If $aH=bH implies Ha=Hb$ for a subgroup H having *finite index*, then $gH=Hg$ for all $g in G$?$aHa^{-1}...
How do I extrude a face to a single vertex
A social experiment. What is the worst that can happen?
Do Legal Documents Require Signing In Standard Pen Colors?
What major Native American tribes were around Santa Fe during the late 1850s?
How to align and center standalone amsmath equations?
What is the difference between "Do you interest" and "...interested in" something?
Should I install hardwood flooring or cabinets first?
Why does Async/Await work properly when the loop is inside the async function and not the other way around?
Is it improper etiquette to ask your opponent what his/her rating is before the game?
What's the difference between 違法 and 不法?
Flux received by a negative charge
Does the Mind Blank spell prevent the target from being frightened?
Can I use my Chinese passport to enter China after I acquired another citizenship?
Sampling Theorem and reconstruction
Should I stop contributing to retirement accounts?
Transformation of random variables and joint distributions
Will adding a BY-SA image to a blog post make the entire post BY-SA?
Diode in opposite direction?
How much character growth crosses the line into breaking the character
How do I repair my stair bannister?
Can I sign legal documents with a smiley face?
Does having a TSA Pre-Check member in your flight reservation increase the chances that everyone gets Pre-Check?
Remove Expired Scratch Orgs From VSCode
Is possible to search in vim history?
If $aH=bH implies Ha=Hb$ for a subgroup H having *finite index*, then $gH=Hg$ for all $g in G$?
$aHa^{-1} subset H$ without $aHa^{-1} = H$If $aH=bH$ implies $Ha=Hb$ then $aHa^{-1}=H$ for all $ain G$.what is exactly the index of a subgroup?$H$ has finite index $implies aH$ has finite indexIndex of a subgroup is finite iff two given indexes are finiteProve that if $H$ is a subgroup of index $2$ in a finite group $G$, then $gH = Hg forall g in G$.Prove $g^2 = e$ if there is a subgroup of index 2 that does not contain $g$ for every $g in G$.If $H$ is a normal subgroup of $G$, prove $g^{n} in H$Subgroups of finite index have finitely many conjugatesProblem with proof of $H cap K $ is of finite index if $ H,K$ are finite index subgroupsIf $H$ is subgroup $G$ of index 2 then $H$ is normal subgroup of $G$Subgroup of Finite Index
$begingroup$
Problem 2.5.9 of Herstein's Topics in Algebra asks us to prove that if $H$ is a subgroup of a group $G$ such that $Ha not = Hb implies aH not = bH$, then $gHg^{-1} subset H$ for all $g in G$, which is equivalent to $gH subset Hg.$ Suppose I have proved this (this result is of course true). I want to prove more, namely that, under the same premise, $gH = Hg$ for all $g in G$. I think I can prove it if I further assume that the index of $H$ in $G$ is finite. Is this true? And if so, is this additional hypothesis necessary as well as sufficient? I.e. can you give an example of a subgroup $H$ of infinite index such that $aH subset Ha$ but not vice versa? Thanks in advance. [DISCLAIMER: I am not yet familiar with normal subgroups]
Here's my attempt at a proof:
Suppose that $|G:H|=n in mathbb{N}$ and that $aH subset Ha$ for all $a in G$. Now, suppose that there is an element $x$ such that $x in Ha$ but $x not in aH$. Then $x$ must be contained in a different left coset, say $bH$, because the left cosets form a partition of $G$. Then, by our hypothesis, $x in Hb$, which implies $Hb = Ha$, because the right cosets form a partition of $G$ as well. So far we have shown that $aH, bH subset Ha=Hb$. But now we can prove that there are more left cosets than right ones! In fact, the remaining $n-1$ right cosets distinct from $Ha$ must contain at least one left coset, by our hypothesis, but $Ha$ contains two left cosets. So there are at least $n+1$ left cosets, a contradiction. Therefore $aH=Ha. square$
abstract-algebra group-theory normal-subgroups
asked Mar 14 at 11:43
The FootprintThe Footprint
877
$endgroup$
add a comment |
$begingroup$
