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Homomorphisms between a finite group and $(mathbb R,+)$
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I'm trying to prove that any homomorphism from a finite group to $(mathbb R,+)$ and from $(mathbb R,+)$ to a finite group must be trivial.
What I have so far: let $G$ be a finite group and $f: Gto mathbb R$ a homomorphism. Since $G$ is finite, for every element $xin G$ there is a positive $n$ such that $x^n=e$. Then $0=f(e)=f(x^n)=f(x)cdot nimplies f(x)=0$. So $f$ is trivial. Is that correct?
For the other statement, that's what I have (not sure if it's the right direction). Let $f: mathbb Rto G$ be a homomorphism. Let $a=f(1)$. Since $G$ is finite, $a^n=e$ for $nin mathbb N$. Then $e=a^n=[f(1)]^n=f(n)$. This implies $f(nmathbb Z)=e$. is this helpful at all?
abstract-algebra group-theory group-homomorphism
asked Jul 19 at 5:12
user437309
556212
add a comment |Â
up vote
2
down vote
favorite
I'm trying to prove that any homomorphism from a finite group to $(mathbb R,+)$ and from $(mathbb R,+)$ to a finite group must be trivial.
What I have so far: let $G$ be a finite group and $f: Gto mathbb R$ a homomorphism. Since $G$ is finite, for every element $xin G$ there is a positive $n$ such that $x^n=e$. Then $0=f(e)=f(x^n)=f(x)cdot nimplies f(x)=0$. So $f$ is trivial. Is that correct?
For the other statement, that's what I have (not sure if it's the right direction). Let $f: mathbb Rto G$ be a homomorphism. Let $a=f(1)$. Since $G$ is finite, $a^n=e$ for $nin mathbb N$. Then $e=a^n=[f(1)]^n=f(n)$. This implies $f(nmathbb Z)=e$. is this helpful at all?
abstract-algebra group-theory group-homomorphism
asked Jul 19 at 5:12
user437309
556212
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm trying to prove that any homomorphism from a finite group to $(mathbb R,+)$ and from $(mathbb R,+)$ to a finite group must be trivial.
What I have so far: let $G$ be a finite group and $f: Gto mathbb R$ a homomorphism. Since $G$ is finite, for every element $xin G$ there is a positive $n$ such that $x^n=e$. Then $0=f(e)=f(x^n)=f(x)cdot nimplies f(x)=0$. So $f$ is trivial. Is that correct?
For the other statement, that's what I have (not sure if it's the right direction). Let $f: mathbb Rto G$ be a homomorphism. Let $a=f(1)$. Since $G$ is finite, $a^n=e$ for $nin mathbb N$. Then $e=a^n=[f(1)]^n=f(n)$. This implies $f(nmathbb Z)=e$. is this helpful at all?
abstract-algebra group-theory group-homomorphism
asked Jul 19 at 5:12
user437309
556212
I'm trying to prove that any homomorphism from a finite group to $(mathbb R,+)$ and from $(mathbb R,+)$ to a finite group must be trivial.
What I have so far: let $G$ be a finite group and $f: Gto mathbb R$ a homomorphism. Since $G$ is finite, for every element $xin G$ there is a positive $n$ such that $x^n=e$. Then $0=f(e)=f(x^n)=f(x)cdot nimplies f(x)=0$. So $f$ is trivial. Is that correct?
For the other statement, that's what I have (not sure if it's the right direction). Let $f: mathbb Rto G$ be a homomorphism. Let $a=f(1)$. Since $G$ is finite, $a^n=e$ for $nin mathbb N$. Then $e=a^n=[f(1)]^n=f(n)$. This implies $f(nmathbb Z)=e$. is this helpful at all?
abstract-algebra group-theory group-homomorphism
asked Jul 19 at 5:12
user437309
556212
asked Jul 19 at 5:12
user437309
556212
asked Jul 19 at 5:12
user437309
556212
asked Jul 19 at 5:12
user437309
556212
556212
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
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up vote
3
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accepted
Your first proof is correct.
To prove the second statement, notice indeed that by setting $n=|G|$, we find out that for any $xin mathbbR$, $f(nx)=f(x)^n=e$. Then, taking any $yin mathbbR$ and setting $x=fracyn$, we obtain $$f(y)=f(nx)=e$$ so that $f$ is the trivial morphism.
(As Lord Shark the Unknown states it, this trick is due to the fact that $(mathbbR,+)$ is divisible).
answered Jul 19 at 5:22
Suzet
2,216527
add a comment |Â
up vote
2
down vote
The first way round is fine.
For the converse, the group $Bbb R^+$ has a property called divisibility. An additive (resp. multiplicative) group $G$ is divisible if
whenever $ain G$ and $ninBbb N$, then there is $bin G$ with
$nb=a$ (resp. $b^n=a$).
Prove (i) $Bbb R^+$ is divisible, (ii) the image of a divisible group under a homomorphism is divisible.
answered Jul 19 at 5:20
Lord Shark the Unknown
85.5k951112
What does (ii) give? Is the only non-divisible finite group the trivial group? If so, then this would imply the result I believe.
– user437309
Jul 19 at 5:29
@user437309 Let's add (iii) a non-trivial finite group is not divisible.
– Lord Shark the Unknown
Jul 19 at 5:31
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Your first proof is correct.
To prove the second statement, notice indeed that by setting $n=|G|$, we find out that for any $xin mathbbR$, $f(nx)=f(x)^n=e$. Then, taking any $yin mathbbR$ and setting $x=fracyn$, we obtain $$f(y)=f(nx)=e$$ so that $f$ is the trivial morphism.
(As Lord Shark the Unknown states it, this trick is due to the fact that $(mathbbR,+)$ is divisible).
answered Jul 19 at 5:22
Suzet
2,216527
add a comment |Â
up vote
3
down vote
accepted
Your first proof is correct.
