Show $Ktimes Hcong G$
$begingroup$
Let $G$ be an abelian group and $phi :Gto H$ is a surjective homomorphism with kernel $K$. Suppose there is a homomorphism $psi :Hto G $ such that $phipsi$ is identity map on $H$. Show $Ktimes Hcong G$
This is a bit like first homomorphism theorem , we already know $G/Kcong H$. What confuse me is to define a proper isomorphism.
abstract-algebra
asked Dec 6 '18 at 14:16
Jaqen ChouJaqen Chou
441110
$endgroup$
|
show 1 more comment
$begingroup$
Let $G$ be an abelian group and $phi :Gto H$ is a surjective homomorphism with kernel $K$. Suppose there is a homomorphism $psi :Hto G $ such that $phipsi$ is identity map on $H$. Show $Ktimes Hcong G$
This is a bit like first homomorphism theorem , we already know $G/Kcong H$. What confuse me is to define a proper isomorphism.
abstract-algebra
asked Dec 6 '18 at 14:16
Jaqen ChouJaqen Chou
441110
$endgroup$
1
$begingroup$
Map $(k,h)mapsto kpsi(h)$ for all $kin K, hin H$, and check everything that needs to be checked.
$endgroup$
– Jyrki Lahtonen
Dec 6 '18 at 14:18
$begingroup$
You have a canonical inclusion map $Khookrightarrow G$, and you're given a homomorphism $psi:Hto G$. By definition of $times$, this gives you a homomorphism $Ktimes Hto G$ (a priori this is not necessarily the same homomorphism as Jyrki's above, but it will turn out to be). Check everything that needs to be checked.
$endgroup$
– Arthur
Dec 6 '18 at 14:20
2
$begingroup$
Take $gin G$. Let $h = phi(g)$ and $k = gpsi(h)^{-1}$, which makes $kpsi(h) = g$. You need to show that $kin K$, and here I suspect it is crucial that $K$ is the kernel of $phi$.
$endgroup$
– Arthur
Dec 6 '18 at 14:46
$begingroup$
That's an awesome statement! I'm wondering where can we use this result?
$endgroup$
– nafhgood
Dec 6 '18 at 15:05
1
$begingroup$
@mathnoob This result is called the splitting lemma and since it has a name, it's important.
$endgroup$
– Arthur
Dec 6 '18 at 15:06
|
show 1 more comment
$begingroup$
Let $G$ be an abelian group and $phi :Gto H$ is a surjective homomorphism with kernel $K$. Suppose there is a homomorphism $psi :Hto G $ such that $phipsi$ is identity map on $H$. Show $Ktimes Hcong G$
This is a bit like first homomorphism theorem , we already know $G/Kcong H$. What confuse me is to define a proper isomorphism.
abstract-algebra
asked Dec 6 '18 at 14:16
Jaqen ChouJaqen Chou
441110
$endgroup$
Let $G$ be an abelian group and $phi :Gto H$ is a surjective homomorphism with kernel $K$. Suppose there is a homomorphism $psi :Hto G $ such that $phipsi$ is identity map on $H$. Show $Ktimes Hcong G$
This is a bit like first homomorphism theorem , we already know $G/Kcong H$. What confuse me is to define a proper isomorphism.
abstract-algebra
abstract-algebra
asked Dec 6 '18 at 14:16
Jaqen ChouJaqen Chou
441110
asked Dec 6 '18 at 14:16
Jaqen ChouJaqen Chou
441110
asked Dec 6 '18 at 14:16
Jaqen ChouJaqen Chou
441110
asked Dec 6 '18 at 14:16
Jaqen ChouJaqen Chou
441110
asked Dec 6 '18 at 14:16
Jaqen ChouJaqen Chou
441110
441110
1
$begingroup$
Map $(k,h)mapsto kpsi(h)$ for all $kin K, hin H$, and check everything that needs to be checked.
$endgroup$
– Jyrki Lahtonen
Dec 6 '18 at 14:18
$begingroup$
You have a canonical inclusion map $Khookrightarrow G$, and you're given a homomorphism $psi:Hto G$. By definition of $times$, this gives you a homomorphism $Ktimes Hto G$ (a priori this is not necessarily the same homomorphism as Jyrki's above, but it will turn out to be). Check everything that needs to be checked.