Problem 2.5.9 of Herstein's Topics in Algebra asks us to prove that if $H$ is a subgroup of a group $G$ such that $Ha not = Hb implies aH not = bH$, then $gHg^{-1} subset H$ for all $g in G$, which is equivalent to $gH subset Hg.$ Suppose I have proved this (this result is of course true). I want to prove more, namely that, under the same premise, $gH = Hg$ for all $g in G$. I think I can prove it if I further assume that the index of $H$ in $G$ is finite. Is this true? And if so, is this additional hypothesis necessary as well as sufficient? I.e. can you give an example of a subgroup $H$ of infinite index such that $aH subset Ha$ but not vice versa? Thanks in advance. [DISCLAIMER: I am not yet familiar with normal subgroups]
Here's my attempt at a proof:
Suppose that $|G:H|=n in mathbb{N}$ and that $aH subset Ha$ for all $a in G$. Now, suppose that there is an element $x$ such that $x in Ha$ but $x not in aH$. Then $x$ must be contained in a different left coset, say $bH$, because the left cosets form a partition of $G$. Then, by our hypothesis, $x in Hb$, which implies $Hb = Ha$, because the right cosets form a partition of $G$ as well. So far we have shown that $aH, bH subset Ha=Hb$. But now we can prove that there are more left cosets than right ones! In fact, the remaining $n-1$ right cosets distinct from $Ha$ must contain at least one left coset, by our hypothesis, but $Ha$ contains two left cosets. So there are at least $n+1$ left cosets, a contradiction. Therefore $aH=Ha. square$
abstract-algebra group-theory normal-subgroups
asked Mar 14 at 11:43
The FootprintThe Footprint
877
$endgroup$
$begingroup$
Compare with this duplicate and its answers.
$endgroup$
– Dietrich Burde
Mar 14 at 12:00
2
$begingroup$
@DietrichBurde The answers in your linked question only prove $gHg^{-1} subset H$ and not equality. So I don't think they address the OP’s problem.
$endgroup$
– Claudius
Mar 14 at 12:15
$begingroup$
@Claudius This is not true. They prove equality, e.g., see Deven's answer.
$endgroup$
– Dietrich Burde
Mar 14 at 12:18
2
$begingroup$
@DietrichBurde Yes, Deven claims to have shown $gHg^{-1} =H$. But what he really did was to prove $gHg^{-1} subset H$. At least, I don't see how he proved the other inclusion.
$endgroup$
– Claudius
Mar 14 at 12:20
add a comment |
$begingroup$
Problem 2.5.9 of Herstein's Topics in Algebra asks us to prove that if $H$ is a subgroup of a group $G$ such that $Ha not = Hb implies aH not = bH$, then $gHg^{-1} subset H$ for all $g in G$, which is equivalent to $gH subset Hg.$ Suppose I have proved this (this result is of course true). I want to prove more, namely that, under the same premise, $gH = Hg$ for all $g in G$. I think I can prove it if I further assume that the index of $H$ in $G$ is finite. Is this true? And if so, is this additional hypothesis necessary as well as sufficient? I.e. can you give an example of a subgroup $H$ of infinite index such that $aH subset Ha$ but not vice versa? Thanks in advance. [DISCLAIMER: I am not yet familiar with normal subgroups]
Here's my attempt at a proof:
Suppose that $|G:H|=n in mathbb{N}$ and that $aH subset Ha$ for all $a in G$. Now, suppose that there is an element $x$ such that $x in Ha$ but $x not in aH$. Then $x$ must be contained in a different left coset, say $bH$, because the left cosets form a partition of $G$. Then, by our hypothesis, $x in Hb$, which implies $Hb = Ha$, because the right cosets form a partition of $G$ as well. So far we have shown that $aH, bH subset Ha=Hb$. But now we can prove that there are more left cosets than right ones! In fact, the remaining $n-1$ right cosets distinct from $Ha$ must contain at least one left coset, by our hypothesis, but $Ha$ contains two left cosets. So there are at least $n+1$ left cosets, a contradiction. Therefore $aH=Ha. square$
abstract-algebra group-theory normal-subgroups
asked Mar 14 at 11:43
The FootprintThe Footprint
877
$endgroup$