To prove the second statement, notice indeed that by setting $n=|G|$, we find out that for any $xin mathbbR$, $f(nx)=f(x)^n=e$. Then, taking any $yin mathbbR$ and setting $x=fracyn$, we obtain $$f(y)=f(nx)=e$$ so that $f$ is the trivial morphism.
(As Lord Shark the Unknown states it, this trick is due to the fact that $(mathbbR,+)$ is divisible).
answered Jul 19 at 5:22
Suzet
2,216527
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Your first proof is correct.
To prove the second statement, notice indeed that by setting $n=|G|$, we find out that for any $xin mathbbR$, $f(nx)=f(x)^n=e$. Then, taking any $yin mathbbR$ and setting $x=fracyn$, we obtain $$f(y)=f(nx)=e$$ so that $f$ is the trivial morphism.
(As Lord Shark the Unknown states it, this trick is due to the fact that $(mathbbR,+)$ is divisible).
answered Jul 19 at 5:22
Suzet
2,216527
Your first proof is correct.
To prove the second statement, notice indeed that by setting $n=|G|$, we find out that for any $xin mathbbR$, $f(nx)=f(x)^n=e$. Then, taking any $yin mathbbR$ and setting $x=fracyn$, we obtain $$f(y)=f(nx)=e$$ so that $f$ is the trivial morphism.
(As Lord Shark the Unknown states it, this trick is due to the fact that $(mathbbR,+)$ is divisible).
answered Jul 19 at 5:22
Suzet
2,216527
answered Jul 19 at 5:22
Suzet
2,216527
answered Jul 19 at 5:22
Suzet
2,216527
answered Jul 19 at 5:22
Suzet
2,216527
2,216527
add a comment |Â
add a comment |Â
up vote
2
down vote
The first way round is fine.
For the converse, the group $Bbb R^+$ has a property called divisibility. An additive (resp. multiplicative) group $G$ is divisible if
whenever $ain G$ and $ninBbb N$, then there is $bin G$ with
$nb=a$ (resp. $b^n=a$).
Prove (i) $Bbb R^+$ is divisible, (ii) the image of a divisible group under a homomorphism is divisible.
answered Jul 19 at 5:20
Lord Shark the Unknown
85.5k951112
What does (ii) give? Is the only non-divisible finite group the trivial group? If so, then this would imply the result I believe.
– user437309
Jul 19 at 5:29
@user437309 Let's add (iii) a non-trivial finite group is not divisible.
– Lord Shark the Unknown
Jul 19 at 5:31
add a comment |Â
up vote
2
down vote
The first way round is fine.
For the converse, the group $Bbb R^+$ has a property called divisibility. An additive (resp. multiplicative) group $G$ is divisible if
whenever $ain G$ and $ninBbb N$, then there is $bin G$ with
$nb=a$ (resp. $b^n=a$).
Prove (i) $Bbb R^+$ is divisible, (ii) the image of a divisible group under a homomorphism is divisible.
answered Jul 19 at 5:20
Lord Shark the Unknown
85.5k951112
What does (ii) give? Is the only non-divisible finite group the trivial group? If so, then this would imply the result I believe.
– user437309
Jul 19 at 5:29
@user437309 Let's add (iii) a non-trivial finite group is not divisible.
– Lord Shark the Unknown
Jul 19 at 5:31
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The first way round is fine.
For the converse, the group $Bbb R^+$ has a property called divisibility. An additive (resp. multiplicative) group $G$ is divisible if
whenever $ain G$ and $ninBbb N$, then there is $bin G$ with
$nb=a$ (resp. $b^n=a$).
Prove (i) $Bbb R^+$ is divisible, (ii) the image of a divisible group under a homomorphism is divisible.
answered Jul 19 at 5:20
Lord Shark the Unknown
85.5k951112
The first way round is fine.
For the converse, the group $Bbb R^+$ has a property called divisibility. An additive (resp. multiplicative) group $G$ is divisible if
whenever $ain G$ and $ninBbb N$, then there is $bin G$ with
$nb=a$ (resp. $b^n=a$).
Prove (i) $Bbb R^+$ is divisible, (ii) the image of a divisible group under a homomorphism is divisible.
answered Jul 19 at 5:20
Lord Shark the Unknown
85.5k951112
answered Jul 19 at 5:20
Lord Shark the Unknown
85.5k951112
answered Jul 19 at 5:20
Lord Shark the Unknown
85.5k951112
answered Jul 19 at 5:20
Lord Shark the Unknown
85.5k951112
85.5k951112
What does (ii) give? Is the only non-divisible finite group the trivial group? If so, then this would imply the result I believe.
– user437309
Jul 19 at 5:29
@user437309 Let's add (iii) a non-trivial finite group is not divisible.
– Lord Shark the Unknown
Jul 19 at 5:31
add a comment |Â
What does (ii) give? Is the only non-divisible finite group the trivial group? If so, then this would imply the result I believe.
– user437309
Jul 19 at 5:29
@user437309 Let's add (iii) a non-trivial finite group is not divisible.
– Lord Shark the Unknown
Jul 19 at 5:31
What does (ii) give? Is the only non-divisible finite group the trivial group? If so, then this would imply the result I believe.
– user437309
Jul 19 at 5:29
What does (ii) give? Is the only non-divisible finite group the trivial group? If so, then this would imply the result I believe.
– user437309
Jul 19 at 5:29
@user437309 Let's add (iii) a non-trivial finite group is not divisible.
– Lord Shark the Unknown
Jul 19 at 5:31
@user437309 Let's add (iii) a non-trivial finite group is not divisible.
– Lord Shark the Unknown
Jul 19 at 5:31
add a comment |Â
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