Showing $Ha=H$ if and only if $a$ belongs to $H$. The Next CEO of Stack Overflow$gN = hN$ if...
Is it safe to use c_str() on a temporary string?
Why here is plural "We went to the movies last night."
Can the Reverse Gravity spell affect the Meteor Swarm spell?
How did people program for Consoles with multiple CPUs?
Describing a person. What needs to be mentioned?
Failed to fetch jessie backports repository
Can a caster that cast Polymorph on themselves stop concentrating at any point even if their Int is low?
How to be diplomatic in refusing to write code that breaches the privacy of our users
Opposite of a diet
Inappropriate reference requests from Journal reviewers
Why does standard notation not preserve intervals (visually)
Which organization defines CJK Unified Ideographs?
If I blow insulation everywhere in my attic except the door trap, will heat escape through it?
How do I solve this limit?
Is it okay to store user locations?
How do I get the green key off the shelf in the Dobby level of Lego Harry Potter 2?
Customer Requests (Sometimes) Drive Me Bonkers!
How long to clear the 'suck zone' of a turbofan after start is initiated?
How do we know the LHC results are robust?
Is HostGator storing my password in plaintext?
How to count occurrences of text in a file?
Science fiction (dystopian) short story set after WWIII
Grabbing quick drinks
How can I quit an app using Terminal?
Showing $Ha=H$ if and only if $a$ belongs to $H$.
The Next CEO of Stack Overflow$gN = hN$ if and only if $g^{-1} h in N$$H$ is a maximal normal subgroup of $G$ if and only if $G/H$ is simple.Show that there is just one subgroup $H subset S_4$ such that $[S_4:H] = 2$If a finite set $G$ is closed under an associative product and both cancellation laws hold, then it is a groupShowing that the intersection of 2 subgroups is a subgroupShow that $G/N$ acts faithfully on $S$ if and only if $N=kerphi$Prove that it is impossible that every non-identity element of $G$ has an order of $2$.Explanations on the proof of Theorem 2.5 in Hungerford's algebraShow $H$ is the only subgroup in $G$ of index 2, when $|G| not = 4$ and $[G:H]=2$A group $G$ is abelian if and only if a certain subset of the direct product is a subgroupNon-identity element in a group has infinite order
$begingroup$
Let be $H$ a subgroup. Show $Ha=H$ if and only if $a$ belongs to $H$.
Here what I have understand of the proof: $a$ must belong to $Ha$ because $ea=a$ where $e$ is the identity in $H$. So the only way $H=Ha$ is that a belongs to $H$.
Is that correct?
Now suppose that a belongs to H. for every element h in H, $h=(ha^{-1})a$. This implies that h belongs to Ha so also a belongs to Ha. This implies H=Ha.
group-theory
asked Mar 16 at 8:12
homunculushomunculus
1227
$endgroup$
add a comment |
$begingroup$
Let be $H$ a subgroup. Show $Ha=H$ if and only if $a$ belongs to $H$.
Here what I have understand of the proof: $a$ must belong to $Ha$ because $ea=a$ where $e$ is the identity in $H$. So the only way $H=Ha$ is that a belongs to $H$.
Is that correct?
Now suppose that a belongs to H. for every element h in H, $h=(ha^{-1})a$. This implies that h belongs to Ha so also a belongs to Ha. This implies H=Ha.
group-theory
asked Mar 16 at 8:12
homunculushomunculus
1227
$endgroup$
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Shaun
Mar 16 at 8:33
$begingroup$
not enough reputation i have
$endgroup$
– homunculus
Mar 16 at 8:34
$begingroup$
Try again, please. It was probably the 15 minute wait timer that stopped you, not lack of reputation.
$endgroup$
– Shaun
Mar 16 at 8:54
$begingroup$
no it is lack of Rep only 15
$endgroup$
– homunculus
Mar 16 at 8:55
add a comment |
$begingroup$
Let be $H$ a subgroup. Show $Ha=H$ if and only if $a$ belongs to $H$.
