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
125k16150261
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
125k16150261
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
125k16150261
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
125k16150261
asked Nov 30 '18 at 22:33
JoeyFJoeyF
62
edited Nov 30 '18 at 22:45
Davide Giraudo
125k16150261
asked Nov 30 '18 at 22:33
JoeyFJoeyF
62
edited Nov 30 '18 at 22:45
Davide Giraudo
125k16150261
edited Nov 30 '18 at 22:45
Davide Giraudo
125k16150261
edited Nov 30 '18 at 22:45
Davide Giraudo
125k16150261
125k16150261
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
11.8k1629
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020735%2fproving-isomorphism-using-cayley%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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
11.8k1629
$endgroup$
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
11.8k1629
$endgroup$
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
11.8k1629
$endgroup$
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
11.8k1629
answered Nov 30 '18 at 22:55
freakishfreakish
11.8k1629
answered Nov 30 '18 at 22:55
freakishfreakish
11.8k1629
answered Nov 30 '18 at 22:55
freakishfreakish
11.8k1629
11.8k1629
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020735%2fproving-isomorphism-using-cayley%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
