Proof using ONLY Group Theoretic Techniques
$begingroup$
Prove that the total number of distinct group homomorphism between $mathbb{Z}_m$ and $mathbb{Z}_n$ is $gcd(m,n)$.
Proof:
Let $xinmathbb{Z}_m$, and let $f:mathbb{Z}_mtomathbb{Z}_n$ be a homomorphism. Then $f(x)=xcdot f(1)$. Thus we just have to determine the possible values of $f(1)$.
We show that $f(x)=acdot x$ for some fixed $ainmathbb{Z}_n$ is a homomorphism iff $maequiv0$ (mod $n$). For that, assume that $f(x)=acdot x$ is a homomorphism for some fixed $ainmathbb{Z}_n$. Then $0=f(0)=f(m)=acdot m$. Since our codomain is $mathbb{Z}_n$, $maequiv0$ (mod $n$). Conversely assume that $maequiv0$ (mod $n$). Let $x,yinmathbb{Z}$. Since $mathbb{Z}$ is an Euclidean ring, there are integers $q,r$ such that $x+y=mq+r$ where $0leq r<m$. Now using our assumption, we get $f(x+y)=f(mq+r)=f(r)=acdot r=acdot (x+y-mq)=acdot x+acdot y-acdot mq=acdot x+acdot y=f(x)+f(y)$. Hence $f:mathbb{Z}_mtomathbb{Z}_n$ defined by $f(x)=acdot x$ for some fixed $ainmathbb{Z}_n$ is $indeed$ homomorphism provided $maequiv0$ (mod $n$).
Now by the virtue of techniques developed in Number Theory, We know that the congruence $maequiv0$ (mod $n$) has $gcd(m,n)$ distinct solutions.
Here proof is completely understood by me but I want to prove the theorem by using $only$ Group Theoretic techniques. Can someone suggest me an another proof ?
alternative-proof group-homomorphism
edited Dec 23 '18 at 3:53
Shaun
9,246113684
asked Jul 26 '17 at 3:29
UDAY PATELUDAY PATEL
236213
$endgroup$
add a comment |
$begingroup$
Prove that the total number of distinct group homomorphism between $mathbb{Z}_m$ and $mathbb{Z}_n$ is $gcd(m,n)$.
Proof:
Let $xinmathbb{Z}_m$, and let $f:mathbb{Z}_mtomathbb{Z}_n$ be a homomorphism. Then $f(x)=xcdot f(1)$. Thus we just have to determine the possible values of $f(1)$.
We show that $f(x)=acdot x$ for some fixed $ainmathbb{Z}_n$ is a homomorphism iff $maequiv0$ (mod $n$). For that, assume that $f(x)=acdot x$ is a homomorphism for some fixed $ainmathbb{Z}_n$. Then $0=f(0)=f(m)=acdot m$. Since our codomain is $mathbb{Z}_n$, $maequiv0$ (mod $n$). Conversely assume that $maequiv0$ (mod $n$). Let $x,yinmathbb{Z}$. Since $mathbb{Z}$ is an Euclidean ring, there are integers $q,r$ such that $x+y=mq+r$ where $0leq r<m$. Now using our assumption, we get $f(x+y)=f(mq+r)=f(r)=acdot r=acdot (x+y-mq)=acdot x+acdot y-acdot mq=acdot x+acdot y=f(x)+f(y)$. Hence $f:mathbb{Z}_mtomathbb{Z}_n$ defined by $f(x)=acdot x$ for some fixed $ainmathbb{Z}_n$ is $indeed$ homomorphism provided $maequiv0$ (mod $n$).
Now by the virtue of techniques developed in Number Theory, We know that the congruence $maequiv0$ (mod $n$) has $gcd(m,n)$ distinct solutions.
Here proof is completely understood by me but I want to prove the theorem by using $only$ Group Theoretic techniques. Can someone suggest me an another proof ?
alternative-proof group-homomorphism
edited Dec 23 '18 at 3:53
Shaun
9,246113684
asked Jul 26 '17 at 3:29
UDAY PATELUDAY PATEL
236213
$endgroup$
2
$begingroup$
If $G$ and $H$ are groups with normal subgroups $M$ and $N$, respectively, then the homomorphisms from $G/M$ to $H/N$ are in natural bijection with those homorphisms $f$ from $G$ to $H$ such that $f(M)$ is contained in $N$. You can apply that with $G=H=Bbb Z$, $M=mBbb Z$, and $N=nBbb Z$.
$endgroup$
– Greg Martin
Jul 26 '17 at 4:19
add a comment |
$begingroup$
Prove that the total number of distinct group homomorphism between $mathbb{Z}_m$ and $mathbb{Z}_n$ is $gcd(m,n)$.
Proof:
Let $xinmathbb{Z}_m$, and let $f:mathbb{Z}_mtomathbb{Z}_n$ be a homomorphism. Then $f(x)=xcdot f(1)$. Thus we just have to determine the possible values of $f(1)$.
We show that $f(x)=acdot x$ for some fixed $ainmathbb{Z}_n$ is a homomorphism iff $maequiv0$ (mod $n$). For that, assume that $f(x)=acdot x$ is a homomorphism for some fixed $ainmathbb{Z}_n$. Then $0=f(0)=f(m)=acdot m$. Since our codomain is $mathbb{Z}_n$, $maequiv0$ (mod $n$). Conversely assume that $maequiv0$ (mod $n$). Let $x,yinmathbb{Z}$. Since $mathbb{Z}$ is an Euclidean ring, there are integers $q,r$ such that $x+y=mq+r$ where $0leq r<m$. Now using our assumption, we get $f(x+y)=f(mq+r)=f(r)=acdot r=acdot (x+y-mq)=acdot x+acdot y-acdot mq=acdot x+acdot y=f(x)+f(y)$. Hence $f:mathbb{Z}_mtomathbb{Z}_n$ defined by $f(x)=acdot x$ for some fixed $ainmathbb{Z}_n$ is $indeed$ homomorphism provided $maequiv0$ (mod $n$).
