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How many group homomorphisms we can get from $mathbb Z_20$ to $mathbb Z_10$?
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up vote
4
down vote
favorite
How many group homomorphisms we can get from $mathbb Z_20$ to $ mathbb Z_10$?
My Try: I think $4$. Because $1 , 2 , 5 , 10$ are the only possible order of image of $mathbb Z_20 $.
Can anyone please help me?
group-theory finite-groups cyclic-groups
edited Jul 29 at 8:29
Resident Dementor
57.3k649139
asked Jul 29 at 7:45
cmi
5999
add a comment |Â
up vote
4
down vote
favorite
How many group homomorphisms we can get from $mathbb Z_20$ to $ mathbb Z_10$?
My Try: I think $4$. Because $1 , 2 , 5 , 10$ are the only possible order of image of $mathbb Z_20 $.
Can anyone please help me?
group-theory finite-groups cyclic-groups
edited Jul 29 at 8:29
Resident Dementor
57.3k649139
asked Jul 29 at 7:45
cmi
5999
According to theorem it is gcd(10,20)=10 so 10 group homomorphisam.
– Ninja hatori
Jul 29 at 7:55
Hint: there is a 1-1 correspondence between normal subgroups of $mathbbZ/20 mathbbZ$ and kernels of homomorphisms from $mathbbZ/20 mathbbZ$. Think how you could use this fact coupled with the first isomorphism theorem.
– Oiler
Jul 29 at 8:23
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
How many group homomorphisms we can get from $mathbb Z_20$ to $ mathbb Z_10$?
My Try: I think $4$. Because $1 , 2 , 5 , 10$ are the only possible order of image of $mathbb Z_20 $.
Can anyone please help me?
group-theory finite-groups cyclic-groups
edited Jul 29 at 8:29
Resident Dementor
57.3k649139
asked Jul 29 at 7:45
cmi
5999
How many group homomorphisms we can get from $mathbb Z_20$ to $ mathbb Z_10$?
My Try: I think $4$. Because $1 , 2 , 5 , 10$ are the only possible order of image of $mathbb Z_20 $.
Can anyone please help me?
group-theory finite-groups cyclic-groups
edited Jul 29 at 8:29
Resident Dementor
57.3k649139
asked Jul 29 at 7:45
cmi
5999
edited Jul 29 at 8:29
Resident Dementor
57.3k649139
edited Jul 29 at 8:29
Resident Dementor
57.3k649139
edited Jul 29 at 8:29
Resident Dementor
57.3k649139
57.3k649139
asked Jul 29 at 7:45
cmi
5999
asked Jul 29 at 7:45
cmi
5999
asked Jul 29 at 7:45
cmi
5999
5999
According to theorem it is gcd(10,20)=10 so 10 group homomorphisam.
– Ninja hatori
Jul 29 at 7:55
Hint: there is a 1-1 correspondence between normal subgroups of $mathbbZ/20 mathbbZ$ and kernels of homomorphisms from $mathbbZ/20 mathbbZ$. Think how you could use this fact coupled with the first isomorphism theorem.
– Oiler
Jul 29 at 8:23
add a comment |Â
According to theorem it is gcd(10,20)=10 so 10 group homomorphisam.
– Ninja hatori
Jul 29 at 7:55
Hint: there is a 1-1 correspondence between normal subgroups of $mathbbZ/20 mathbbZ$ and kernels of homomorphisms from $mathbbZ/20 mathbbZ$. Think how you could use this fact coupled with the first isomorphism theorem.
– Oiler
Jul 29 at 8:23
According to theorem it is gcd(10,20)=10 so 10 group homomorphisam.
– Ninja hatori
Jul 29 at 7:55
According to theorem it is gcd(10,20)=10 so 10 group homomorphisam.
– Ninja hatori
Jul 29 at 7:55
Hint: there is a 1-1 correspondence between normal subgroups of $mathbbZ/20 mathbbZ$ and kernels of homomorphisms from $mathbbZ/20 mathbbZ$. Think how you could use this fact coupled with the first isomorphism theorem.
– Oiler
Jul 29 at 8:23
Hint: there is a 1-1 correspondence between normal subgroups of $mathbbZ/20 mathbbZ$ and kernels of homomorphisms from $mathbbZ/20 mathbbZ$. Think how you could use this fact coupled with the first isomorphism theorem.
– Oiler
Jul 29 at 8:23
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Any homomorphism is completely determined by the image of $1$. That is, if $1 mapsto a$, then $x mapsto xa$.