Here what I have understand of the proof: $a$ must belong to $Ha$ because $ea=a$ where $e$ is the identity in $H$. So the only way $H=Ha$ is that a belongs to $H$.
Is that correct?
Now suppose that a belongs to H. for every element h in H, $h=(ha^{-1})a$. This implies that h belongs to Ha so also a belongs to Ha. This implies H=Ha.
group-theory
asked Mar 16 at 8:12
homunculushomunculus
1227
$endgroup$
Let be $H$ a subgroup. Show $Ha=H$ if and only if $a$ belongs to $H$.
Here what I have understand of the proof: $a$ must belong to $Ha$ because $ea=a$ where $e$ is the identity in $H$. So the only way $H=Ha$ is that a belongs to $H$.
Is that correct?
Now suppose that a belongs to H. for every element h in H, $h=(ha^{-1})a$. This implies that h belongs to Ha so also a belongs to Ha. This implies H=Ha.
group-theory
group-theory
asked Mar 16 at 8:12
homunculushomunculus
1227
asked Mar 16 at 8:12
homunculushomunculus
1227
edited Mar 16 at 8:29
homunculus
asked Mar 16 at 8:12
homunculushomunculus
1227
asked Mar 16 at 8:12
homunculushomunculus
1227
asked Mar 16 at 8:12
homunculushomunculus
1227
1227
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Shaun
Mar 16 at 8:33
$begingroup$
not enough reputation i have
$endgroup$
– homunculus
Mar 16 at 8:34
$begingroup$
Try again, please. It was probably the 15 minute wait timer that stopped you, not lack of reputation.
$endgroup$
– Shaun
Mar 16 at 8:54
$begingroup$
no it is lack of Rep only 15
$endgroup$
– homunculus
Mar 16 at 8:55
add a comment |
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Shaun
Mar 16 at 8:33
$begingroup$
not enough reputation i have
$endgroup$
– homunculus
Mar 16 at 8:34
$begingroup$
Try again, please. It was probably the 15 minute wait timer that stopped you, not lack of reputation.
$endgroup$
– Shaun
Mar 16 at 8:54
$begingroup$
no it is lack of Rep only 15
$endgroup$
– homunculus
Mar 16 at 8:55
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Shaun
Mar 16 at 8:33
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Shaun
Mar 16 at 8:33
$begingroup$
not enough reputation i have
$endgroup$
– homunculus
Mar 16 at 8:34
$begingroup$
not enough reputation i have
$endgroup$
– homunculus
Mar 16 at 8:34
$begingroup$
Try again, please. It was probably the 15 minute wait timer that stopped you, not lack of reputation.
$endgroup$
– Shaun
Mar 16 at 8:54
$begingroup$
Try again, please. It was probably the 15 minute wait timer that stopped you, not lack of reputation.
$endgroup$
– Shaun
Mar 16 at 8:54
$begingroup$
no it is lack of Rep only 15
$endgroup$
– homunculus
Mar 16 at 8:55
$begingroup$
no it is lack of Rep only 15
$endgroup$
– homunculus
Mar 16 at 8:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $ain H$, then every element of $H$ can be written in the form $(ha^{-1})ain Ha$ and due to closure, $Ha=H$. Contrarily, if $Ha=H$ then $a=eain Ha=H$.
answered Mar 16 at 8:28
TheSimpliFireTheSimpliFire
12.9k62462
$endgroup$
$begingroup$
$ha^{-1}$ always belongs to H because by definition of group, the composition of two elements of a group h and a belongs always to the same group H right?
$endgroup$
– homunculus
Mar 16 at 8:46
$begingroup$
Correct, in general for a group $(G,circ)$ and elements $x,yin G$, $xcirc yin G$ to satisfy the closure axiom. Furthermore, $x^{-1}$ exists to satisfy the inverse axiom.