Problem 2.5.9 of Herstein's Topics in Algebra asks us to prove that if $H$ is a subgroup of a group $G$ such that $Ha not = Hb implies aH not = bH$, then $gHg^{-1} subset H$ for all $g in G$, which is equivalent to $gH subset Hg.$ Suppose I have proved this (this result is of course true). I want to prove more, namely that, under the same premise, $gH = Hg$ for all $g in G$. I think I can prove it if I further assume that the index of $H$ in $G$ is finite. Is this true? And if so, is this additional hypothesis necessary as well as sufficient? I.e. can you give an example of a subgroup $H$ of infinite index such that $aH subset Ha$ but not vice versa? Thanks in advance. [DISCLAIMER: I am not yet familiar with normal subgroups]
Here's my attempt at a proof:
Suppose that $|G:H|=n in mathbb{N}$ and that $aH subset Ha$ for all $a in G$. Now, suppose that there is an element $x$ such that $x in Ha$ but $x not in aH$. Then $x$ must be contained in a different left coset, say $bH$, because the left cosets form a partition of $G$. Then, by our hypothesis, $x in Hb$, which implies $Hb = Ha$, because the right cosets form a partition of $G$ as well. So far we have shown that $aH, bH subset Ha=Hb$. But now we can prove that there are more left cosets than right ones! In fact, the remaining $n-1$ right cosets distinct from $Ha$ must contain at least one left coset, by our hypothesis, but $Ha$ contains two left cosets. So there are at least $n+1$ left cosets, a contradiction. Therefore $aH=Ha. square$
abstract-algebra group-theory normal-subgroups
abstract-algebra group-theory normal-subgroups
asked Mar 14 at 11:43
The FootprintThe Footprint
877
asked Mar 14 at 11:43
The FootprintThe Footprint
877
asked Mar 14 at 11:43
The FootprintThe Footprint
877
asked Mar 14 at 11:43
The FootprintThe Footprint
877
asked Mar 14 at 11:43
The FootprintThe Footprint
877
877
$begingroup$
Compare with this duplicate and its answers.
$endgroup$
– Dietrich Burde
Mar 14 at 12:00
2
$begingroup$
@DietrichBurde The answers in your linked question only prove $gHg^{-1} subset H$ and not equality. So I don't think they address the OP’s problem.
$endgroup$
– Claudius
Mar 14 at 12:15
$begingroup$
@Claudius This is not true. They prove equality, e.g., see Deven's answer.
$endgroup$
– Dietrich Burde
Mar 14 at 12:18
2
$begingroup$
@DietrichBurde Yes, Deven claims to have shown $gHg^{-1} =H$. But what he really did was to prove $gHg^{-1} subset H$. At least, I don't see how he proved the other inclusion.
$endgroup$
– Claudius
Mar 14 at 12:20
add a comment |
$begingroup$
Compare with this duplicate and its answers.
$endgroup$
– Dietrich Burde
Mar 14 at 12:00
2
$begingroup$
@DietrichBurde The answers in your linked question only prove $gHg^{-1} subset H$ and not equality. So I don't think they address the OP’s problem.
$endgroup$
– Claudius
Mar 14 at 12:15
$begingroup$
@Claudius This is not true. They prove equality, e.g., see Deven's answer.
$endgroup$
– Dietrich Burde
Mar 14 at 12:18
2
$begingroup$
@DietrichBurde Yes, Deven claims to have shown $gHg^{-1} =H$. But what he really did was to prove $gHg^{-1} subset H$. At least, I don't see how he proved the other inclusion.
$endgroup$
– Claudius
Mar 14 at 12:20
$begingroup$
Compare with this duplicate and its answers.
$endgroup$
– Dietrich Burde
Mar 14 at 12:00
$begingroup$
Compare with this duplicate and its answers.
$endgroup$
– Dietrich Burde
Mar 14 at 12:00
2
2
$begingroup$
@DietrichBurde The answers in your linked question only prove $gHg^{-1} subset H$ and not equality. So I don't think they address the OP’s problem.
$endgroup$
– Claudius
Mar 14 at 12:15
$begingroup$
@DietrichBurde The answers in your linked question only prove $gHg^{-1} subset H$ and not equality. So I don't think they address the OP’s problem.
$endgroup$
– Claudius
Mar 14 at 12:15
$begingroup$
@Claudius This is not true. They prove equality, e.g., see Deven's answer.
$endgroup$
– Dietrich Burde
Mar 14 at 12:18
$begingroup$
@Claudius This is not true. They prove equality, e.g., see Deven's answer.
$endgroup$
– Dietrich Burde
Mar 14 at 12:18
2
2
$begingroup$
@DietrichBurde Yes, Deven claims to have shown $gHg^{-1} =H$. But what he really did was to prove $gHg^{-1} subset H$. At least, I don't see how he proved the other inclusion.