So assume $1 mapsto a$. Then by Lagrange's theorem $vert a vert$ divides $10$. Also by this result$Bigl( textIf; vert g vert$ is finite, then $vert phi(g) vert$ divides $vert g vert Bigr)$,we have $vert a vert$ divides $20$ ,since $vert 1 vert =20$ is finite.
Thus $vert a vert$ divides both $20$ and $10$. So $vert a vert=1,2,5,10$
Thus $a=0,5,2,8,4,6,1,3,7,9$. ( Actually the range of this "$a$" is all the elements of $BbbZ_10 $, but I am writing in the order mentioned above). This give a list of $10$
homomorphisms.
That is
$$x mapsto ax$$ where $a=0,1,2,...,9$ are all the required ten homomorphisms.
In General, the number of group homomorphisms from $BbbZ_m$ to $BbbZ_n$ is $textgcd(m,n)$
answered Jul 29 at 9:29
Chinnapparaj R
1,489315
The proof of the general result is available in the pdf format.That pdf also explains how many ring homomorphisms are there from $BbbZ_m$ to $BbbZ_n$ .I don't know, how to attach pdf files. So if you want this paper, kindly inform me and I send it via email!
– Chinnapparaj R
Jul 29 at 9:34
@cmi: Are you a CMI (Chennai Mathematical Institute) Student?
– Chinnapparaj R
Jul 29 at 9:42
add a comment |Â
up vote
3
down vote
Let's consider the more general problem of determining the homomorphisms from $Bbb Z/mBbb Z$ to $BbbZ/nBbb Z$. Consider the short exact sequence:
$$0longrightarrow Bbb Zxrightarrow~times~ menspaceBbb Zxrightarrowqquad Bbb Z/mBbb Zlongrightarrow 0, $$
from which we deduce the exact sequence of $DeclareMathOperatorHomHomHom$s:
$$0longrightarrowHom(Bbb Z/mBbb Z,Bbb Z/nBbb Z)xrightarrowqquadHom(Bbb Z,Bbb Z/nBbb Z)xrightarrow~times~ menspaceHom(Bbb Z,Bbb Z/nBbb Z).$$
Thus $Hom(Bbb Z/mBbb Z,Bbb Z/nBbb Z)$ is the kernel of the map
beginalign
Hom(Bbb Z,Bbb Z/nBbb Z)simeqBbb Z/nBbb Z&xrightarrowqquadBbb Z/nBbb Z, newline
f(1)=bar x& longmapsto f(m)=mbar x,
endalign
i.e. it corresponds to the set $;bar xinBbb Z/nBbb Zmid mxin nBbb Z$.
Nenote $d=gcd(m,n)$, $:m'=dfrac md$, $:n'=dfrac nd$. As $m'$ and $n'$ are coprime, :
$$mxin nBbb Ziff m'xin n' Bbb Z iff xin n' Bbb Z ,$$
so that $; Hom(Bbb Z/mBbb Z,Bbb Z/nBbb Z)$ corresponds to the ideal in $Bbb Z/nBbb Z$ generated by the congruence class of $;n'=dfrac nd$. This ideal is isomorphic to $;Bbb Z/dBbb Z$.
answered Jul 29 at 8:53
Bernard
110k635102
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Any homomorphism is completely determined by the image of $1$. That is, if $1 mapsto a$, then $x mapsto xa$.
So assume $1 mapsto a$. Then by Lagrange's theorem $vert a vert$ divides $10$. Also by this result$Bigl( textIf; vert g vert$ is finite, then $vert phi(g) vert$ divides $vert g vert Bigr)$,we have $vert a vert$ divides $20$ ,since $vert 1 vert =20$ is finite.
Thus $vert a vert$ divides both $20$ and $10$. So $vert a vert=1,2,5,10$
Thus $a=0,5,2,8,4,6,1,3,7,9$. ( Actually the range of this "$a$" is all the elements of $BbbZ_10 $, but I am writing in the order mentioned above). This give a list of $10$
homomorphisms.
That is
$$x mapsto ax$$ where $a=0,1,2,...,9$ are all the required ten homomorphisms.
In General, the number of group homomorphisms from $BbbZ_m$ to $BbbZ_n$ is $textgcd(m,n)$
answered Jul 29 at 9:29
Chinnapparaj R
1,489315
The proof of the general result is available in the pdf format.That pdf also explains how many ring homomorphisms are there from $BbbZ_m$ to $BbbZ_n$ .I don't know, how to attach pdf files. So if you want this paper, kindly inform me and I send it via email!
– Chinnapparaj R
Jul 29 at 9:34
@cmi: Are you a CMI (Chennai Mathematical Institute) Student?
