Proving that any given element $hk$ appears $|H cap K|$ times as a product in the list of $HK$
$begingroup$
I have been trying to prove that if $H, K$ are finite subgroups of $G$ then $|HK|=dfrac{|H| |K|}{|H cap K|}$. I saw the proofs in Herstein and Gallian textbook and they are essentially the same. However, it confuses me. The claim is that any given element $hk$ appears $|H cap K|$ times as a product in the list of $HK$ and what follows is my attempt of the proving the same but in a little different way and I wonder if it is correct.
We define the set $mathcal{R}(hk)={ (h',k') in Htimes K , | , hk=h'k' }$. Now, we define the map $fcolon Hcap K to mathcal{R}(hk)$ by $f(x)=(hx, x^{-1}k)$. Clearly, $f$ is well defined and we will show that $f$ is bijective.
Let $x,y in Hcap K$ and suppose that $f(x)=f(y)$. Then $(hx, x^{-1}k)=(hy, y^{-1}k)$ and hence $x=y$.
Let $(h' , k') in mathcal{R}(hk)$. Then $hk=h'k'$ and let $x = h^{-1} h' = kk'^{-1}$. Clearly, $xin H cap K$ and we have that $h'=hx$ and $k'=x^{-1}k$. Thus, $f(x)=(h',k')$. This shows $f$ is surjective.
Hence, we have that $|mathcal{R}(hk)|=|H cap K|$. This proves our claim.
I wonder if this is okay. Are Herstein and Gallian doing the same thing?
Here's the proof from Herstein:
group-theory proof-verification proof-explanation alternative-proof
edited Dec 21 '18 at 16:54
Shaun
10.1k113685
asked Dec 19 '18 at 12:09
user561733user561733
334
$endgroup$
add a comment |
$begingroup$
I have been trying to prove that if $H, K$ are finite subgroups of $G$ then $|HK|=dfrac{|H| |K|}{|H cap K|}$. I saw the proofs in Herstein and Gallian textbook and they are essentially the same. However, it confuses me. The claim is that any given element $hk$ appears $|H cap K|$ times as a product in the list of $HK$ and what follows is my attempt of the proving the same but in a little different way and I wonder if it is correct.
We define the set $mathcal{R}(hk)={ (h',k') in Htimes K , | , hk=h'k' }$. Now, we define the map $fcolon Hcap K to mathcal{R}(hk)$ by $f(x)=(hx, x^{-1}k)$. Clearly, $f$ is well defined and we will show that $f$ is bijective.
Let $x,y in Hcap K$ and suppose that $f(x)=f(y)$. Then $(hx, x^{-1}k)=(hy, y^{-1}k)$ and hence $x=y$.
Let $(h' , k') in mathcal{R}(hk)$. Then $hk=h'k'$ and let $x = h^{-1} h' = kk'^{-1}$. Clearly, $xin H cap K$ and we have that $h'=hx$ and $k'=x^{-1}k$. Thus, $f(x)=(h',k')$. This shows $f$ is surjective.
Hence, we have that $|mathcal{R}(hk)|=|H cap K|$. This proves our claim.
I wonder if this is okay. Are Herstein and Gallian doing the same thing?
Here's the proof from Herstein:
group-theory proof-verification proof-explanation alternative-proof
edited Dec 21 '18 at 16:54
Shaun
10.1k113685
asked Dec 19 '18 at 12:09
user561733user561733
334
$endgroup$
3
$begingroup$
We cannot compare your proof with Herstein and Gallian's version, unless you post their proof here. Not everybody has the book. But from where I stand, your proof looks ok.
$endgroup$
– user593746
Dec 19 '18 at 18:17
1
$begingroup$
@Zvi i have updated the question
$endgroup$
– user561733
Dec 19 '18 at 19:00
3
$begingroup$
Yes, your proof and the textbook's proof are doing the same thing, phrased differently.
$endgroup$
– user593746
Dec 20 '18 at 15:53
add a comment |
$begingroup$
I have been trying to prove that if $H, K$ are finite subgroups of $G$ then $|HK|=dfrac{|H| |K|}{|H cap K|}$. I saw the proofs in Herstein and Gallian textbook and they are essentially the same. However, it confuses me. The claim is that any given element $hk$ appears $|H cap K|$ times as a product in the list of $HK$ and what follows is my attempt of the proving the same but in a little different way and I wonder if it is correct.
We define the set $mathcal{R}(hk)={ (h',k') in Htimes K , | , hk=h'k' }$. Now, we define the map $fcolon Hcap K to mathcal{R}(hk)$ by $f(x)=(hx, x^{-1}k)$. Clearly, $f$ is well defined and we will show that $f$ is bijective.
Let $x,y in Hcap K$ and suppose that $f(x)=f(y)$. Then $(hx, x^{-1}k)=(hy, y^{-1}k)$ and hence $x=y$.
Let $(h' , k') in mathcal{R}(hk)$. Then $hk=h'k'$ and let $x = h^{-1} h' = kk'^{-1}$. Clearly, $xin H cap K$ and we have that $h'=hx$ and $k'=x^{-1}k$. Thus, $f(x)=(h',k')$. This shows $f$ is surjective.
Hence, we have that $|mathcal{R}(hk)|=|H cap K|$. This proves our claim.
I wonder if this is okay. Are Herstein and Gallian doing the same thing?
