If $G$ is a finite group of odd order, show there is only one homomorphism between $G$ and...
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If $G$ is a finite group of odd order, show there is only one homomorphism between $G$ and $mathbb{R}^{times}$ which is the trivial one.
I know there is a similar case if the order of G and H is coprime. then the order of image of $G$ is 1. But in this case, order of $mathbb{R}^{times}$ is infinite. Any hints would be helpful. Thanks in advance.
abstract-algebra group-theory finite-groups group-homomorphism
edited Nov 13 at 7:26
Batominovski
31.5k23187
asked Nov 13 at 6:22
LOIS
3558
add a comment |
up vote
2
down vote
favorite
If $G$ is a finite group of odd order, show there is only one homomorphism between $G$ and $mathbb{R}^{times}$ which is the trivial one.
I know there is a similar case if the order of G and H is coprime. then the order of image of $G$ is 1. But in this case, order of $mathbb{R}^{times}$ is infinite. Any hints would be helpful. Thanks in advance.
abstract-algebra group-theory finite-groups group-homomorphism
edited Nov 13 at 7:26
Batominovski
31.5k23187
asked Nov 13 at 6:22
LOIS
3558
3
The only real solution to $x^n=1$ when $n$ is odd, is $x=1$.
– Lord Shark the Unknown
Nov 13 at 6:25
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
If $G$ is a finite group of odd order, show there is only one homomorphism between $G$ and $mathbb{R}^{times}$ which is the trivial one.
I know there is a similar case if the order of G and H is coprime. then the order of image of $G$ is 1. But in this case, order of $mathbb{R}^{times}$ is infinite. Any hints would be helpful. Thanks in advance.
abstract-algebra group-theory finite-groups group-homomorphism
edited Nov 13 at 7:26
Batominovski
31.5k23187
asked Nov 13 at 6:22
LOIS
3558
If $G$ is a finite group of odd order, show there is only one homomorphism between $G$ and $mathbb{R}^{times}$ which is the trivial one.
I know there is a similar case if the order of G and H is coprime. then the order of image of $G$ is 1. But in this case, order of $mathbb{R}^{times}$ is infinite. Any hints would be helpful. Thanks in advance.
abstract-algebra group-theory finite-groups group-homomorphism
abstract-algebra group-theory finite-groups group-homomorphism
edited Nov 13 at 7:26
Batominovski
31.5k23187
asked Nov 13 at 6:22
LOIS
3558
edited Nov 13 at 7:26
Batominovski
31.5k23187
asked Nov 13 at 6:22
LOIS
3558
edited Nov 13 at 7:26
Batominovski
31.5k23187
edited Nov 13 at 7:26
Batominovski
31.5k23187
edited Nov 13 at 7:26
Batominovski
31.5k23187
31.5k23187
asked Nov 13 at 6:22
LOIS
3558
asked Nov 13 at 6:22
LOIS
3558
asked Nov 13 at 6:22
LOIS
3558
3558
3
The only real solution to $x^n=1$ when $n$ is odd, is $x=1$.
– Lord Shark the Unknown
Nov 13 at 6:25
add a comment |
3
The only real solution to $x^n=1$ when $n$ is odd, is $x=1$.
– Lord Shark the Unknown
Nov 13 at 6:25
3
3
The only real solution to $x^n=1$ when $n$ is odd, is $x=1$.
– Lord Shark the Unknown
Nov 13 at 6:25
The only real solution to $x^n=1$ when $n$ is odd, is $x=1$.
– Lord Shark the Unknown
Nov 13 at 6:25
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
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A homomorphism from $G$ to $mathbb R^times$ satisfies $phi(a^n) = (phi(a))^n$ for all $a in G$ and $n in mathbb N$.
With this in mind, note that $a in G$ always satisfies $a^{|G|} = e_G$. Applying $phi$ on both sides yields $(phi(a))^{|G|} = 1$.
That is, $phi(a)$ is a real number, which satisfies $phi(a)^{|G|} = 1$, where $|G|$ is odd. This forces $phi(a) = 1$ for all $a$. So $phi$ is trivial.
Note that non-trivial homomorphisms in the reverse direction i.e. from $mathbb R^times$ to $G$ may very much exist. Can you think of an obvious map from $mathbb R^times$ to the multiplicative group $pm 1$ of two elements, for example?