$endgroup$
– TheSimpliFire
Mar 16 at 8:48
$begingroup$
ok thanks very much
$endgroup$
– homunculus
Mar 16 at 8:50
add a comment |
$begingroup$
Let $ain G$. Then
begin{align}
Ha=H &iff Ha=He \
&iff ae^{-1}in H \
&iff ain H.
end{align}
This is due to the more general theorem that $Hx=Hy$ if and only if $xy^{-1}in H$.
answered Mar 16 at 8:22
ShaunShaun
9,804113684
$endgroup$
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%2f3150187%2fshowing-ha-h-if-and-only-if-a-belongs-to-h%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $ain H$, then every element of $H$ can be written in the form $(ha^{-1})ain Ha$ and due to closure, $Ha=H$. Contrarily, if $Ha=H$ then $a=eain Ha=H$.
answered Mar 16 at 8:28
TheSimpliFireTheSimpliFire
12.9k62462
$endgroup$
$begingroup$
$ha^{-1}$ always belongs to H because by definition of group, the composition of two elements of a group h and a belongs always to the same group H right?
$endgroup$
– homunculus
Mar 16 at 8:46
$begingroup$
Correct, in general for a group $(G,circ)$ and elements $x,yin G$, $xcirc yin G$ to satisfy the closure axiom. Furthermore, $x^{-1}$ exists to satisfy the inverse axiom.
$endgroup$
– TheSimpliFire
Mar 16 at 8:48
$begingroup$
ok thanks very much
$endgroup$
– homunculus
Mar 16 at 8:50
add a comment |
$begingroup$
If $ain H$, then every element of $H$ can be written in the form $(ha^{-1})ain Ha$ and due to closure, $Ha=H$. Contrarily, if $Ha=H$ then $a=eain Ha=H$.
answered Mar 16 at 8:28
TheSimpliFireTheSimpliFire
12.9k62462
$endgroup$
$begingroup$
$ha^{-1}$ always belongs to H because by definition of group, the composition of two elements of a group h and a belongs always to the same group H right?
$endgroup$
– homunculus
Mar 16 at 8:46
$begingroup$
Correct, in general for a group $(G,circ)$ and elements $x,yin G$, $xcirc yin G$ to satisfy the closure axiom. Furthermore, $x^{-1}$ exists to satisfy the inverse axiom.
$endgroup$
– TheSimpliFire
Mar 16 at 8:48
$begingroup$
ok thanks very much
$endgroup$
– homunculus
Mar 16 at 8:50
add a comment |
$begingroup$
If $ain H$, then every element of $H$ can be written in the form $(ha^{-1})ain Ha$ and due to closure, $Ha=H$. Contrarily, if $Ha=H$ then $a=eain Ha=H$.
answered Mar 16 at 8:28
TheSimpliFireTheSimpliFire
12.9k62462
$endgroup$
If $ain H$, then every element of $H$ can be written in the form $(ha^{-1})ain Ha$ and due to closure, $Ha=H$. Contrarily, if $Ha=H$ then $a=eain Ha=H$.
answered Mar 16 at 8:28
TheSimpliFireTheSimpliFire
12.9k62462
answered Mar 16 at 8:28
TheSimpliFireTheSimpliFire
12.9k62462
answered Mar 16 at 8:28
TheSimpliFireTheSimpliFire
12.9k62462
answered Mar 16 at 8:28
TheSimpliFireTheSimpliFire
12.9k62462
12.9k62462
$begingroup$
$ha^{-1}$ always belongs to H because by definition of group, the composition of two elements of a group h and a belongs always to the same group H right?
$endgroup$
– homunculus
Mar 16 at 8:46
$begingroup$
Correct, in general for a group $(G,circ)$ and elements $x,yin G$, $xcirc yin G$ to satisfy the closure axiom. Furthermore, $x^{-1}$ exists to satisfy the inverse axiom.
$endgroup$
– TheSimpliFire
Mar 16 at 8:48
$begingroup$
ok thanks very much
$endgroup$
– homunculus
Mar 16 at 8:50
add a comment |
$begingroup$
$ha^{-1}$ always belongs to H because by definition of group, the composition of two elements of a group h and a belongs always to the same group H right?
