Endomorphism of $(mathbbZ, +)$ maps $z$ to multiple of $z$?
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Why if I map some $zin mathbbZ$ I must get multiple of $kz$ where $k,zin mathbbZ$? Couldn't be that I map $4$ to $2$?
abstract-algebra
edited Sep 3 at 9:15
Arnaud D.
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asked Sep 3 at 8:33
bilanush
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Why if I map some $zin mathbbZ$ I must get multiple of $kz$ where $k,zin mathbbZ$? Couldn't be that I map $4$ to $2$?
abstract-algebra
edited Sep 3 at 9:15
Arnaud D.
14.9k52142
asked Sep 3 at 8:33
bilanush
1347
If you map $4$ to $2$, then you must map $1$ to some integer which satisfies $4x = 2$, so that then $phi(4) = 4 times phi(1) = 2$, but no such integer exists.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 3 at 8:37
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up vote
2
down vote
favorite
up vote
2
down vote
favorite
Why if I map some $zin mathbbZ$ I must get multiple of $kz$ where $k,zin mathbbZ$? Couldn't be that I map $4$ to $2$?
abstract-algebra
edited Sep 3 at 9:15
Arnaud D.
14.9k52142
asked Sep 3 at 8:33
bilanush
1347
Why if I map some $zin mathbbZ$ I must get multiple of $kz$ where $k,zin mathbbZ$? Couldn't be that I map $4$ to $2$?
abstract-algebra
abstract-algebra
edited Sep 3 at 9:15
Arnaud D.
14.9k52142
asked Sep 3 at 8:33
bilanush
1347
edited Sep 3 at 9:15
Arnaud D.
14.9k52142
asked Sep 3 at 8:33
bilanush
1347
edited Sep 3 at 9:15
Arnaud D.
14.9k52142
edited Sep 3 at 9:15
Arnaud D.
14.9k52142
edited Sep 3 at 9:15
Arnaud D.
14.9k52142
14.9k52142
asked Sep 3 at 8:33
bilanush
1347
asked Sep 3 at 8:33
bilanush
1347
asked Sep 3 at 8:33
bilanush
1347
1347
If you map $4$ to $2$, then you must map $1$ to some integer which satisfies $4x = 2$, so that then $phi(4) = 4 times phi(1) = 2$, but no such integer exists.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 3 at 8:37
add a comment |Â
If you map $4$ to $2$, then you must map $1$ to some integer which satisfies $4x = 2$, so that then $phi(4) = 4 times phi(1) = 2$, but no such integer exists.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 3 at 8:37
If you map $4$ to $2$, then you must map $1$ to some integer which satisfies $4x = 2$, so that then $phi(4) = 4 times phi(1) = 2$, but no such integer exists.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 3 at 8:37
If you map $4$ to $2$, then you must map $1$ to some integer which satisfies $4x = 2$, so that then $phi(4) = 4 times phi(1) = 2$, but no such integer exists.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 3 at 8:37
add a comment |Â
2 Answers
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That is because any integer $z$ can be written as $overbrace1+ldots+1^z, texttimes$, so that by the additive property of homomorphisms, we must have $$f(z)=zf(1)$$
for every homomorphism $f:(mathbb Z,+)rightarrow (mathbb Z,+)$ and every $zin mathbb Z$. This integer $f(1)$ corresponds to the $k$ you are referring to.
answered Sep 3 at 8:37
Suzet
2,426527
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We can directly exclude the case that $f(4)=2$, because
$$
f(4)=f(1+1+1+1)=f(1)+f(1)+f(1)+f(1)=4f(1)
$$
so $f(4)=2$ would imply $2f(1)=1$.
More generally, the group $mathbbZ$ (with respect to addition) is cyclic, generated by $1$. Hence a homomorphism $fcolonmathbbZtomathbbZ$ is completely determined by $f(1)$.
Let $f(1)=k$. Then, by standard properties of homomorphisms, for every $zinmathbbZ$,
$$
f(z)=f(z1)=zf(1)=zk=kz
$$
Conversely, for every $kinmathbbZ$, $zmapsto kz$ is a homomorphism $mathbbZtomathbbZ$.
answered Sep 3 at 9:35
egreg
167k1180189
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
That is because any integer $z$ can be written as $overbrace1+ldots+1^z, texttimes$, so that by the additive property of homomorphisms, we must have $$f(z)=zf(1)$$
for every homomorphism $f:(mathbb Z,+)rightarrow (mathbb Z,+)$ and every $zin mathbb Z$. This integer $f(1)$ corresponds to the $k$ you are referring to.
answered Sep 3 at 8:37
Suzet
2,426527
add a comment |Â
up vote
4
down vote
That is because any integer $z$ can be written as $overbrace1+ldots+1^z, texttimes$, so that by the additive property of homomorphisms, we must have $$f(z)=zf(1)$$
for every homomorphism $f:(mathbb Z,+)rightarrow (mathbb Z,+)$ and every $zin mathbb Z$. This integer $f(1)$ corresponds to the $k$ you are referring to.