$endgroup$
– Arthur
Dec 6 '18 at 14:20
2
$begingroup$
Take $gin G$. Let $h = phi(g)$ and $k = gpsi(h)^{-1}$, which makes $kpsi(h) = g$. You need to show that $kin K$, and here I suspect it is crucial that $K$ is the kernel of $phi$.
$endgroup$
– Arthur
Dec 6 '18 at 14:46
$begingroup$
That's an awesome statement! I'm wondering where can we use this result?
$endgroup$
– nafhgood
Dec 6 '18 at 15:05
1
$begingroup$
@mathnoob This result is called the splitting lemma and since it has a name, it's important.
$endgroup$
– Arthur
Dec 6 '18 at 15:06
|
show 1 more comment
1
$begingroup$
Map $(k,h)mapsto kpsi(h)$ for all $kin K, hin H$, and check everything that needs to be checked.
$endgroup$
– Jyrki Lahtonen
Dec 6 '18 at 14:18
$begingroup$
You have a canonical inclusion map $Khookrightarrow G$, and you're given a homomorphism $psi:Hto G$. By definition of $times$, this gives you a homomorphism $Ktimes Hto G$ (a priori this is not necessarily the same homomorphism as Jyrki's above, but it will turn out to be). Check everything that needs to be checked.
$endgroup$
– Arthur
Dec 6 '18 at 14:20
2
$begingroup$
Take $gin G$. Let $h = phi(g)$ and $k = gpsi(h)^{-1}$, which makes $kpsi(h) = g$. You need to show that $kin K$, and here I suspect it is crucial that $K$ is the kernel of $phi$.
$endgroup$
– Arthur
Dec 6 '18 at 14:46
$begingroup$
That's an awesome statement! I'm wondering where can we use this result?
$endgroup$
– nafhgood
Dec 6 '18 at 15:05
1
$begingroup$
@mathnoob This result is called the splitting lemma and since it has a name, it's important.
$endgroup$
– Arthur
Dec 6 '18 at 15:06
1
1
$begingroup$
Map $(k,h)mapsto kpsi(h)$ for all $kin K, hin H$, and check everything that needs to be checked.
$endgroup$
– Jyrki Lahtonen
Dec 6 '18 at 14:18
$begingroup$
Map $(k,h)mapsto kpsi(h)$ for all $kin K, hin H$, and check everything that needs to be checked.
$endgroup$
– Jyrki Lahtonen
Dec 6 '18 at 14:18
$begingroup$
You have a canonical inclusion map $Khookrightarrow G$, and you're given a homomorphism $psi:Hto G$. By definition of $times$, this gives you a homomorphism $Ktimes Hto G$ (a priori this is not necessarily the same homomorphism as Jyrki's above, but it will turn out to be). Check everything that needs to be checked.
$endgroup$
– Arthur
Dec 6 '18 at 14:20
$begingroup$
You have a canonical inclusion map $Khookrightarrow G$, and you're given a homomorphism $psi:Hto G$. By definition of $times$, this gives you a homomorphism $Ktimes Hto G$ (a priori this is not necessarily the same homomorphism as Jyrki's above, but it will turn out to be). Check everything that needs to be checked.
$endgroup$
– Arthur
Dec 6 '18 at 14:20
2
2
$begingroup$
Take $gin G$. Let $h = phi(g)$ and $k = gpsi(h)^{-1}$, which makes $kpsi(h) = g$. You need to show that $kin K$, and here I suspect it is crucial that $K$ is the kernel of $phi$.
$endgroup$
– Arthur
Dec 6 '18 at 14:46
$begingroup$
Take $gin G$. Let $h = phi(g)$ and $k = gpsi(h)^{-1}$, which makes $kpsi(h) = g$. You need to show that $kin K$, and here I suspect it is crucial that $K$ is the kernel of $phi$.
$endgroup$
– Arthur
Dec 6 '18 at 14:46
$begingroup$
That's an awesome statement! I'm wondering where can we use this result?
