Suppose I have a group $G cong mathbb{Z}$ and a homomorphism $f : G to G$, is $f$ always multiplication by...
$begingroup$
Suppose I have a group $G cong mathbb{Z}$ and a homomorphism $f : G to G$, is $f$ always multiplication by $m$ for some $m in mathbb{Z}$? Stated more precisely, is $f$ defined by $f(x) = mx$ for all $x in G$?
The reason why $f$ would be multiplication by $m$, is probably because all homomorphisms from $mathbb{Z}$ to $mathbb{Z}$ are multiplication by $m$ for some $m in mathbb{Z}$.
Let me make this question a bit more rigorous. Choose a isomorpshism $psi : G to mathbb{Z}$. Now I'm assuming (but I don't know how to show), that $f$ induces a homomorphism $gamma : mathbb{Z} to mathbb{Z}$ and produces the following commutative diagram
$$
newcommand{ra}[1]{!!!!!!!!!!!!xrightarrow{quad#1quad}!!!!!!!!}
newcommand{da}[1]{leftdownarrow{scriptstyle#1}vphantom{displaystyleint_0^1}right.}
%
begin{array}{llllllllllll}
G & ra{f} & G \
da{psi} & & da{psi^{-1}} \
mathbb{Z} & ra{gamma} & mathbb{Z} \
end{array}
$$
Then if the above is true, then since $gamma$ is a homomorphism from $mathbb{Z}$ to $mathbb{Z}$ then we have $gamma(x) = mx$ for all $x in mathbb{Z}$ so we have
begin{align*}
f(x) &= psi^{-1}(gamma(psi(x))) \
&= psi^{-1}(mcdot psi(x)) \
&= m left( psi^{-1}(psi(x)right) \
&= mx
end{align*}
for all $x in G$ and so $f$ is multiplication by $m$ in this case. But in the above construction (which I don't even know if it's true), I don't know if $f$ induces a unique homomorphism $gamma$ making the diagram commute, and if $m$ is independent of the choice of isomorphism from $G$ to $mathbb{Z}$.
It could also be the case that I'm viewing this wrong and a homomorphism from $mathbb{Z}$ to $mathbb{Z}$ induces a unique homomorphism $f : G to G$ (which I guess seems a bit more natural in this context), and it's just that case that all homomorphisms from $G$ to $G$ are induced by homomorphisms from $mathbb{Z}$ to $mathbb{Z}$.
group-theory
asked Dec 8 '18 at 15:19
PerturbativePerturbative
4,26811551
$endgroup$
add a comment |
$begingroup$
Suppose I have a group $G cong mathbb{Z}$ and a homomorphism $f : G to G$, is $f$ always multiplication by $m$ for some $m in mathbb{Z}$? Stated more precisely, is $f$ defined by $f(x) = mx$ for all $x in G$?
The reason why $f$ would be multiplication by $m$, is probably because all homomorphisms from $mathbb{Z}$ to $mathbb{Z}$ are multiplication by $m$ for some $m in mathbb{Z}$.
Let me make this question a bit more rigorous. Choose a isomorpshism $psi : G to mathbb{Z}$. Now I'm assuming (but I don't know how to show), that $f$ induces a homomorphism $gamma : mathbb{Z} to mathbb{Z}$ and produces the following commutative diagram
$$
newcommand{ra}[1]{!!!!!!!!!!!!xrightarrow{quad#1quad}!!!!!!!!}
newcommand{da}[1]{leftdownarrow{scriptstyle#1}vphantom{displaystyleint_0^1}right.}
%
begin{array}{llllllllllll}
G & ra{f} & G \
da{psi} & & da{psi^{-1}} \
mathbb{Z} & ra{gamma} & mathbb{Z} \
end{array}
$$
Then if the above is true, then since $gamma$ is a homomorphism from $mathbb{Z}$ to $mathbb{Z}$ then we have $gamma(x) = mx$ for all $x in mathbb{Z}$ so we have
begin{align*}
f(x) &= psi^{-1}(gamma(psi(x))) \
&= psi^{-1}(mcdot psi(x)) \
&= m left( psi^{-1}(psi(x)right) \
&= mx
end{align*}
for all $x in G$ and so $f$ is multiplication by $m$ in this case. But in the above construction (which I don't even know if it's true), I don't know if $f$ induces a unique homomorphism $gamma$ making the diagram commute, and if $m$ is independent of the choice of isomorphism from $G$ to $mathbb{Z}$.
