(For any group $H$) Bijection between $H$ and the group consists of all the homomorphisms between $(mathbb Z,...
$begingroup$
My question is, for any group $H$, how to prove there exists a bijective function between $Z$ and the group consists of all the homomorphisms which I will call IS later on for convenience
I just started learning abstract algebra and soon I encountered this problem.
Find a group $G$ such that, for any group $H$, there is always a bijection between the set of group homomorphisms $phi : G rightarrow H$ and the set of elements of $H$. Note: $G$ is independent of $H$.
I have known the $G$ is $Z$ and what I need to do is to construct a function and prove it bijective.
Define $i_h(n) = h^n$ and obviously $i_h$ is a homomorphism.
So I can construct a function named $b$ which is from $H$ to $IS$. Thence $mathrm { b } : mathrm { H } rightarrow mathrm { IS }$
$$
b(h) = i_h
$$
And finally the last two tasks.
a) injective
$forall h _ { 1 } , h _ { 2 } in H text { if } b left( h _ { 1 } right) = b left( h _ { 2 } right) text { we will have } b left( h _ { 1 } right) ( 1 ) = b left( h _ { 2 } right) ( 1 ) text { which will result in the fact } h _ { 1 } = h _ { 2 }$
b) surjective
But I don't know how to prove the surjective part.
abstract-algebra group-theory functions group-homomorphism
edited Dec 28 '18 at 9:56
Shaun
10.6k113687
asked Dec 28 '18 at 9:04
ShuumatsuShuumatsu
637
$endgroup$
add a comment |
$begingroup$
My question is, for any group $H$, how to prove there exists a bijective function between $Z$ and the group consists of all the homomorphisms which I will call IS later on for convenience
I just started learning abstract algebra and soon I encountered this problem.
Find a group $G$ such that, for any group $H$, there is always a bijection between the set of group homomorphisms $phi : G rightarrow H$ and the set of elements of $H$. Note: $G$ is independent of $H$.
I have known the $G$ is $Z$ and what I need to do is to construct a function and prove it bijective.
Define $i_h(n) = h^n$ and obviously $i_h$ is a homomorphism.
So I can construct a function named $b$ which is from $H$ to $IS$. Thence $mathrm { b } : mathrm { H } rightarrow mathrm { IS }$
$$
b(h) = i_h
$$
And finally the last two tasks.
a) injective
$forall h _ { 1 } , h _ { 2 } in H text { if } b left( h _ { 1 } right) = b left( h _ { 2 } right) text { we will have } b left( h _ { 1 } right) ( 1 ) = b left( h _ { 2 } right) ( 1 ) text { which will result in the fact } h _ { 1 } = h _ { 2 }$
b) surjective
But I don't know how to prove the surjective part.
abstract-algebra group-theory functions group-homomorphism
edited Dec 28 '18 at 9:56
Shaun
10.6k113687
asked Dec 28 '18 at 9:04
ShuumatsuShuumatsu
637
$endgroup$
$begingroup$
Define a function $c: IS to H$, by $c(f) = f(1)$. Can you show that $b$ and $c$ are mutually inverse?
$endgroup$
– Joppy
Dec 28 '18 at 9:15
add a comment |
$begingroup$
My question is, for any group $H$, how to prove there exists a bijective function between $Z$ and the group consists of all the homomorphisms which I will call IS later on for convenience
I just started learning abstract algebra and soon I encountered this problem.
Find a group $G$ such that, for any group $H$, there is always a bijection between the set of group homomorphisms $phi : G rightarrow H$ and the set of elements of $H$. Note: $G$ is independent of $H$.
I have known the $G$ is $Z$ and what I need to do is to construct a function and prove it bijective.
Define $i_h(n) = h^n$ and obviously $i_h$ is a homomorphism.
So I can construct a function named $b$ which is from $H$ to $IS$. Thence $mathrm { b } : mathrm { H } rightarrow mathrm { IS }$
$$
b(h) = i_h
$$
And finally the last two tasks.
a) injective
$forall h _ { 1 } , h _ { 2 } in H text { if } b left( h _ { 1 } right) = b left( h _ { 2 } right) text { we will have } b left( h _ { 1 } right) ( 1 ) = b left( h _ { 2 } right) ( 1 ) text { which will result in the fact } h _ { 1 } = h _ { 2 }$
b) surjective
But I don't know how to prove the surjective part.
abstract-algebra group-theory functions group-homomorphism
edited Dec 28 '18 at 9:56
Shaun
10.6k113687
asked Dec 28 '18 at 9:04
ShuumatsuShuumatsu
637
$endgroup$
My question is, for any group $H$, how to prove there exists a bijective function between $Z$ and the group consists of all the homomorphisms which I will call IS later on for convenience
I just started learning abstract algebra and soon I encountered this problem.