$endgroup$
– nafhgood
Dec 6 '18 at 15:05
$begingroup$
That's an awesome statement! I'm wondering where can we use this result?
$endgroup$
– nafhgood
Dec 6 '18 at 15:05
1
1
$begingroup$
@mathnoob This result is called the splitting lemma and since it has a name, it's important.
$endgroup$
– Arthur
Dec 6 '18 at 15:06
$begingroup$
@mathnoob This result is called the splitting lemma and since it has a name, it's important.
$endgroup$
– Arthur
Dec 6 '18 at 15:06
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
So consider a map $f :K times H rightarrow G$, $f((k,h))=k+psi(h)$. Now we show that this map is an isomorphism:
$f((k_1,h_1)+(k_2,h_2))=f((k_1+k_2,h_1+h_2))=(k_1+k_2)+psi(h_1+h_2)=(k_1+k_2)+(psi(h_1)+psi(h_2))=k_1+psi(h_1)+k_2+psi(h_2)=f((k_1,h_1))+f((k_2+h_2))$. So this show $f$ is an homomorphism.
Now for bijectivity, Consider $f((k_1,h_1))=f((k_2,h_2))$, this means $k_1+psi(h_1)=k_2+psi(h_2)$. Apply $phi$ to both side of the equation to get $phi(psi(h_1))=phi(psi(h_2))$ which implies $h_1=h_2$ which implies $k_1+h_1=k_2+h_2$ which implies $k_1=k_2$ which implies $(k_1,h_1)=(k_2,h_2)$. So this shows injectivity.
Now for surjectivity, Let $gin G$, $k=gpsi(phi(g))^{-1}$, then $phi(k)=phi(g)phi(g)^{-1}=1$ so $kin K$ and then $g=kpsi(phi(g))$. So $(k, phi(g))$ maps to $g$.
answered Dec 6 '18 at 14:44
nafhgoodnafhgood
1,803422
$endgroup$
add a comment |
$begingroup$
It is a well known result that if $K, L$ are normal in some group $G$ and $KL = G$, $K cap L = 1$, then $G simeq K times L$.
From the fact that $phipsi = 1_H$, we know that $psi$ is injective (Hint: take elements that map to the same image and apply $phi$). Hence $im psi simeq H$ and so $im psi times K simeq H times K$. Therefore we can use the aforementioned result with $L = im psi subset G$ and $K = ker phi subset G$, because $G$ is abelian and so every subgroup is normal. We will have then proved that $G simeq im psi times K simeq H times K$, as desired. In effect,
let $x in K cap L$. Then $x = psi(h)$ for some $h in H$ and also
$$
1 = phi(x) = phipsi(h) = id(h) = h
$$
which proves that $x = psi(h) = psi(1) = 1$ and thus $K cap L = 1$.take $g in G$. Now $g = g(psiphi(g))^{-1}psiphi(g)$. Since
$$
phi(g(psiphi(g))^{-1}) = phi(g)phi(psiphi(g))^{-1}) =\ phi(g)phi(psiphi(g)))^{-1} = phi(g)(phipsiphi)(g)^{-1} = phi(g)phi(g)^{-1} = 1,
$$
as $phipsiphi = 1_Hphi = phi$, then $g in KL$, which concludes the proof.
answered Dec 6 '18 at 15:15
Guido A.Guido A.
7,8631730
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
So consider a map $f :K times H rightarrow G$, $f((k,h))=k+psi(h)$. Now we show that this map is an isomorphism:
$f((k_1,h_1)+(k_2,h_2))=f((k_1+k_2,h_1+h_2))=(k_1+k_2)+psi(h_1+h_2)=(k_1+k_2)+(psi(h_1)+psi(h_2))=k_1+psi(h_1)+k_2+psi(h_2)=f((k_1,h_1))+f((k_2+h_2))$. So this show $f$ is an homomorphism.
