Proving isomorphism using Cayley
$begingroup$
Prove that $theta$ is a group isomorphism.
Let
$theta :mathbb{Q}^* rightarrow operatorname{Aut}(mathbb{Q})$
$w mapsto f_w$
and
$f_w(x) = wx$ where $w in mathbb{Q}^* $ and $x in mathbb{Q}$
Is there a way to prove this without resorting to proving that $theta$ is a homomorphism followed by proving the surjectivity and injectivity?
Maybe using Cayley's theorem since $f_w$ is a permutation.
Thanks for the help
group-theory group-isomorphism automorphism-group
edited Nov 30 '18 at 22:45
Davide Giraudo
126k16150261
asked Nov 30 '18 at 22:33
JoeyFJoeyF
62
$endgroup$
add a comment |
$begingroup$
Prove that $theta$ is a group isomorphism.
Let
$theta :mathbb{Q}^* rightarrow operatorname{Aut}(mathbb{Q})$
$w mapsto f_w$
and
$f_w(x) = wx$ where $w in mathbb{Q}^* $ and $x in mathbb{Q}$
Is there a way to prove this without resorting to proving that $theta$ is a homomorphism followed by proving the surjectivity and injectivity?
Maybe using Cayley's theorem since $f_w$ is a permutation.
Thanks for the help
group-theory group-isomorphism automorphism-group
edited Nov 30 '18 at 22:45
Davide Giraudo
126k16150261
asked Nov 30 '18 at 22:33
JoeyFJoeyF
62
$endgroup$
add a comment |
$begingroup$
Prove that $theta$ is a group isomorphism.
Let
$theta :mathbb{Q}^* rightarrow operatorname{Aut}(mathbb{Q})$
$w mapsto f_w$
and
$f_w(x) = wx$ where $w in mathbb{Q}^* $ and $x in mathbb{Q}$
Is there a way to prove this without resorting to proving that $theta$ is a homomorphism followed by proving the surjectivity and injectivity?
Maybe using Cayley's theorem since $f_w$ is a permutation.
Thanks for the help
group-theory group-isomorphism automorphism-group
edited Nov 30 '18 at 22:45
Davide Giraudo
126k16150261
asked Nov 30 '18 at 22:33
JoeyFJoeyF
62
$endgroup$
Prove that $theta$ is a group isomorphism.
Let
$theta :mathbb{Q}^* rightarrow operatorname{Aut}(mathbb{Q})$
$w mapsto f_w$
and
$f_w(x) = wx$ where $w in mathbb{Q}^* $ and $x in mathbb{Q}$
Is there a way to prove this without resorting to proving that $theta$ is a homomorphism followed by proving the surjectivity and injectivity?
Maybe using Cayley's theorem since $f_w$ is a permutation.
Thanks for the help
group-theory group-isomorphism automorphism-group
group-theory group-isomorphism automorphism-group
edited Nov 30 '18 at 22:45
Davide Giraudo
126k16150261
asked Nov 30 '18 at 22:33
JoeyFJoeyF
62
edited Nov 30 '18 at 22:45
Davide Giraudo
126k16150261
asked Nov 30 '18 at 22:33
JoeyFJoeyF
62
edited Nov 30 '18 at 22:45
Davide Giraudo
126k16150261
edited Nov 30 '18 at 22:45
Davide Giraudo
126k16150261
edited Nov 30 '18 at 22:45
Davide Giraudo
126k16150261
126k16150261
asked Nov 30 '18 at 22:33
JoeyFJoeyF
62
asked Nov 30 '18 at 22:33
JoeyFJoeyF
62
asked Nov 30 '18 at 22:33
JoeyFJoeyF
62
62
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
While $f_w$ is a permutation, it doesn't really lead anywhere in my opinion.
So yes, showing that $theta$ is injective and surjective is a good way to do that. Together with some linear algebra: you just have to realize that if $f:mathbb{Q}tomathbb{Q}$ is a group homomorphism, meaning it preserves addition, then it also preserves multiplication automatically. This is because in $mathbb{Q}$ multiplication comes from addition. I encourage you to show that.
This implies that every group homomorphism $f:mathbb{Q}tomathbb{Q}$ is actually a linear map with $mathbb{Q}$ treated as a vector space over $mathbb{Q}$. General linear algebra applies and so $f(x)=lambda x$ for some unique $lambdainmathbb{Q}$. Note that $f$ is invertible if and only if $lambdaneq 0$.