answered Sep 3 at 8:37
Suzet
2,426527
add a comment |Â
up vote
4
down vote
up vote
4
down vote
That is because any integer $z$ can be written as $overbrace1+ldots+1^z, texttimes$, so that by the additive property of homomorphisms, we must have $$f(z)=zf(1)$$
for every homomorphism $f:(mathbb Z,+)rightarrow (mathbb Z,+)$ and every $zin mathbb Z$. This integer $f(1)$ corresponds to the $k$ you are referring to.
answered Sep 3 at 8:37
Suzet
2,426527
That is because any integer $z$ can be written as $overbrace1+ldots+1^z, texttimes$, so that by the additive property of homomorphisms, we must have $$f(z)=zf(1)$$
for every homomorphism $f:(mathbb Z,+)rightarrow (mathbb Z,+)$ and every $zin mathbb Z$. This integer $f(1)$ corresponds to the $k$ you are referring to.
answered Sep 3 at 8:37
Suzet
2,426527
answered Sep 3 at 8:37
Suzet
2,426527
answered Sep 3 at 8:37
Suzet
2,426527
answered Sep 3 at 8:37
Suzet
2,426527
2,426527
add a comment |Â
add a comment |Â
up vote
2
down vote
We can directly exclude the case that $f(4)=2$, because
$$
f(4)=f(1+1+1+1)=f(1)+f(1)+f(1)+f(1)=4f(1)
$$
so $f(4)=2$ would imply $2f(1)=1$.
More generally, the group $mathbbZ$ (with respect to addition) is cyclic, generated by $1$. Hence a homomorphism $fcolonmathbbZtomathbbZ$ is completely determined by $f(1)$.
Let $f(1)=k$. Then, by standard properties of homomorphisms, for every $zinmathbbZ$,
$$
f(z)=f(z1)=zf(1)=zk=kz
$$
Conversely, for every $kinmathbbZ$, $zmapsto kz$ is a homomorphism $mathbbZtomathbbZ$.
answered Sep 3 at 9:35
egreg
167k1180189
add a comment |Â
up vote
2
down vote
We can directly exclude the case that $f(4)=2$, because
$$
f(4)=f(1+1+1+1)=f(1)+f(1)+f(1)+f(1)=4f(1)
$$
so $f(4)=2$ would imply $2f(1)=1$.
More generally, the group $mathbbZ$ (with respect to addition) is cyclic, generated by $1$. Hence a homomorphism $fcolonmathbbZtomathbbZ$ is completely determined by $f(1)$.
Let $f(1)=k$. Then, by standard properties of homomorphisms, for every $zinmathbbZ$,
$$
f(z)=f(z1)=zf(1)=zk=kz
$$
Conversely, for every $kinmathbbZ$, $zmapsto kz$ is a homomorphism $mathbbZtomathbbZ$.
answered Sep 3 at 9:35
egreg
167k1180189
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We can directly exclude the case that $f(4)=2$, because
$$
f(4)=f(1+1+1+1)=f(1)+f(1)+f(1)+f(1)=4f(1)
$$
so $f(4)=2$ would imply $2f(1)=1$.
More generally, the group $mathbbZ$ (with respect to addition) is cyclic, generated by $1$. Hence a homomorphism $fcolonmathbbZtomathbbZ$ is completely determined by $f(1)$.
Let $f(1)=k$. Then, by standard properties of homomorphisms, for every $zinmathbbZ$,
$$
f(z)=f(z1)=zf(1)=zk=kz
$$
Conversely, for every $kinmathbbZ$, $zmapsto kz$ is a homomorphism $mathbbZtomathbbZ$.
answered Sep 3 at 9:35
egreg
167k1180189
We can directly exclude the case that $f(4)=2$, because
$$
f(4)=f(1+1+1+1)=f(1)+f(1)+f(1)+f(1)=4f(1)
$$
so $f(4)=2$ would imply $2f(1)=1$.
More generally, the group $mathbbZ$ (with respect to addition) is cyclic, generated by $1$. Hence a homomorphism $fcolonmathbbZtomathbbZ$ is completely determined by $f(1)$.
Let $f(1)=k$. Then, by standard properties of homomorphisms, for every $zinmathbbZ$,
$$
f(z)=f(z1)=zf(1)=zk=kz
$$
Conversely, for every $kinmathbbZ$, $zmapsto kz$ is a homomorphism $mathbbZtomathbbZ$.
answered Sep 3 at 9:35
egreg
167k1180189
answered Sep 3 at 9:35
egreg
167k1180189
answered Sep 3 at 9:35
egreg
167k1180189
answered Sep 3 at 9:35
egreg
167k1180189
167k1180189
add a comment |Â
add a comment |Â
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If you map $4$ to $2$, then you must map $1$ to some integer which satisfies $4x = 2$, so that then $phi(4) = 4 times phi(1) = 2$, but no such integer exists.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 3 at 8:37