Here's the proof from Herstein:
group-theory proof-verification proof-explanation alternative-proof
edited Dec 21 '18 at 16:54
Shaun
10.1k113685
asked Dec 19 '18 at 12:09
user561733user561733
334
$endgroup$
I have been trying to prove that if $H, K$ are finite subgroups of $G$ then $|HK|=dfrac{|H| |K|}{|H cap K|}$. I saw the proofs in Herstein and Gallian textbook and they are essentially the same. However, it confuses me. The claim is that any given element $hk$ appears $|H cap K|$ times as a product in the list of $HK$ and what follows is my attempt of the proving the same but in a little different way and I wonder if it is correct.
We define the set $mathcal{R}(hk)={ (h',k') in Htimes K , | , hk=h'k' }$. Now, we define the map $fcolon Hcap K to mathcal{R}(hk)$ by $f(x)=(hx, x^{-1}k)$. Clearly, $f$ is well defined and we will show that $f$ is bijective.
Let $x,y in Hcap K$ and suppose that $f(x)=f(y)$. Then $(hx, x^{-1}k)=(hy, y^{-1}k)$ and hence $x=y$.
Let $(h' , k') in mathcal{R}(hk)$. Then $hk=h'k'$ and let $x = h^{-1} h' = kk'^{-1}$. Clearly, $xin H cap K$ and we have that $h'=hx$ and $k'=x^{-1}k$. Thus, $f(x)=(h',k')$. This shows $f$ is surjective.
Hence, we have that $|mathcal{R}(hk)|=|H cap K|$. This proves our claim.
I wonder if this is okay. Are Herstein and Gallian doing the same thing?
Here's the proof from Herstein:
group-theory proof-verification proof-explanation alternative-proof
group-theory proof-verification proof-explanation alternative-proof
edited Dec 21 '18 at 16:54
Shaun
10.1k113685
asked Dec 19 '18 at 12:09
user561733user561733
334
edited Dec 21 '18 at 16:54
Shaun
10.1k113685
asked Dec 19 '18 at 12:09
user561733user561733
334
edited Dec 21 '18 at 16:54
Shaun
10.1k113685
edited Dec 21 '18 at 16:54
Shaun
10.1k113685
edited Dec 21 '18 at 16:54
Shaun
10.1k113685
10.1k113685
asked Dec 19 '18 at 12:09
user561733user561733
334
asked Dec 19 '18 at 12:09
user561733user561733
334
asked Dec 19 '18 at 12:09
user561733user561733
334
334
3
$begingroup$
We cannot compare your proof with Herstein and Gallian's version, unless you post their proof here. Not everybody has the book. But from where I stand, your proof looks ok.
$endgroup$
– user593746
Dec 19 '18 at 18:17
1
$begingroup$
@Zvi i have updated the question
$endgroup$
– user561733
Dec 19 '18 at 19:00
3
$begingroup$
Yes, your proof and the textbook's proof are doing the same thing, phrased differently.
$endgroup$
– user593746
Dec 20 '18 at 15:53
add a comment |
3
$begingroup$
We cannot compare your proof with Herstein and Gallian's version, unless you post their proof here. Not everybody has the book. But from where I stand, your proof looks ok.
$endgroup$
– user593746
Dec 19 '18 at 18:17
1
$begingroup$
@Zvi i have updated the question
$endgroup$
– user561733
Dec 19 '18 at 19:00
3
$begingroup$
Yes, your proof and the textbook's proof are doing the same thing, phrased differently.
$endgroup$
– user593746
Dec 20 '18 at 15:53
3
3
$begingroup$
We cannot compare your proof with Herstein and Gallian's version, unless you post their proof here. Not everybody has the book. But from where I stand, your proof looks ok.
$endgroup$
– user593746
Dec 19 '18 at 18:17
$begingroup$
We cannot compare your proof with Herstein and Gallian's version, unless you post their proof here. Not everybody has the book. But from where I stand, your proof looks ok.
$endgroup$
– user593746
Dec 19 '18 at 18:17
1
1
$begingroup$
@Zvi i have updated the question
$endgroup$
– user561733
Dec 19 '18 at 19:00
$begingroup$
@Zvi i have updated the question
$endgroup$
– user561733
Dec 19 '18 at 19:00
3
3
$begingroup$
Yes, your proof and the textbook's proof are doing the same thing, phrased differently.
$endgroup$
– user593746
Dec 20 '18 at 15:53
$begingroup$
Yes, your proof and the textbook's proof are doing the same thing, phrased differently.
$endgroup$
– user593746
Dec 20 '18 at 15:53
add a comment |
0
active
oldest
votes
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%2f3046313%2fproving-that-any-given-element-hk-appears-h-cap-k-times-as-a-product-in-t%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3046313%2fproving-that-any-given-element-hk-appears-h-cap-k-times-as-a-product-in-t%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
3
$begingroup$
We cannot compare your proof with Herstein and Gallian's version, unless you post their proof here. Not everybody has the book. But from where I stand, your proof looks ok.
$endgroup$
– user593746
Dec 19 '18 at 18:17
1
$begingroup$
@Zvi i have updated the question
$endgroup$
– user561733
Dec 19 '18 at 19:00
3
$begingroup$
Yes, your proof and the textbook's proof are doing the same thing, phrased differently.
$endgroup$
– user593746
Dec 20 '18 at 15:53