Now for bijectivity, Consider $f((k_1,h_1))=f((k_2,h_2))$, this means $k_1+psi(h_1)=k_2+psi(h_2)$. Apply $phi$ to both side of the equation to get $phi(psi(h_1))=phi(psi(h_2))$ which implies $h_1=h_2$ which implies $k_1+h_1=k_2+h_2$ which implies $k_1=k_2$ which implies $(k_1,h_1)=(k_2,h_2)$. So this shows injectivity.
Now for surjectivity, Let $gin G$, $k=gpsi(phi(g))^{-1}$, then $phi(k)=phi(g)phi(g)^{-1}=1$ so $kin K$ and then $g=kpsi(phi(g))$. So $(k, phi(g))$ maps to $g$.
answered Dec 6 '18 at 14:44
nafhgoodnafhgood
1,803422
$endgroup$
add a comment |
$begingroup$
So consider a map $f :K times H rightarrow G$, $f((k,h))=k+psi(h)$. Now we show that this map is an isomorphism:
$f((k_1,h_1)+(k_2,h_2))=f((k_1+k_2,h_1+h_2))=(k_1+k_2)+psi(h_1+h_2)=(k_1+k_2)+(psi(h_1)+psi(h_2))=k_1+psi(h_1)+k_2+psi(h_2)=f((k_1,h_1))+f((k_2+h_2))$. So this show $f$ is an homomorphism.
Now for bijectivity, Consider $f((k_1,h_1))=f((k_2,h_2))$, this means $k_1+psi(h_1)=k_2+psi(h_2)$. Apply $phi$ to both side of the equation to get $phi(psi(h_1))=phi(psi(h_2))$ which implies $h_1=h_2$ which implies $k_1+h_1=k_2+h_2$ which implies $k_1=k_2$ which implies $(k_1,h_1)=(k_2,h_2)$. So this shows injectivity.
Now for surjectivity, Let $gin G$, $k=gpsi(phi(g))^{-1}$, then $phi(k)=phi(g)phi(g)^{-1}=1$ so $kin K$ and then $g=kpsi(phi(g))$. So $(k, phi(g))$ maps to $g$.
answered Dec 6 '18 at 14:44
nafhgoodnafhgood
1,803422
$endgroup$
add a comment |
$begingroup$
So consider a map $f :K times H rightarrow G$, $f((k,h))=k+psi(h)$. Now we show that this map is an isomorphism:
$f((k_1,h_1)+(k_2,h_2))=f((k_1+k_2,h_1+h_2))=(k_1+k_2)+psi(h_1+h_2)=(k_1+k_2)+(psi(h_1)+psi(h_2))=k_1+psi(h_1)+k_2+psi(h_2)=f((k_1,h_1))+f((k_2+h_2))$. So this show $f$ is an homomorphism.
Now for bijectivity, Consider $f((k_1,h_1))=f((k_2,h_2))$, this means $k_1+psi(h_1)=k_2+psi(h_2)$. Apply $phi$ to both side of the equation to get $phi(psi(h_1))=phi(psi(h_2))$ which implies $h_1=h_2$ which implies $k_1+h_1=k_2+h_2$ which implies $k_1=k_2$ which implies $(k_1,h_1)=(k_2,h_2)$. So this shows injectivity.
Now for surjectivity, Let $gin G$, $k=gpsi(phi(g))^{-1}$, then $phi(k)=phi(g)phi(g)^{-1}=1$ so $kin K$ and then $g=kpsi(phi(g))$. So $(k, phi(g))$ maps to $g$.
answered Dec 6 '18 at 14:44
nafhgoodnafhgood
1,803422
$endgroup$
So consider a map $f :K times H rightarrow G$, $f((k,h))=k+psi(h)$. Now we show that this map is an isomorphism:
$f((k_1,h_1)+(k_2,h_2))=f((k_1+k_2,h_1+h_2))=(k_1+k_2)+psi(h_1+h_2)=(k_1+k_2)+(psi(h_1)+psi(h_2))=k_1+psi(h_1)+k_2+psi(h_2)=f((k_1,h_1))+f((k_2+h_2))$. So this show $f$ is an homomorphism.
