cosets of $ operatorname{Ker} f$ is:
$$
G/ operatorname{ker}f:=left{ a operatorname{Ker}f, space forall a in G right}
$$
Now, since $ operatorname{ker}f triangleleft G$, $G/ operatorname{Ker}f$ is a group. Moreover, the fundamental theorem of homomorphisms gives:
$$
G/ operatorname{Ker}f cong operatorname{Im}f
$$
and so there exists a bijection $varphi: space G/ operatorname{Ker}f longrightarrow operatorname{Im}f;$ s.t.
$$
varphi(a operatorname{Ker}f)= operatorname{Im}f, space forall a in G
$$
Given that, we must show:
$$
f^{-1}[varphi(a operatorname{Ker}f)]=a operatorname{Ker}f
$$
Question:
Which argument about the relation of $f$ and $varphi$ should one use to prove the initial statement?
abstract-algebra group-theory group-isomorphism group-homomorphism
abstract-algebra group-theory group-isomorphism group-homomorphism
edited Feb 1 at 17:19
Shaun
10.5k113687
asked Feb 1 at 15:44
JevautJevaut
1,168312
edited Feb 1 at 17:19
Shaun
10.5k113687
asked Feb 1 at 15:44
JevautJevaut
1,168312
edited Feb 1 at 17:19
Shaun
10.5k113687
edited Feb 1 at 17:19
Shaun
10.5k113687
edited Feb 1 at 17:19
Shaun
10.5k113687
10.5k113687
asked Feb 1 at 15:44
JevautJevaut
1,168312
asked Feb 1 at 15:44
JevautJevaut
1,168312
asked Feb 1 at 15:44
JevautJevaut
1,168312
1,168312
5
$begingroup$
Are you imposed to use all this armada of theorems? I mean a direct proof of $f^{-1}{f(a)}=atext{Ker}f$ is quite simple.
$endgroup$
– mathcounterexamples.net
Feb 1 at 15:52
add a comment |
5
$begingroup$
Are you imposed to use all this armada of theorems? I mean a direct proof of $f^{-1}{f(a)}=atext{Ker}f$ is quite simple.
$endgroup$
– mathcounterexamples.net
Feb 1 at 15:52
5
5
$begingroup$
Are you imposed to use all this armada of theorems? I mean a direct proof of $f^{-1}{f(a)}=atext{Ker}f$ is quite simple.
$endgroup$
– mathcounterexamples.net
Feb 1 at 15:52
$begingroup$
Are you imposed to use all this armada of theorems? I mean a direct proof of $f^{-1}{f(a)}=atext{Ker}f$ is quite simple.
$endgroup$
– mathcounterexamples.net
Feb 1 at 15:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You're overthinking the problem.
Just prove that $f(b)=f(a)$ iff $b=ak$ with $k in ker f$.
answered Feb 1 at 15:53
lhflhf
168k11172404
$endgroup$
$begingroup$
Indeed, I overcomplicated this. Simple proof: $f(ak)=f(a) cdot f(k)=f(a) cdot e'=f(a)$
$endgroup$
– Jevaut
Feb 1 at 16:55
1
$begingroup$
@Jevaut, yes, that's one side. But the other side is equally easy.
$endgroup$
– lhf
Feb 1 at 17:37
add a comment |
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$begingroup$
You're overthinking the problem.
Just prove that $f(b)=f(a)$ iff $b=ak$ with $k in ker f$.
answered Feb 1 at 15:53
lhflhf
168k11172404
$endgroup$
$begingroup$
Indeed, I overcomplicated this. Simple proof: $f(ak)=f(a) cdot f(k)=f(a) cdot e'=f(a)$
$endgroup$
– Jevaut
Feb 1 at 16:55
1
$begingroup$
@Jevaut, yes, that's one side. But the other side is equally easy.
$endgroup$
– lhf
Feb 1 at 17:37
add a comment |
$begingroup$
You're overthinking the problem.
Just prove that $f(b)=f(a)$ iff $b=ak$ with $k in ker f$.
answered Feb 1 at 15:53
lhflhf
168k11172404
$endgroup$
$begingroup$
Indeed, I overcomplicated this. Simple proof: $f(ak)=f(a) cdot f(k)=f(a) cdot e'=f(a)$
$endgroup$
– Jevaut
Feb 1 at 16:55
1
$begingroup$
@Jevaut, yes, that's one side. But the other side is equally easy.
$endgroup$
– lhf
Feb 1 at 17:37
add a comment |
$begingroup$
You're overthinking the problem.
Just prove that $f(b)=f(a)$ iff $b=ak$ with $k in ker f$.
answered Feb 1 at 15:53
lhflhf
168k11172404
$endgroup$
You're overthinking the problem.
Just prove that $f(b)=f(a)$ iff $b=ak$ with $k in ker f$.
answered Feb 1 at 15:53
lhflhf
168k11172404
answered Feb 1 at 15:53
lhflhf
168k11172404
answered Feb 1 at 15:53
lhflhf
168k11172404
answered Feb 1 at 15:53
lhflhf
168k11172404
168k11172404
$begingroup$
Indeed, I overcomplicated this. Simple proof: $f(ak)=f(a) cdot f(k)=f(a) cdot e'=f(a)$
$endgroup$
– Jevaut
Feb 1 at 16:55
1
$begingroup$
@Jevaut, yes, that's one side. But the other side is equally easy.
$endgroup$
– lhf
Feb 1 at 17:37
add a comment |
$begingroup$
Indeed, I overcomplicated this. Simple proof: $f(ak)=f(a) cdot f(k)=f(a) cdot e'=f(a)$
$endgroup$
– Jevaut
Feb 1 at 16:55
1
$begingroup$
@Jevaut, yes, that's one side. But the other side is equally easy.
$endgroup$
– lhf
Feb 1 at 17:37
$begingroup$
Indeed, I overcomplicated this. Simple proof: $f(ak)=f(a) cdot f(k)=f(a) cdot e'=f(a)$
$endgroup$
– Jevaut
Feb 1 at 16:55
$begingroup$
Indeed, I overcomplicated this. Simple proof: $f(ak)=f(a) cdot f(k)=f(a) cdot e'=f(a)$
$endgroup$
– Jevaut
Feb 1 at 16:55
1
1
$begingroup$
@Jevaut, yes, that's one side. But the other side is equally easy.
$endgroup$
– lhf
Feb 1 at 17:37
$begingroup$
@Jevaut, yes, that's one side. But the other side is equally easy.
$endgroup$
– lhf
Feb 1 at 17:37
add a comment |
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5
$begingroup$
Are you imposed to use all this armada of theorems? I mean a direct proof of $f^{-1}{f(a)}=atext{Ker}f$ is quite simple.
$endgroup$
– mathcounterexamples.net
Feb 1 at 15:52