It could also be the case that I'm viewing this wrong and a homomorphism from $mathbb{Z}$ to $mathbb{Z}$ induces a unique homomorphism $f : G to G$ (which I guess seems a bit more natural in this context), and it's just that case that all homomorphisms from $G$ to $G$ are induced by homomorphisms from $mathbb{Z}$ to $mathbb{Z}$.
group-theory
asked Dec 8 '18 at 15:19
PerturbativePerturbative
4,26811551
$endgroup$
$begingroup$
You need to consider the image of the unit element.
$endgroup$
– Wuestenfux
Dec 8 '18 at 15:22
$begingroup$
Of course $gamma$ is unique. $gamma = psi^{-1} circ f circ psi$ there is only one such a function.
$endgroup$
– Yanko
Dec 8 '18 at 15:47
add a comment |
$begingroup$
Suppose I have a group $G cong mathbb{Z}$ and a homomorphism $f : G to G$, is $f$ always multiplication by $m$ for some $m in mathbb{Z}$? Stated more precisely, is $f$ defined by $f(x) = mx$ for all $x in G$?
The reason why $f$ would be multiplication by $m$, is probably because all homomorphisms from $mathbb{Z}$ to $mathbb{Z}$ are multiplication by $m$ for some $m in mathbb{Z}$.
Let me make this question a bit more rigorous. Choose a isomorpshism $psi : G to mathbb{Z}$. Now I'm assuming (but I don't know how to show), that $f$ induces a homomorphism $gamma : mathbb{Z} to mathbb{Z}$ and produces the following commutative diagram
$$
newcommand{ra}[1]{!!!!!!!!!!!!xrightarrow{quad#1quad}!!!!!!!!}
newcommand{da}[1]{leftdownarrow{scriptstyle#1}vphantom{displaystyleint_0^1}right.}
%
begin{array}{llllllllllll}
G & ra{f} & G \
da{psi} & & da{psi^{-1}} \
mathbb{Z} & ra{gamma} & mathbb{Z} \
end{array}
$$
Then if the above is true, then since $gamma$ is a homomorphism from $mathbb{Z}$ to $mathbb{Z}$ then we have $gamma(x) = mx$ for all $x in mathbb{Z}$ so we have
begin{align*}
f(x) &= psi^{-1}(gamma(psi(x))) \
&= psi^{-1}(mcdot psi(x)) \
&= m left( psi^{-1}(psi(x)right) \
&= mx
end{align*}
for all $x in G$ and so $f$ is multiplication by $m$ in this case. But in the above construction (which I don't even know if it's true), I don't know if $f$ induces a unique homomorphism $gamma$ making the diagram commute, and if $m$ is independent of the choice of isomorphism from $G$ to $mathbb{Z}$.
It could also be the case that I'm viewing this wrong and a homomorphism from $mathbb{Z}$ to $mathbb{Z}$ induces a unique homomorphism $f : G to G$ (which I guess seems a bit more natural in this context), and it's just that case that all homomorphisms from $G$ to $G$ are induced by homomorphisms from $mathbb{Z}$ to $mathbb{Z}$.
group-theory
asked Dec 8 '18 at 15:19
PerturbativePerturbative
4,26811551
$endgroup$
Suppose I have a group $G cong mathbb{Z}$ and a homomorphism $f : G to G$, is $f$ always multiplication by $m$ for some $m in mathbb{Z}$? Stated more precisely, is $f$ defined by $f(x) = mx$ for all $x in G$?
The reason why $f$ would be multiplication by $m$, is probably because all homomorphisms from $mathbb{Z}$ to $mathbb{Z}$ are multiplication by $m$ for some $m in mathbb{Z}$.