$endgroup$
– homunculus
Mar 16 at 8:46
$begingroup$
Correct, in general for a group $(G,circ)$ and elements $x,yin G$, $xcirc yin G$ to satisfy the closure axiom. Furthermore, $x^{-1}$ exists to satisfy the inverse axiom.
$endgroup$
– TheSimpliFire
Mar 16 at 8:48
$begingroup$
ok thanks very much
$endgroup$
– homunculus
Mar 16 at 8:50
$begingroup$
$ha^{-1}$ always belongs to H because by definition of group, the composition of two elements of a group h and a belongs always to the same group H right?
$endgroup$
– homunculus
Mar 16 at 8:46
$begingroup$
$ha^{-1}$ always belongs to H because by definition of group, the composition of two elements of a group h and a belongs always to the same group H right?
$endgroup$
– homunculus
Mar 16 at 8:46
$begingroup$
Correct, in general for a group $(G,circ)$ and elements $x,yin G$, $xcirc yin G$ to satisfy the closure axiom. Furthermore, $x^{-1}$ exists to satisfy the inverse axiom.
$endgroup$
– TheSimpliFire
Mar 16 at 8:48
$begingroup$
Correct, in general for a group $(G,circ)$ and elements $x,yin G$, $xcirc yin G$ to satisfy the closure axiom. Furthermore, $x^{-1}$ exists to satisfy the inverse axiom.
$endgroup$
– TheSimpliFire
Mar 16 at 8:48
$begingroup$
ok thanks very much
$endgroup$
– homunculus
Mar 16 at 8:50
$begingroup$
ok thanks very much
$endgroup$
– homunculus
Mar 16 at 8:50
add a comment |
$begingroup$
Let $ain G$. Then
begin{align}
Ha=H &iff Ha=He \
&iff ae^{-1}in H \
&iff ain H.
end{align}
This is due to the more general theorem that $Hx=Hy$ if and only if $xy^{-1}in H$.
answered Mar 16 at 8:22
ShaunShaun
9,804113684
$endgroup$
add a comment |
$begingroup$
Let $ain G$. Then
begin{align}
Ha=H &iff Ha=He \
&iff ae^{-1}in H \
&iff ain H.
end{align}
This is due to the more general theorem that $Hx=Hy$ if and only if $xy^{-1}in H$.
answered Mar 16 at 8:22
ShaunShaun
9,804113684
$endgroup$
add a comment |
$begingroup$
Let $ain G$. Then
begin{align}
Ha=H &iff Ha=He \
&iff ae^{-1}in H \
&iff ain H.
end{align}
This is due to the more general theorem that $Hx=Hy$ if and only if $xy^{-1}in H$.
answered Mar 16 at 8:22
ShaunShaun
9,804113684
$endgroup$
Let $ain G$. Then
begin{align}
Ha=H &iff Ha=He \
&iff ae^{-1}in H \
&iff ain H.
end{align}
This is due to the more general theorem that $Hx=Hy$ if and only if $xy^{-1}in H$.
answered Mar 16 at 8:22
ShaunShaun
9,804113684
edited Mar 16 at 8:32
answered Mar 16 at 8:22
ShaunShaun
9,804113684
answered Mar 16 at 8:22
ShaunShaun
9,804113684
answered Mar 16 at 8:22
ShaunShaun
9,804113684
9,804113684
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%2f3150187%2fshowing-ha-h-if-and-only-if-a-belongs-to-h%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
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Shaun
Mar 16 at 8:33
$begingroup$
not enough reputation i have
$endgroup$
– homunculus
Mar 16 at 8:34
$begingroup$
Try again, please. It was probably the 15 minute wait timer that stopped you, not lack of reputation.
$endgroup$
– Shaun
Mar 16 at 8:54
$begingroup$
no it is lack of Rep only 15
$endgroup$
– homunculus
Mar 16 at 8:55