Now for bijectivity, Consider $f((k_1,h_1))=f((k_2,h_2))$, this means $k_1+psi(h_1)=k_2+psi(h_2)$. Apply $phi$ to both side of the equation to get $phi(psi(h_1))=phi(psi(h_2))$ which implies $h_1=h_2$ which implies $k_1+h_1=k_2+h_2$ which implies $k_1=k_2$ which implies $(k_1,h_1)=(k_2,h_2)$. So this shows injectivity.
Now for surjectivity, Let $gin G$, $k=gpsi(phi(g))^{-1}$, then $phi(k)=phi(g)phi(g)^{-1}=1$ so $kin K$ and then $g=kpsi(phi(g))$. So $(k, phi(g))$ maps to $g$.
answered Dec 6 '18 at 14:44
nafhgoodnafhgood
1,803422
edited Dec 6 '18 at 15:03
answered Dec 6 '18 at 14:44
nafhgoodnafhgood
1,803422
answered Dec 6 '18 at 14:44
nafhgoodnafhgood
1,803422
answered Dec 6 '18 at 14:44
nafhgoodnafhgood
1,803422
1,803422
add a comment |
add a comment |
$begingroup$
It is a well known result that if $K, L$ are normal in some group $G$ and $KL = G$, $K cap L = 1$, then $G simeq K times L$.
From the fact that $phipsi = 1_H$, we know that $psi$ is injective (Hint: take elements that map to the same image and apply $phi$). Hence $im psi simeq H$ and so $im psi times K simeq H times K$. Therefore we can use the aforementioned result with $L = im psi subset G$ and $K = ker phi subset G$, because $G$ is abelian and so every subgroup is normal. We will have then proved that $G simeq im psi times K simeq H times K$, as desired. In effect,
let $x in K cap L$. Then $x = psi(h)$ for some $h in H$ and also
$$
1 = phi(x) = phipsi(h) = id(h) = h
$$
which proves that $x = psi(h) = psi(1) = 1$ and thus $K cap L = 1$.take $g in G$. Now $g = g(psiphi(g))^{-1}psiphi(g)$. Since
$$
phi(g(psiphi(g))^{-1}) = phi(g)phi(psiphi(g))^{-1}) =\ phi(g)phi(psiphi(g)))^{-1} = phi(g)(phipsiphi)(g)^{-1} = phi(g)phi(g)^{-1} = 1,
$$
as $phipsiphi = 1_Hphi = phi$, then $g in KL$, which concludes the proof.
answered Dec 6 '18 at 15:15
Guido A.Guido A.
7,8631730
$endgroup$
add a comment |
$begingroup$
It is a well known result that if $K, L$ are normal in some group $G$ and $KL = G$, $K cap L = 1$, then $G simeq K times L$.
From the fact that $phipsi = 1_H$, we know that $psi$ is injective (Hint: take elements that map to the same image and apply $phi$). Hence $im psi simeq H$ and so $im psi times K simeq H times K$. Therefore we can use the aforementioned result with $L = im psi subset G$ and $K = ker phi subset G$, because $G$ is abelian and so every subgroup is normal. We will have then proved that $G simeq im psi times K simeq H times K$, as desired. In effect,
let $x in K cap L$. Then $x = psi(h)$ for some $h in H$ and also
$$
1 = phi(x) = phipsi(h) = id(h) = h
$$
which proves that $x = psi(h) = psi(1) = 1$ and thus $K cap L = 1$.take $g in G$. Now $g = g(psiphi(g))^{-1}psiphi(g)$. Since
$$
phi(g(psiphi(g))^{-1}) = phi(g)phi(psiphi(g))^{-1}) =\ phi(g)phi(psiphi(g)))^{-1} = phi(g)(phipsiphi)(g)^{-1} = phi(g)phi(g)^{-1} = 1,
$$
as $phipsiphi = 1_Hphi = phi$, then $g in KL$, which concludes the proof.
answered Dec 6 '18 at 15:15
Guido A.Guido A.
7,8631730
$endgroup$
add a comment |
$begingroup$
It is a well known result that if $K, L$ are normal in some group $G$ and $KL = G$, $K cap L = 1$, then $G simeq K times L$.