Note that if $|G|$ is even, then the statement is not true. Can you think of a non-trivial homomorphism from $pm 1$ to $mathbb R^times$? This is even more obvious than the previous question.
answered Nov 13 at 9:00
астон вілла олоф мэллбэрг
36.3k33375
Thanks . Can't image the first map. The second is identical?
– LOIS
Nov 13 at 16:03
The first map is the sign map. So we map $a$ to $1$ if it is positive, and $-1$ if it is negative(it cannot be zero, since $0 notin mathbb R^times$). This is a homomorphism, obviously. The answer to the second one is the identity map, as you said : take $1$ to $1$ and $-1$ to $-1$, this is obviously a homomorphism again.
– астон вілла олоф мэллбэрг
Nov 13 at 16:17
Thanks again for your training.
– LOIS
Nov 14 at 7:26
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
A homomorphism from $G$ to $mathbb R^times$ satisfies $phi(a^n) = (phi(a))^n$ for all $a in G$ and $n in mathbb N$.
With this in mind, note that $a in G$ always satisfies $a^{|G|} = e_G$. Applying $phi$ on both sides yields $(phi(a))^{|G|} = 1$.
That is, $phi(a)$ is a real number, which satisfies $phi(a)^{|G|} = 1$, where $|G|$ is odd. This forces $phi(a) = 1$ for all $a$. So $phi$ is trivial.
Note that non-trivial homomorphisms in the reverse direction i.e. from $mathbb R^times$ to $G$ may very much exist. Can you think of an obvious map from $mathbb R^times$ to the multiplicative group $pm 1$ of two elements, for example?
Note that if $|G|$ is even, then the statement is not true. Can you think of a non-trivial homomorphism from $pm 1$ to $mathbb R^times$? This is even more obvious than the previous question.
answered Nov 13 at 9:00
астон вілла олоф мэллбэрг
36.3k33375
Thanks . Can't image the first map. The second is identical?
– LOIS
Nov 13 at 16:03
The first map is the sign map. So we map $a$ to $1$ if it is positive, and $-1$ if it is negative(it cannot be zero, since $0 notin mathbb R^times$). This is a homomorphism, obviously. The answer to the second one is the identity map, as you said : take $1$ to $1$ and $-1$ to $-1$, this is obviously a homomorphism again.
– астон вілла олоф мэллбэрг
Nov 13 at 16:17
Thanks again for your training.
– LOIS
Nov 14 at 7:26
add a comment |
up vote
4
down vote
accepted
A homomorphism from $G$ to $mathbb R^times$ satisfies $phi(a^n) = (phi(a))^n$ for all $a in G$ and $n in mathbb N$.
With this in mind, note that $a in G$ always satisfies $a^{|G|} = e_G$. Applying $phi$ on both sides yields $(phi(a))^{|G|} = 1$.
That is, $phi(a)$ is a real number, which satisfies $phi(a)^{|G|} = 1$, where $|G|$ is odd. This forces $phi(a) = 1$ for all $a$. So $phi$ is trivial.
Note that non-trivial homomorphisms in the reverse direction i.e. from $mathbb R^times$ to $G$ may very much exist. Can you think of an obvious map from $mathbb R^times$ to the multiplicative group $pm 1$ of two elements, for example?
Note that if $|G|$ is even, then the statement is not true. Can you think of a non-trivial homomorphism from $pm 1$ to $mathbb R^times$? This is even more obvious than the previous question.
answered Nov 13 at 9:00
астон вілла олоф мэллбэрг
36.3k33375
Thanks . Can't image the first map. The second is identical?
– LOIS
Nov 13 at 16:03
The first map is the sign map. So we map $a$ to $1$ if it is positive, and $-1$ if it is negative(it cannot be zero, since $0 notin mathbb R^times$). This is a homomorphism, obviously. The answer to the second one is the identity map, as you said : take $1$ to $1$ and $-1$ to $-1$, this is obviously a homomorphism again.
– астон вілла олоф мэллбэрг
Nov 13 at 16:17
Thanks again for your training.
– LOIS
Nov 14 at 7:26
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
A homomorphism from $G$ to $mathbb R^times$ satisfies $phi(a^n) = (phi(a))^n$ for all $a in G$ and $n in mathbb N$.
With this in mind, note that $a in G$ always satisfies $a^{|G|} = e_G$. Applying $phi$ on both sides yields $(phi(a))^{|G|} = 1$.