Let me make this question a bit more rigorous. Choose a isomorpshism $psi : G to mathbb{Z}$. Now I'm assuming (but I don't know how to show), that $f$ induces a homomorphism $gamma : mathbb{Z} to mathbb{Z}$ and produces the following commutative diagram
$$
newcommand{ra}[1]{!!!!!!!!!!!!xrightarrow{quad#1quad}!!!!!!!!}
newcommand{da}[1]{leftdownarrow{scriptstyle#1}vphantom{displaystyleint_0^1}right.}
%
begin{array}{llllllllllll}
G & ra{f} & G \
da{psi} & & da{psi^{-1}} \
mathbb{Z} & ra{gamma} & mathbb{Z} \
end{array}
$$
Then if the above is true, then since $gamma$ is a homomorphism from $mathbb{Z}$ to $mathbb{Z}$ then we have $gamma(x) = mx$ for all $x in mathbb{Z}$ so we have
begin{align*}
f(x) &= psi^{-1}(gamma(psi(x))) \
&= psi^{-1}(mcdot psi(x)) \
&= m left( psi^{-1}(psi(x)right) \
&= mx
end{align*}
for all $x in G$ and so $f$ is multiplication by $m$ in this case. But in the above construction (which I don't even know if it's true), I don't know if $f$ induces a unique homomorphism $gamma$ making the diagram commute, and if $m$ is independent of the choice of isomorphism from $G$ to $mathbb{Z}$.
It could also be the case that I'm viewing this wrong and a homomorphism from $mathbb{Z}$ to $mathbb{Z}$ induces a unique homomorphism $f : G to G$ (which I guess seems a bit more natural in this context), and it's just that case that all homomorphisms from $G$ to $G$ are induced by homomorphisms from $mathbb{Z}$ to $mathbb{Z}$.
group-theory
group-theory
asked Dec 8 '18 at 15:19
PerturbativePerturbative
4,26811551
asked Dec 8 '18 at 15:19
PerturbativePerturbative
4,26811551
asked Dec 8 '18 at 15:19
PerturbativePerturbative
4,26811551
asked Dec 8 '18 at 15:19
PerturbativePerturbative
4,26811551
asked Dec 8 '18 at 15:19
PerturbativePerturbative
4,26811551
4,26811551
$begingroup$
You need to consider the image of the unit element.
$endgroup$
– Wuestenfux
Dec 8 '18 at 15:22
$begingroup$
Of course $gamma$ is unique. $gamma = psi^{-1} circ f circ psi$ there is only one such a function.
$endgroup$
– Yanko
Dec 8 '18 at 15:47
add a comment |
$begingroup$
You need to consider the image of the unit element.
$endgroup$
– Wuestenfux
Dec 8 '18 at 15:22
$begingroup$
Of course $gamma$ is unique. $gamma = psi^{-1} circ f circ psi$ there is only one such a function.
$endgroup$
– Yanko
Dec 8 '18 at 15:47
$begingroup$
You need to consider the image of the unit element.
$endgroup$
– Wuestenfux
Dec 8 '18 at 15:22
$begingroup$
You need to consider the image of the unit element.
$endgroup$
– Wuestenfux
Dec 8 '18 at 15:22
$begingroup$
Of course $gamma$ is unique. $gamma = psi^{-1} circ f circ psi$ there is only one such a function.
$endgroup$
– Yanko
Dec 8 '18 at 15:47
$begingroup$
Of course $gamma$ is unique. $gamma = psi^{-1} circ f circ psi$ there is only one such a function.
$endgroup$
– Yanko
Dec 8 '18 at 15:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let me suggest more elementary approach without commutative diagrams (First look at my comment it might be clear from there).
Let $varphi:mathbb{Z}rightarrow G$ be the isomorphism.
Let $f:Grightarrow G$ be an homomorphism. Look at $varphi circ f circ varphi^{-1}:mathbb{Z}rightarrow mathbb{Z}$ is a homomorphism. Hence multiplying by some $minmathbb{Z}$ in other words
$varphi circ f circ varphi^{-1}(n) = mn$ for all $ninmathbb{Z}$.
Now for all $gin G$, let $n:=varphi(g)$ then from the above we conclude that
$varphi (f(g)) = m varphi(g)$
Let $hin H$ be such that $varphi(h)=m$ we conclude that $varphi(f(g))=varphi(hg)$ for all $gin G$. Since $varphi$ is injective we have $f(g)=hg$. This completes the proof.
answered Dec 8 '18 at 15:45
YankoYanko
6,5571529
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Let me suggest more elementary approach without commutative diagrams (First look at my comment it might be clear from there).
Let $varphi:mathbb{Z}rightarrow G$ be the isomorphism.