$endgroup$
– Claudius
Mar 14 at 12:20
$begingroup$
@DietrichBurde Yes, Deven claims to have shown $gHg^{-1} =H$. But what he really did was to prove $gHg^{-1} subset H$. At least, I don't see how he proved the other inclusion.
$endgroup$
– Claudius
Mar 14 at 12:20
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your proof seems correct to me.
In fact, you can remove the finite index hypothesis. If $gHg^{-1} subset H$ for all $gin G$, then for each $gin G$ you also have
$$
H = g^{-1}(gHg^{-1})g subset g^{-1}Hg = g^{-1}H(g^{-1})^{-1} subset H,
$$
so we must have equality throughout, i. e. $H = g^{-1}Hg$ (for each $gin G$).
More generally, for any subgroup $H$ of $G$ the following holds: the set $N:= { gin G mid gHg^{-1}subset H}$ is a subgroup of $G$ if and only if $gHg^{-1} = H$ for all $gin N$. (In that case $N$ is the normalizer of $H$ in $G$.)
answered Mar 14 at 12:08
ClaudiusClaudius
3,9041616
$endgroup$
$begingroup$
That's brilliant, thank you. Now I feel stupid for not having come up with that immediately! But why would Herstein keep this result from the reader and only ask to prove one inclusion? And what about this thread math.stackexchange.com/questions/217601/… in which there seems to be a counterexample?
$endgroup$
– The Footprint
Mar 14 at 12:36
1
$begingroup$
In the answer of Brian M. Scott the set ${gin Gmid gHg^{-1} subset H}$ is not stable under inversion, hence not a group. It is only closed under composition.
$endgroup$
– Claudius
Mar 14 at 12:40
add a comment |
$begingroup$
You don't need any additional hypotheses. Multiplying both sides by $g^{-1}$ on the left and $g$ on the right yields
$$gHg^{-1} subset H implies H subset g^{-1}Hg$$
Since this is true for all $g in G$, we can substitute $g$ for $g^{-1}$, concluding
$$gHg^{-1} subset H subset gHg^{-1}$$
$$H = gHg^{-1}$$
This is equivalent to
$$gH = Hg$$
answered Mar 14 at 12:21
Alex G.Alex G.
6,2581128
$endgroup$
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%2f3147904%2fif-ah-bh-implies-ha-hb-for-a-subgroup-h-having-finite-index-then-gh-hg-f%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
$begingroup$
Your proof seems correct to me.
In fact, you can remove the finite index hypothesis. If $gHg^{-1} subset H$ for all $gin G$, then for each $gin G$ you also have
$$
H = g^{-1}(gHg^{-1})g subset g^{-1}Hg = g^{-1}H(g^{-1})^{-1} subset H,
$$
so we must have equality throughout, i. e. $H = g^{-1}Hg$ (for each $gin G$).
More generally, for any subgroup $H$ of $G$ the following holds: the set $N:= { gin G mid gHg^{-1}subset H}$ is a subgroup of $G$ if and only if $gHg^{-1} = H$ for all $gin N$. (In that case $N$ is the normalizer of $H$ in $G$.)
answered Mar 14 at 12:08
ClaudiusClaudius
3,9041616
$endgroup$
$begingroup$
That's brilliant, thank you. Now I feel stupid for not having come up with that immediately! But why would Herstein keep this result from the reader and only ask to prove one inclusion? And what about this thread math.stackexchange.com/questions/217601/… in which there seems to be a counterexample?
$endgroup$
– The Footprint
Mar 14 at 12:36
1
$begingroup$
In the answer of Brian M. Scott the set ${gin Gmid gHg^{-1} subset H}$ is not stable under inversion, hence not a group. It is only closed under composition.
$endgroup$
– Claudius
Mar 14 at 12:40
add a comment |
$begingroup$
Your proof seems correct to me.
In fact, you can remove the finite index hypothesis. If $gHg^{-1} subset H$ for all $gin G$, then for each $gin G$ you also have
$$
H = g^{-1}(gHg^{-1})g subset g^{-1}Hg = g^{-1}H(g^{-1})^{-1} subset H,
$$
so we must have equality throughout, i. e. $H = g^{-1}Hg$ (for each $gin G$).