– Chinnapparaj R
Jul 29 at 9:42
add a comment |Â
up vote
4
down vote
accepted
Any homomorphism is completely determined by the image of $1$. That is, if $1 mapsto a$, then $x mapsto xa$.
So assume $1 mapsto a$. Then by Lagrange's theorem $vert a vert$ divides $10$. Also by this result$Bigl( textIf; vert g vert$ is finite, then $vert phi(g) vert$ divides $vert g vert Bigr)$,we have $vert a vert$ divides $20$ ,since $vert 1 vert =20$ is finite.
Thus $vert a vert$ divides both $20$ and $10$. So $vert a vert=1,2,5,10$
Thus $a=0,5,2,8,4,6,1,3,7,9$. ( Actually the range of this "$a$" is all the elements of $BbbZ_10 $, but I am writing in the order mentioned above). This give a list of $10$
homomorphisms.
That is
$$x mapsto ax$$ where $a=0,1,2,...,9$ are all the required ten homomorphisms.
In General, the number of group homomorphisms from $BbbZ_m$ to $BbbZ_n$ is $textgcd(m,n)$
answered Jul 29 at 9:29
Chinnapparaj R
1,489315
The proof of the general result is available in the pdf format.That pdf also explains how many ring homomorphisms are there from $BbbZ_m$ to $BbbZ_n$ .I don't know, how to attach pdf files. So if you want this paper, kindly inform me and I send it via email!
– Chinnapparaj R
Jul 29 at 9:34
@cmi: Are you a CMI (Chennai Mathematical Institute) Student?
– Chinnapparaj R
Jul 29 at 9:42
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Any homomorphism is completely determined by the image of $1$. That is, if $1 mapsto a$, then $x mapsto xa$.
So assume $1 mapsto a$. Then by Lagrange's theorem $vert a vert$ divides $10$. Also by this result$Bigl( textIf; vert g vert$ is finite, then $vert phi(g) vert$ divides $vert g vert Bigr)$,we have $vert a vert$ divides $20$ ,since $vert 1 vert =20$ is finite.
Thus $vert a vert$ divides both $20$ and $10$. So $vert a vert=1,2,5,10$
Thus $a=0,5,2,8,4,6,1,3,7,9$. ( Actually the range of this "$a$" is all the elements of $BbbZ_10 $, but I am writing in the order mentioned above). This give a list of $10$
homomorphisms.
That is
$$x mapsto ax$$ where $a=0,1,2,...,9$ are all the required ten homomorphisms.
In General, the number of group homomorphisms from $BbbZ_m$ to $BbbZ_n$ is $textgcd(m,n)$
answered Jul 29 at 9:29
Chinnapparaj R
1,489315
Any homomorphism is completely determined by the image of $1$. That is, if $1 mapsto a$, then $x mapsto xa$.
So assume $1 mapsto a$. Then by Lagrange's theorem $vert a vert$ divides $10$. Also by this result$Bigl( textIf; vert g vert$ is finite, then $vert phi(g) vert$ divides $vert g vert Bigr)$,we have $vert a vert$ divides $20$ ,since $vert 1 vert =20$ is finite.
Thus $vert a vert$ divides both $20$ and $10$. So $vert a vert=1,2,5,10$
Thus $a=0,5,2,8,4,6,1,3,7,9$. ( Actually the range of this "$a$" is all the elements of $BbbZ_10 $, but I am writing in the order mentioned above). This give a list of $10$
homomorphisms.
That is
$$x mapsto ax$$ where $a=0,1,2,...,9$ are all the required ten homomorphisms.
In General, the number of group homomorphisms from $BbbZ_m$ to $BbbZ_n$ is $textgcd(m,n)$
answered Jul 29 at 9:29
Chinnapparaj R
1,489315
edited Jul 29 at 9:36
answered Jul 29 at 9:29
Chinnapparaj R
1,489315
answered Jul 29 at 9:29
Chinnapparaj R
1,489315
answered Jul 29 at 9:29
Chinnapparaj R
1,489315
1,489315
The proof of the general result is available in the pdf format.That pdf also explains how many ring homomorphisms are there from $BbbZ_m$ to $BbbZ_n$ .I don't know, how to attach pdf files. So if you want this paper, kindly inform me and I send it via email!
– Chinnapparaj R
Jul 29 at 9:34
@cmi: Are you a CMI (Chennai Mathematical Institute) Student?
– Chinnapparaj R
Jul 29 at 9:42
add a comment |Â
The proof of the general result is available in the pdf format.That pdf also explains how many ring homomorphisms are there from $BbbZ_m$ to $BbbZ_n$ .I don't know, how to attach pdf files. So if you want this paper, kindly inform me and I send it via email!