Let $f:Grightarrow G$ be an homomorphism. Look at $varphi circ f circ varphi^{-1}:mathbb{Z}rightarrow mathbb{Z}$ is a homomorphism. Hence multiplying by some $minmathbb{Z}$ in other words
$varphi circ f circ varphi^{-1}(n) = mn$ for all $ninmathbb{Z}$.
Now for all $gin G$, let $n:=varphi(g)$ then from the above we conclude that
$varphi (f(g)) = m varphi(g)$
Let $hin H$ be such that $varphi(h)=m$ we conclude that $varphi(f(g))=varphi(hg)$ for all $gin G$. Since $varphi$ is injective we have $f(g)=hg$. This completes the proof.
answered Dec 8 '18 at 15:45
YankoYanko
6,5571529
$endgroup$
add a comment |
$begingroup$
Let me suggest more elementary approach without commutative diagrams (First look at my comment it might be clear from there).
Let $varphi:mathbb{Z}rightarrow G$ be the isomorphism.
Let $f:Grightarrow G$ be an homomorphism. Look at $varphi circ f circ varphi^{-1}:mathbb{Z}rightarrow mathbb{Z}$ is a homomorphism. Hence multiplying by some $minmathbb{Z}$ in other words
$varphi circ f circ varphi^{-1}(n) = mn$ for all $ninmathbb{Z}$.
Now for all $gin G$, let $n:=varphi(g)$ then from the above we conclude that
$varphi (f(g)) = m varphi(g)$
Let $hin H$ be such that $varphi(h)=m$ we conclude that $varphi(f(g))=varphi(hg)$ for all $gin G$. Since $varphi$ is injective we have $f(g)=hg$. This completes the proof.
answered Dec 8 '18 at 15:45
YankoYanko
6,5571529
$endgroup$
add a comment |
$begingroup$
Let me suggest more elementary approach without commutative diagrams (First look at my comment it might be clear from there).
Let $varphi:mathbb{Z}rightarrow G$ be the isomorphism.
Let $f:Grightarrow G$ be an homomorphism. Look at $varphi circ f circ varphi^{-1}:mathbb{Z}rightarrow mathbb{Z}$ is a homomorphism. Hence multiplying by some $minmathbb{Z}$ in other words
$varphi circ f circ varphi^{-1}(n) = mn$ for all $ninmathbb{Z}$.
Now for all $gin G$, let $n:=varphi(g)$ then from the above we conclude that
$varphi (f(g)) = m varphi(g)$
Let $hin H$ be such that $varphi(h)=m$ we conclude that $varphi(f(g))=varphi(hg)$ for all $gin G$. Since $varphi$ is injective we have $f(g)=hg$. This completes the proof.
answered Dec 8 '18 at 15:45
YankoYanko
6,5571529
$endgroup$
Let me suggest more elementary approach without commutative diagrams (First look at my comment it might be clear from there).
Let $varphi:mathbb{Z}rightarrow G$ be the isomorphism.
Let $f:Grightarrow G$ be an homomorphism. Look at $varphi circ f circ varphi^{-1}:mathbb{Z}rightarrow mathbb{Z}$ is a homomorphism. Hence multiplying by some $minmathbb{Z}$ in other words
$varphi circ f circ varphi^{-1}(n) = mn$ for all $ninmathbb{Z}$.
Now for all $gin G$, let $n:=varphi(g)$ then from the above we conclude that
$varphi (f(g)) = m varphi(g)$
Let $hin H$ be such that $varphi(h)=m$ we conclude that $varphi(f(g))=varphi(hg)$ for all $gin G$. Since $varphi$ is injective we have $f(g)=hg$. This completes the proof.
answered Dec 8 '18 at 15:45
YankoYanko
6,5571529
answered Dec 8 '18 at 15:45
YankoYanko
6,5571529
answered Dec 8 '18 at 15:45
YankoYanko
6,5571529
answered Dec 8 '18 at 15:45
YankoYanko
6,5571529
6,5571529
add a comment |
add a comment |
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$begingroup$
You need to consider the image of the unit element.
$endgroup$
– Wuestenfux
Dec 8 '18 at 15:22
$begingroup$
Of course $gamma$ is unique. $gamma = psi^{-1} circ f circ psi$ there is only one such a function.
$endgroup$
– Yanko
Dec 8 '18 at 15:47