So surjectivity of $theta$ is equivalent to existance of $lambda$. Injectivety of $theta$ to a simple observation that different linear maps have to have different $lambda$ coefficient.
answered Nov 30 '18 at 22:55
freakishfreakish
12.3k1630
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
While $f_w$ is a permutation, it doesn't really lead anywhere in my opinion.
So yes, showing that $theta$ is injective and surjective is a good way to do that. Together with some linear algebra: you just have to realize that if $f:mathbb{Q}tomathbb{Q}$ is a group homomorphism, meaning it preserves addition, then it also preserves multiplication automatically. This is because in $mathbb{Q}$ multiplication comes from addition. I encourage you to show that.
This implies that every group homomorphism $f:mathbb{Q}tomathbb{Q}$ is actually a linear map with $mathbb{Q}$ treated as a vector space over $mathbb{Q}$. General linear algebra applies and so $f(x)=lambda x$ for some unique $lambdainmathbb{Q}$. Note that $f$ is invertible if and only if $lambdaneq 0$.
So surjectivity of $theta$ is equivalent to existance of $lambda$. Injectivety of $theta$ to a simple observation that different linear maps have to have different $lambda$ coefficient.
answered Nov 30 '18 at 22:55
freakishfreakish
12.3k1630
$endgroup$
add a comment |
$begingroup$
While $f_w$ is a permutation, it doesn't really lead anywhere in my opinion.
So yes, showing that $theta$ is injective and surjective is a good way to do that. Together with some linear algebra: you just have to realize that if $f:mathbb{Q}tomathbb{Q}$ is a group homomorphism, meaning it preserves addition, then it also preserves multiplication automatically. This is because in $mathbb{Q}$ multiplication comes from addition. I encourage you to show that.
This implies that every group homomorphism $f:mathbb{Q}tomathbb{Q}$ is actually a linear map with $mathbb{Q}$ treated as a vector space over $mathbb{Q}$. General linear algebra applies and so $f(x)=lambda x$ for some unique $lambdainmathbb{Q}$. Note that $f$ is invertible if and only if $lambdaneq 0$.
So surjectivity of $theta$ is equivalent to existance of $lambda$. Injectivety of $theta$ to a simple observation that different linear maps have to have different $lambda$ coefficient.
answered Nov 30 '18 at 22:55
freakishfreakish
12.3k1630
$endgroup$
add a comment |
$begingroup$
While $f_w$ is a permutation, it doesn't really lead anywhere in my opinion.
So yes, showing that $theta$ is injective and surjective is a good way to do that. Together with some linear algebra: you just have to realize that if $f:mathbb{Q}tomathbb{Q}$ is a group homomorphism, meaning it preserves addition, then it also preserves multiplication automatically. This is because in $mathbb{Q}$ multiplication comes from addition. I encourage you to show that.
This implies that every group homomorphism $f:mathbb{Q}tomathbb{Q}$ is actually a linear map with $mathbb{Q}$ treated as a vector space over $mathbb{Q}$. General linear algebra applies and so $f(x)=lambda x$ for some unique $lambdainmathbb{Q}$. Note that $f$ is invertible if and only if $lambdaneq 0$.
So surjectivity of $theta$ is equivalent to existance of $lambda$. Injectivety of $theta$ to a simple observation that different linear maps have to have different $lambda$ coefficient.
answered Nov 30 '18 at 22:55
freakishfreakish
12.3k1630
$endgroup$
While $f_w$ is a permutation, it doesn't really lead anywhere in my opinion.
So yes, showing that $theta$ is injective and surjective is a good way to do that. Together with some linear algebra: you just have to realize that if $f:mathbb{Q}tomathbb{Q}$ is a group homomorphism, meaning it preserves addition, then it also preserves multiplication automatically. This is because in $mathbb{Q}$ multiplication comes from addition. I encourage you to show that.
This implies that every group homomorphism $f:mathbb{Q}tomathbb{Q}$ is actually a linear map with $mathbb{Q}$ treated as a vector space over $mathbb{Q}$. General linear algebra applies and so $f(x)=lambda x$ for some unique $lambdainmathbb{Q}$. Note that $f$ is invertible if and only if $lambdaneq 0$.
So surjectivity of $theta$ is equivalent to existance of $lambda$. Injectivety of $theta$ to a simple observation that different linear maps have to have different $lambda$ coefficient.
answered Nov 30 '18 at 22:55
freakishfreakish
12.3k1630
answered Nov 30 '18 at 22:55
freakishfreakish
12.3k1630
answered Nov 30 '18 at 22:55
freakishfreakish
12.3k1630
answered Nov 30 '18 at 22:55
freakishfreakish
12.3k1630
12.3k1630
add a comment |
add a comment |
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