Now by the virtue of techniques developed in Number Theory, We know that the congruence $maequiv0$ (mod $n$) has $gcd(m,n)$ distinct solutions.
Here proof is completely understood by me but I want to prove the theorem by using $only$ Group Theoretic techniques. Can someone suggest me an another proof ?
alternative-proof group-homomorphism
edited Dec 23 '18 at 3:53
Shaun
9,246113684
asked Jul 26 '17 at 3:29
UDAY PATELUDAY PATEL
236213
$endgroup$
Prove that the total number of distinct group homomorphism between $mathbb{Z}_m$ and $mathbb{Z}_n$ is $gcd(m,n)$.
Proof:
Let $xinmathbb{Z}_m$, and let $f:mathbb{Z}_mtomathbb{Z}_n$ be a homomorphism. Then $f(x)=xcdot f(1)$. Thus we just have to determine the possible values of $f(1)$.
We show that $f(x)=acdot x$ for some fixed $ainmathbb{Z}_n$ is a homomorphism iff $maequiv0$ (mod $n$). For that, assume that $f(x)=acdot x$ is a homomorphism for some fixed $ainmathbb{Z}_n$. Then $0=f(0)=f(m)=acdot m$. Since our codomain is $mathbb{Z}_n$, $maequiv0$ (mod $n$). Conversely assume that $maequiv0$ (mod $n$). Let $x,yinmathbb{Z}$. Since $mathbb{Z}$ is an Euclidean ring, there are integers $q,r$ such that $x+y=mq+r$ where $0leq r<m$. Now using our assumption, we get $f(x+y)=f(mq+r)=f(r)=acdot r=acdot (x+y-mq)=acdot x+acdot y-acdot mq=acdot x+acdot y=f(x)+f(y)$. Hence $f:mathbb{Z}_mtomathbb{Z}_n$ defined by $f(x)=acdot x$ for some fixed $ainmathbb{Z}_n$ is $indeed$ homomorphism provided $maequiv0$ (mod $n$).
Now by the virtue of techniques developed in Number Theory, We know that the congruence $maequiv0$ (mod $n$) has $gcd(m,n)$ distinct solutions.
Here proof is completely understood by me but I want to prove the theorem by using $only$ Group Theoretic techniques. Can someone suggest me an another proof ?
alternative-proof group-homomorphism
alternative-proof group-homomorphism
edited Dec 23 '18 at 3:53
Shaun
9,246113684
asked Jul 26 '17 at 3:29
UDAY PATELUDAY PATEL
236213
edited Dec 23 '18 at 3:53
Shaun
9,246113684
asked Jul 26 '17 at 3:29
UDAY PATELUDAY PATEL
236213
edited Dec 23 '18 at 3:53
Shaun
9,246113684
edited Dec 23 '18 at 3:53
Shaun
9,246113684
edited Dec 23 '18 at 3:53
Shaun
9,246113684
9,246113684
asked Jul 26 '17 at 3:29
UDAY PATELUDAY PATEL
236213
asked Jul 26 '17 at 3:29
UDAY PATELUDAY PATEL
236213
asked Jul 26 '17 at 3:29
UDAY PATELUDAY PATEL
236213
236213
2
$begingroup$
If $G$ and $H$ are groups with normal subgroups $M$ and $N$, respectively, then the homomorphisms from $G/M$ to $H/N$ are in natural bijection with those homorphisms $f$ from $G$ to $H$ such that $f(M)$ is contained in $N$. You can apply that with $G=H=Bbb Z$, $M=mBbb Z$, and $N=nBbb Z$.
$endgroup$
– Greg Martin
Jul 26 '17 at 4:19
add a comment |
2
$begingroup$
If $G$ and $H$ are groups with normal subgroups $M$ and $N$, respectively, then the homomorphisms from $G/M$ to $H/N$ are in natural bijection with those homorphisms $f$ from $G$ to $H$ such that $f(M)$ is contained in $N$. You can apply that with $G=H=Bbb Z$, $M=mBbb Z$, and $N=nBbb Z$.
$endgroup$
– Greg Martin
Jul 26 '17 at 4:19
2
2
$begingroup$
If $G$ and $H$ are groups with normal subgroups $M$ and $N$, respectively, then the homomorphisms from $G/M$ to $H/N$ are in natural bijection with those homorphisms $f$ from $G$ to $H$ such that $f(M)$ is contained in $N$. You can apply that with $G=H=Bbb Z$, $M=mBbb Z$, and $N=nBbb Z$.
$endgroup$
– Greg Martin
Jul 26 '17 at 4:19
$begingroup$
If $G$ and $H$ are groups with normal subgroups $M$ and $N$, respectively, then the homomorphisms from $G/M$ to $H/N$ are in natural bijection with those homorphisms $f$ from $G$ to $H$ such that $f(M)$ is contained in $N$. You can apply that with $G=H=Bbb Z$, $M=mBbb Z$, and $N=nBbb Z$.
$endgroup$
– Greg Martin
Jul 26 '17 at 4:19
add a comment |
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$begingroup$
If $G$ and $H$ are groups with normal subgroups $M$ and $N$, respectively, then the homomorphisms from $G/M$ to $H/N$ are in natural bijection with those homorphisms $f$ from $G$ to $H$ such that $f(M)$ is contained in $N$. You can apply that with $G=H=Bbb Z$, $M=mBbb Z$, and $N=nBbb Z$.
$endgroup$
– Greg Martin
Jul 26 '17 at 4:19