That is, $phi(a)$ is a real number, which satisfies $phi(a)^{|G|} = 1$, where $|G|$ is odd. This forces $phi(a) = 1$ for all $a$. So $phi$ is trivial.
Note that non-trivial homomorphisms in the reverse direction i.e. from $mathbb R^times$ to $G$ may very much exist. Can you think of an obvious map from $mathbb R^times$ to the multiplicative group $pm 1$ of two elements, for example?
Note that if $|G|$ is even, then the statement is not true. Can you think of a non-trivial homomorphism from $pm 1$ to $mathbb R^times$? This is even more obvious than the previous question.
answered Nov 13 at 9:00
астон вілла олоф мэллбэрг
36.3k33375
A homomorphism from $G$ to $mathbb R^times$ satisfies $phi(a^n) = (phi(a))^n$ for all $a in G$ and $n in mathbb N$.
With this in mind, note that $a in G$ always satisfies $a^{|G|} = e_G$. Applying $phi$ on both sides yields $(phi(a))^{|G|} = 1$.
That is, $phi(a)$ is a real number, which satisfies $phi(a)^{|G|} = 1$, where $|G|$ is odd. This forces $phi(a) = 1$ for all $a$. So $phi$ is trivial.
Note that non-trivial homomorphisms in the reverse direction i.e. from $mathbb R^times$ to $G$ may very much exist. Can you think of an obvious map from $mathbb R^times$ to the multiplicative group $pm 1$ of two elements, for example?
Note that if $|G|$ is even, then the statement is not true. Can you think of a non-trivial homomorphism from $pm 1$ to $mathbb R^times$? This is even more obvious than the previous question.
answered Nov 13 at 9:00
астон вілла олоф мэллбэрг
36.3k33375
answered Nov 13 at 9:00
астон вілла олоф мэллбэрг
36.3k33375
answered Nov 13 at 9:00
астон вілла олоф мэллбэрг
36.3k33375
answered Nov 13 at 9:00
астон вілла олоф мэллбэрг
36.3k33375
36.3k33375
Thanks . Can't image the first map. The second is identical?
– LOIS
Nov 13 at 16:03
The first map is the sign map. So we map $a$ to $1$ if it is positive, and $-1$ if it is negative(it cannot be zero, since $0 notin mathbb R^times$). This is a homomorphism, obviously. The answer to the second one is the identity map, as you said : take $1$ to $1$ and $-1$ to $-1$, this is obviously a homomorphism again.
– астон вілла олоф мэллбэрг
Nov 13 at 16:17
Thanks again for your training.
– LOIS
Nov 14 at 7:26
add a comment |
Thanks . Can't image the first map. The second is identical?
– LOIS
Nov 13 at 16:03
The first map is the sign map. So we map $a$ to $1$ if it is positive, and $-1$ if it is negative(it cannot be zero, since $0 notin mathbb R^times$). This is a homomorphism, obviously. The answer to the second one is the identity map, as you said : take $1$ to $1$ and $-1$ to $-1$, this is obviously a homomorphism again.
– астон вілла олоф мэллбэрг
Nov 13 at 16:17
Thanks again for your training.
– LOIS
Nov 14 at 7:26
Thanks . Can't image the first map. The second is identical?
– LOIS
Nov 13 at 16:03
Thanks . Can't image the first map. The second is identical?
– LOIS
Nov 13 at 16:03
The first map is the sign map. So we map $a$ to $1$ if it is positive, and $-1$ if it is negative(it cannot be zero, since $0 notin mathbb R^times$). This is a homomorphism, obviously. The answer to the second one is the identity map, as you said : take $1$ to $1$ and $-1$ to $-1$, this is obviously a homomorphism again.
– астон вілла олоф мэллбэрг
Nov 13 at 16:17
The first map is the sign map. So we map $a$ to $1$ if it is positive, and $-1$ if it is negative(it cannot be zero, since $0 notin mathbb R^times$). This is a homomorphism, obviously. The answer to the second one is the identity map, as you said : take $1$ to $1$ and $-1$ to $-1$, this is obviously a homomorphism again.
– астон вілла олоф мэллбэрг
Nov 13 at 16:17
Thanks again for your training.
– LOIS
Nov 14 at 7:26
Thanks again for your training.
– LOIS
Nov 14 at 7:26
add a comment |
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The only real solution to $x^n=1$ when $n$ is odd, is $x=1$.
– Lord Shark the Unknown
Nov 13 at 6:25