– Chinnapparaj R
Jul 29 at 9:34
@cmi: Are you a CMI (Chennai Mathematical Institute) Student?
– Chinnapparaj R
Jul 29 at 9:42
The proof of the general result is available in the pdf format.That pdf also explains how many ring homomorphisms are there from $BbbZ_m$ to $BbbZ_n$ .I don't know, how to attach pdf files. So if you want this paper, kindly inform me and I send it via email!
– Chinnapparaj R
Jul 29 at 9:34
The proof of the general result is available in the pdf format.That pdf also explains how many ring homomorphisms are there from $BbbZ_m$ to $BbbZ_n$ .I don't know, how to attach pdf files. So if you want this paper, kindly inform me and I send it via email!
– Chinnapparaj R
Jul 29 at 9:34
@cmi: Are you a CMI (Chennai Mathematical Institute) Student?
– Chinnapparaj R
Jul 29 at 9:42
@cmi: Are you a CMI (Chennai Mathematical Institute) Student?
– Chinnapparaj R
Jul 29 at 9:42
add a comment |Â
up vote
3
down vote
Let's consider the more general problem of determining the homomorphisms from $Bbb Z/mBbb Z$ to $BbbZ/nBbb Z$. Consider the short exact sequence:
$$0longrightarrow Bbb Zxrightarrow~times~ menspaceBbb Zxrightarrowqquad Bbb Z/mBbb Zlongrightarrow 0, $$
from which we deduce the exact sequence of $DeclareMathOperatorHomHomHom$s:
$$0longrightarrowHom(Bbb Z/mBbb Z,Bbb Z/nBbb Z)xrightarrowqquadHom(Bbb Z,Bbb Z/nBbb Z)xrightarrow~times~ menspaceHom(Bbb Z,Bbb Z/nBbb Z).$$
Thus $Hom(Bbb Z/mBbb Z,Bbb Z/nBbb Z)$ is the kernel of the map
beginalign
Hom(Bbb Z,Bbb Z/nBbb Z)simeqBbb Z/nBbb Z&xrightarrowqquadBbb Z/nBbb Z, newline
f(1)=bar x& longmapsto f(m)=mbar x,
endalign
i.e. it corresponds to the set $;bar xinBbb Z/nBbb Zmid mxin nBbb Z$.
Nenote $d=gcd(m,n)$, $:m'=dfrac md$, $:n'=dfrac nd$. As $m'$ and $n'$ are coprime, :
$$mxin nBbb Ziff m'xin n' Bbb Z iff xin n' Bbb Z ,$$
so that $; Hom(Bbb Z/mBbb Z,Bbb Z/nBbb Z)$ corresponds to the ideal in $Bbb Z/nBbb Z$ generated by the congruence class of $;n'=dfrac nd$. This ideal is isomorphic to $;Bbb Z/dBbb Z$.
answered Jul 29 at 8:53
Bernard
110k635102
add a comment |Â
up vote
3
down vote
Let's consider the more general problem of determining the homomorphisms from $Bbb Z/mBbb Z$ to $BbbZ/nBbb Z$. Consider the short exact sequence:
$$0longrightarrow Bbb Zxrightarrow~times~ menspaceBbb Zxrightarrowqquad Bbb Z/mBbb Zlongrightarrow 0, $$
from which we deduce the exact sequence of $DeclareMathOperatorHomHomHom$s:
$$0longrightarrowHom(Bbb Z/mBbb Z,Bbb Z/nBbb Z)xrightarrowqquadHom(Bbb Z,Bbb Z/nBbb Z)xrightarrow~times~ menspaceHom(Bbb Z,Bbb Z/nBbb Z).$$
Thus $Hom(Bbb Z/mBbb Z,Bbb Z/nBbb Z)$ is the kernel of the map
beginalign
Hom(Bbb Z,Bbb Z/nBbb Z)simeqBbb Z/nBbb Z&xrightarrowqquadBbb Z/nBbb Z, newline
f(1)=bar x& longmapsto f(m)=mbar x,
endalign
i.e. it corresponds to the set $;bar xinBbb Z/nBbb Zmid mxin nBbb Z$.