Find a group $G$ such that, for any group $H$, there is always a bijection between the set of group homomorphisms $phi : G rightarrow H$ and the set of elements of $H$. Note: $G$ is independent of $H$.
I have known the $G$ is $Z$ and what I need to do is to construct a function and prove it bijective.
Define $i_h(n) = h^n$ and obviously $i_h$ is a homomorphism.
So I can construct a function named $b$ which is from $H$ to $IS$. Thence $mathrm { b } : mathrm { H } rightarrow mathrm { IS }$
$$
b(h) = i_h
$$
And finally the last two tasks.
a) injective
$forall h _ { 1 } , h _ { 2 } in H text { if } b left( h _ { 1 } right) = b left( h _ { 2 } right) text { we will have } b left( h _ { 1 } right) ( 1 ) = b left( h _ { 2 } right) ( 1 ) text { which will result in the fact } h _ { 1 } = h _ { 2 }$
b) surjective
But I don't know how to prove the surjective part.
abstract-algebra group-theory functions group-homomorphism
abstract-algebra group-theory functions group-homomorphism
edited Dec 28 '18 at 9:56
Shaun
10.6k113687
asked Dec 28 '18 at 9:04
ShuumatsuShuumatsu
637
edited Dec 28 '18 at 9:56
Shaun
10.6k113687
asked Dec 28 '18 at 9:04
ShuumatsuShuumatsu
637
edited Dec 28 '18 at 9:56
Shaun
10.6k113687
edited Dec 28 '18 at 9:56
Shaun
10.6k113687
edited Dec 28 '18 at 9:56
Shaun
10.6k113687
10.6k113687
asked Dec 28 '18 at 9:04
ShuumatsuShuumatsu
637
asked Dec 28 '18 at 9:04
ShuumatsuShuumatsu
637
asked Dec 28 '18 at 9:04
ShuumatsuShuumatsu
637
637
$begingroup$
Define a function $c: IS to H$, by $c(f) = f(1)$. Can you show that $b$ and $c$ are mutually inverse?
$endgroup$
– Joppy
Dec 28 '18 at 9:15
add a comment |
$begingroup$
Define a function $c: IS to H$, by $c(f) = f(1)$. Can you show that $b$ and $c$ are mutually inverse?
$endgroup$
– Joppy
Dec 28 '18 at 9:15
$begingroup$
Define a function $c: IS to H$, by $c(f) = f(1)$. Can you show that $b$ and $c$ are mutually inverse?
$endgroup$
– Joppy
Dec 28 '18 at 9:15
$begingroup$
Define a function $c: IS to H$, by $c(f) = f(1)$. Can you show that $b$ and $c$ are mutually inverse?
$endgroup$
– Joppy
Dec 28 '18 at 9:15
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
For any $phiin IS$, we have $h:=phi(1)in H$. We will claim that $$b(h) = i_h=phi$$ as follows:
$$i_h(n)=h^n=(phi(1))^n=phi(n*1)=phi(n)$$
for all $nin mathbb{Z}$. Thus $i_h=phi$. This means that $phi=b(h)$ is in the image of $b$. So $b$ is surjective.
answered Dec 28 '18 at 9:16
yoyoyoyo
756410
$endgroup$
add a comment |
$begingroup$
You have to prove that $IS={i_{h}, hin H}$, $i_{h}$ defined above. $mathbb{Z}$ is the free group generated from the element 1, so an homomorphism $mathbb{Z}to H$ is defined by the imagine of 1, and in your set of homomorphisms there are all the possible imagines for the element 1.
answered Dec 28 '18 at 9:14
ecrinecrin
3477
$endgroup$
add a comment |
$begingroup$
For surjectivity, let $f in text{IS}$. Now take $h=f(1)$, then $b(h)=i_{h}$. Hence surjective.
answered Dec 28 '18 at 9:17
Anurag AAnurag A
26.4k12351
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For any $phiin IS$, we have $h:=phi(1)in H$. We will claim that $$b(h) = i_h=phi$$ as follows:
$$i_h(n)=h^n=(phi(1))^n=phi(n*1)=phi(n)$$
for all $nin mathbb{Z}$. Thus $i_h=phi$. This means that $phi=b(h)$ is in the image of $b$. So $b$ is surjective.
answered Dec 28 '18 at 9:16
yoyoyoyo
756410
$endgroup$
add a comment |
$begingroup$
For any $phiin IS$, we have $h:=phi(1)in H$. We will claim that $$b(h) = i_h=phi$$ as follows:
$$i_h(n)=h^n=(phi(1))^n=phi(n*1)=phi(n)$$
for all $nin mathbb{Z}$. Thus $i_h=phi$. This means that $phi=b(h)$ is in the image of $b$. So $b$ is surjective.