From the fact that $phipsi = 1_H$, we know that $psi$ is injective (Hint: take elements that map to the same image and apply $phi$). Hence $im psi simeq H$ and so $im psi times K simeq H times K$. Therefore we can use the aforementioned result with $L = im psi subset G$ and $K = ker phi subset G$, because $G$ is abelian and so every subgroup is normal. We will have then proved that $G simeq im psi times K simeq H times K$, as desired. In effect,
let $x in K cap L$. Then $x = psi(h)$ for some $h in H$ and also
$$
1 = phi(x) = phipsi(h) = id(h) = h
$$
which proves that $x = psi(h) = psi(1) = 1$ and thus $K cap L = 1$.take $g in G$. Now $g = g(psiphi(g))^{-1}psiphi(g)$. Since
$$
phi(g(psiphi(g))^{-1}) = phi(g)phi(psiphi(g))^{-1}) =\ phi(g)phi(psiphi(g)))^{-1} = phi(g)(phipsiphi)(g)^{-1} = phi(g)phi(g)^{-1} = 1,
$$
as $phipsiphi = 1_Hphi = phi$, then $g in KL$, which concludes the proof.
answered Dec 6 '18 at 15:15
Guido A.Guido A.
7,8631730
$endgroup$
It is a well known result that if $K, L$ are normal in some group $G$ and $KL = G$, $K cap L = 1$, then $G simeq K times L$.
From the fact that $phipsi = 1_H$, we know that $psi$ is injective (Hint: take elements that map to the same image and apply $phi$). Hence $im psi simeq H$ and so $im psi times K simeq H times K$. Therefore we can use the aforementioned result with $L = im psi subset G$ and $K = ker phi subset G$, because $G$ is abelian and so every subgroup is normal. We will have then proved that $G simeq im psi times K simeq H times K$, as desired. In effect,
let $x in K cap L$. Then $x = psi(h)$ for some $h in H$ and also
$$
1 = phi(x) = phipsi(h) = id(h) = h
$$
which proves that $x = psi(h) = psi(1) = 1$ and thus $K cap L = 1$.take $g in G$. Now $g = g(psiphi(g))^{-1}psiphi(g)$. Since
$$
phi(g(psiphi(g))^{-1}) = phi(g)phi(psiphi(g))^{-1}) =\ phi(g)phi(psiphi(g)))^{-1} = phi(g)(phipsiphi)(g)^{-1} = phi(g)phi(g)^{-1} = 1,
$$
as $phipsiphi = 1_Hphi = phi$, then $g in KL$, which concludes the proof.
answered Dec 6 '18 at 15:15
Guido A.Guido A.
7,8631730
answered Dec 6 '18 at 15:15
Guido A.Guido A.
7,8631730
answered Dec 6 '18 at 15:15
Guido A.Guido A.
7,8631730
answered Dec 6 '18 at 15:15
Guido A.Guido A.
7,8631730
7,8631730
add a comment |
add a comment |
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1
$begingroup$
Map $(k,h)mapsto kpsi(h)$ for all $kin K, hin H$, and check everything that needs to be checked.
$endgroup$
– Jyrki Lahtonen
Dec 6 '18 at 14:18
$begingroup$
You have a canonical inclusion map $Khookrightarrow G$, and you're given a homomorphism $psi:Hto G$. By definition of $times$, this gives you a homomorphism $Ktimes Hto G$ (a priori this is not necessarily the same homomorphism as Jyrki's above, but it will turn out to be). Check everything that needs to be checked.
$endgroup$
– Arthur
Dec 6 '18 at 14:20
2
$begingroup$
Take $gin G$. Let $h = phi(g)$ and $k = gpsi(h)^{-1}$, which makes $kpsi(h) = g$. You need to show that $kin K$, and here I suspect it is crucial that $K$ is the kernel of $phi$.
$endgroup$
– Arthur
Dec 6 '18 at 14:46
$begingroup$
That's an awesome statement! I'm wondering where can we use this result?
$endgroup$
– nafhgood
Dec 6 '18 at 15:05
1
$begingroup$
@mathnoob This result is called the splitting lemma and since it has a name, it's important.
$endgroup$
– Arthur
Dec 6 '18 at 15:06