Nenote $d=gcd(m,n)$, $:m'=dfrac md$, $:n'=dfrac nd$. As $m'$ and $n'$ are coprime, :
$$mxin nBbb Ziff m'xin n' Bbb Z iff xin n' Bbb Z ,$$
so that $; Hom(Bbb Z/mBbb Z,Bbb Z/nBbb Z)$ corresponds to the ideal in $Bbb Z/nBbb Z$ generated by the congruence class of $;n'=dfrac nd$. This ideal is isomorphic to $;Bbb Z/dBbb Z$.
answered Jul 29 at 8:53
Bernard
110k635102
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let's consider the more general problem of determining the homomorphisms from $Bbb Z/mBbb Z$ to $BbbZ/nBbb Z$. Consider the short exact sequence:
$$0longrightarrow Bbb Zxrightarrow~times~ menspaceBbb Zxrightarrowqquad Bbb Z/mBbb Zlongrightarrow 0, $$
from which we deduce the exact sequence of $DeclareMathOperatorHomHomHom$s:
$$0longrightarrowHom(Bbb Z/mBbb Z,Bbb Z/nBbb Z)xrightarrowqquadHom(Bbb Z,Bbb Z/nBbb Z)xrightarrow~times~ menspaceHom(Bbb Z,Bbb Z/nBbb Z).$$
Thus $Hom(Bbb Z/mBbb Z,Bbb Z/nBbb Z)$ is the kernel of the map
beginalign
Hom(Bbb Z,Bbb Z/nBbb Z)simeqBbb Z/nBbb Z&xrightarrowqquadBbb Z/nBbb Z, newline
f(1)=bar x& longmapsto f(m)=mbar x,
endalign
i.e. it corresponds to the set $;bar xinBbb Z/nBbb Zmid mxin nBbb Z$.
Nenote $d=gcd(m,n)$, $:m'=dfrac md$, $:n'=dfrac nd$. As $m'$ and $n'$ are coprime, :
$$mxin nBbb Ziff m'xin n' Bbb Z iff xin n' Bbb Z ,$$
so that $; Hom(Bbb Z/mBbb Z,Bbb Z/nBbb Z)$ corresponds to the ideal in $Bbb Z/nBbb Z$ generated by the congruence class of $;n'=dfrac nd$. This ideal is isomorphic to $;Bbb Z/dBbb Z$.
answered Jul 29 at 8:53
Bernard
110k635102
Let's consider the more general problem of determining the homomorphisms from $Bbb Z/mBbb Z$ to $BbbZ/nBbb Z$. Consider the short exact sequence:
$$0longrightarrow Bbb Zxrightarrow~times~ menspaceBbb Zxrightarrowqquad Bbb Z/mBbb Zlongrightarrow 0, $$
from which we deduce the exact sequence of $DeclareMathOperatorHomHomHom$s:
$$0longrightarrowHom(Bbb Z/mBbb Z,Bbb Z/nBbb Z)xrightarrowqquadHom(Bbb Z,Bbb Z/nBbb Z)xrightarrow~times~ menspaceHom(Bbb Z,Bbb Z/nBbb Z).$$
Thus $Hom(Bbb Z/mBbb Z,Bbb Z/nBbb Z)$ is the kernel of the map
beginalign
Hom(Bbb Z,Bbb Z/nBbb Z)simeqBbb Z/nBbb Z&xrightarrowqquadBbb Z/nBbb Z, newline
f(1)=bar x& longmapsto f(m)=mbar x,
endalign
i.e. it corresponds to the set $;bar xinBbb Z/nBbb Zmid mxin nBbb Z$.
Nenote $d=gcd(m,n)$, $:m'=dfrac md$, $:n'=dfrac nd$. As $m'$ and $n'$ are coprime, :
$$mxin nBbb Ziff m'xin n' Bbb Z iff xin n' Bbb Z ,$$
so that $; Hom(Bbb Z/mBbb Z,Bbb Z/nBbb Z)$ corresponds to the ideal in $Bbb Z/nBbb Z$ generated by the congruence class of $;n'=dfrac nd$. This ideal is isomorphic to $;Bbb Z/dBbb Z$.
answered Jul 29 at 8:53
Bernard
110k635102
answered Jul 29 at 8:53
Bernard
110k635102
answered Jul 29 at 8:53
Bernard
110k635102
answered Jul 29 at 8:53
Bernard
110k635102
110k635102
add a comment |Â
add a comment |Â
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According to theorem it is gcd(10,20)=10 so 10 group homomorphisam.
– Ninja hatori
Jul 29 at 7:55
Hint: there is a 1-1 correspondence between normal subgroups of $mathbbZ/20 mathbbZ$ and kernels of homomorphisms from $mathbbZ/20 mathbbZ$. Think how you could use this fact coupled with the first isomorphism theorem.
– Oiler
Jul 29 at 8:23