answered Dec 28 '18 at 9:16
yoyoyoyo
756410
$endgroup$
add a comment |
$begingroup$
For any $phiin IS$, we have $h:=phi(1)in H$. We will claim that $$b(h) = i_h=phi$$ as follows:
$$i_h(n)=h^n=(phi(1))^n=phi(n*1)=phi(n)$$
for all $nin mathbb{Z}$. Thus $i_h=phi$. This means that $phi=b(h)$ is in the image of $b$. So $b$ is surjective.
answered Dec 28 '18 at 9:16
yoyoyoyo
756410
$endgroup$
For any $phiin IS$, we have $h:=phi(1)in H$. We will claim that $$b(h) = i_h=phi$$ as follows:
$$i_h(n)=h^n=(phi(1))^n=phi(n*1)=phi(n)$$
for all $nin mathbb{Z}$. Thus $i_h=phi$. This means that $phi=b(h)$ is in the image of $b$. So $b$ is surjective.
answered Dec 28 '18 at 9:16
yoyoyoyo
756410
answered Dec 28 '18 at 9:16
yoyoyoyo
756410
answered Dec 28 '18 at 9:16
yoyoyoyo
756410
answered Dec 28 '18 at 9:16
yoyoyoyo
756410
756410
add a comment |
add a comment |
$begingroup$
You have to prove that $IS={i_{h}, hin H}$, $i_{h}$ defined above. $mathbb{Z}$ is the free group generated from the element 1, so an homomorphism $mathbb{Z}to H$ is defined by the imagine of 1, and in your set of homomorphisms there are all the possible imagines for the element 1.
answered Dec 28 '18 at 9:14
ecrinecrin
3477
$endgroup$
add a comment |
$begingroup$
You have to prove that $IS={i_{h}, hin H}$, $i_{h}$ defined above. $mathbb{Z}$ is the free group generated from the element 1, so an homomorphism $mathbb{Z}to H$ is defined by the imagine of 1, and in your set of homomorphisms there are all the possible imagines for the element 1.
answered Dec 28 '18 at 9:14
ecrinecrin
3477
$endgroup$
add a comment |
$begingroup$
You have to prove that $IS={i_{h}, hin H}$, $i_{h}$ defined above. $mathbb{Z}$ is the free group generated from the element 1, so an homomorphism $mathbb{Z}to H$ is defined by the imagine of 1, and in your set of homomorphisms there are all the possible imagines for the element 1.
answered Dec 28 '18 at 9:14
ecrinecrin
3477
$endgroup$
You have to prove that $IS={i_{h}, hin H}$, $i_{h}$ defined above. $mathbb{Z}$ is the free group generated from the element 1, so an homomorphism $mathbb{Z}to H$ is defined by the imagine of 1, and in your set of homomorphisms there are all the possible imagines for the element 1.
answered Dec 28 '18 at 9:14
ecrinecrin
3477
answered Dec 28 '18 at 9:14
ecrinecrin
3477
answered Dec 28 '18 at 9:14
ecrinecrin
3477
answered Dec 28 '18 at 9:14
ecrinecrin
3477
3477
add a comment |
add a comment |
$begingroup$
For surjectivity, let $f in text{IS}$. Now take $h=f(1)$, then $b(h)=i_{h}$. Hence surjective.
answered Dec 28 '18 at 9:17
Anurag AAnurag A
26.4k12351
$endgroup$
add a comment |
$begingroup$
For surjectivity, let $f in text{IS}$. Now take $h=f(1)$, then $b(h)=i_{h}$. Hence surjective.
answered Dec 28 '18 at 9:17
Anurag AAnurag A
26.4k12351
$endgroup$
add a comment |
$begingroup$
For surjectivity, let $f in text{IS}$. Now take $h=f(1)$, then $b(h)=i_{h}$. Hence surjective.
answered Dec 28 '18 at 9:17
Anurag AAnurag A
26.4k12351
$endgroup$
For surjectivity, let $f in text{IS}$. Now take $h=f(1)$, then $b(h)=i_{h}$. Hence surjective.
answered Dec 28 '18 at 9:17
Anurag AAnurag A
26.4k12351
answered Dec 28 '18 at 9:17
Anurag AAnurag A
26.4k12351
answered Dec 28 '18 at 9:17
Anurag AAnurag A
26.4k12351
answered Dec 28 '18 at 9:17
Anurag AAnurag A
26.4k12351
26.4k12351
add a comment |
add a comment |
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$begingroup$
Define a function $c: IS to H$, by $c(f) = f(1)$. Can you show that $b$ and $c$ are mutually inverse?
$endgroup$
– Joppy
Dec 28 '18 at 9:15