More generally, for any subgroup $H$ of $G$ the following holds: the set $N:= { gin G mid gHg^{-1}subset H}$ is a subgroup of $G$ if and only if $gHg^{-1} = H$ for all $gin N$. (In that case $N$ is the normalizer of $H$ in $G$.)
answered Mar 14 at 12:08
ClaudiusClaudius
3,9041616
$endgroup$
$begingroup$
That's brilliant, thank you. Now I feel stupid for not having come up with that immediately! But why would Herstein keep this result from the reader and only ask to prove one inclusion? And what about this thread math.stackexchange.com/questions/217601/… in which there seems to be a counterexample?
$endgroup$
– The Footprint
Mar 14 at 12:36
1
$begingroup$
In the answer of Brian M. Scott the set ${gin Gmid gHg^{-1} subset H}$ is not stable under inversion, hence not a group. It is only closed under composition.
$endgroup$
– Claudius
Mar 14 at 12:40
add a comment |
$begingroup$
Your proof seems correct to me.
In fact, you can remove the finite index hypothesis. If $gHg^{-1} subset H$ for all $gin G$, then for each $gin G$ you also have
$$
H = g^{-1}(gHg^{-1})g subset g^{-1}Hg = g^{-1}H(g^{-1})^{-1} subset H,
$$
so we must have equality throughout, i. e. $H = g^{-1}Hg$ (for each $gin G$).
More generally, for any subgroup $H$ of $G$ the following holds: the set $N:= { gin G mid gHg^{-1}subset H}$ is a subgroup of $G$ if and only if $gHg^{-1} = H$ for all $gin N$. (In that case $N$ is the normalizer of $H$ in $G$.)
answered Mar 14 at 12:08
ClaudiusClaudius
3,9041616
$endgroup$
Your proof seems correct to me.
In fact, you can remove the finite index hypothesis. If $gHg^{-1} subset H$ for all $gin G$, then for each $gin G$ you also have
$$
H = g^{-1}(gHg^{-1})g subset g^{-1}Hg = g^{-1}H(g^{-1})^{-1} subset H,
$$
so we must have equality throughout, i. e. $H = g^{-1}Hg$ (for each $gin G$).
More generally, for any subgroup $H$ of $G$ the following holds: the set $N:= { gin G mid gHg^{-1}subset H}$ is a subgroup of $G$ if and only if $gHg^{-1} = H$ for all $gin N$. (In that case $N$ is the normalizer of $H$ in $G$.)
answered Mar 14 at 12:08
ClaudiusClaudius
3,9041616
edited Mar 14 at 12:24
answered Mar 14 at 12:08
ClaudiusClaudius
3,9041616
answered Mar 14 at 12:08
ClaudiusClaudius
3,9041616
answered Mar 14 at 12:08
ClaudiusClaudius
3,9041616
3,9041616
$begingroup$
That's brilliant, thank you. Now I feel stupid for not having come up with that immediately! But why would Herstein keep this result from the reader and only ask to prove one inclusion? And what about this thread math.stackexchange.com/questions/217601/… in which there seems to be a counterexample?
$endgroup$
– The Footprint
Mar 14 at 12:36
1
$begingroup$
In the answer of Brian M. Scott the set ${gin Gmid gHg^{-1} subset H}$ is not stable under inversion, hence not a group. It is only closed under composition.
$endgroup$
– Claudius
Mar 14 at 12:40
add a comment |
$begingroup$
That's brilliant, thank you. Now I feel stupid for not having come up with that immediately! But why would Herstein keep this result from the reader and only ask to prove one inclusion? And what about this thread math.stackexchange.com/questions/217601/… in which there seems to be a counterexample?
$endgroup$
– The Footprint
Mar 14 at 12:36
1
$begingroup$
In the answer of Brian M. Scott the set ${gin Gmid gHg^{-1} subset H}$ is not stable under inversion, hence not a group. It is only closed under composition.
$endgroup$
– Claudius
Mar 14 at 12:40
$begingroup$
That's brilliant, thank you. Now I feel stupid for not having come up with that immediately! But why would Herstein keep this result from the reader and only ask to prove one inclusion? And what about this thread math.stackexchange.com/questions/217601/… in which there seems to be a counterexample?
$endgroup$
– The Footprint
Mar 14 at 12:36
$begingroup$
That's brilliant, thank you. Now I feel stupid for not having come up with that immediately! But why would Herstein keep this result from the reader and only ask to prove one inclusion? And what about this thread math.stackexchange.com/questions/217601/… in which there seems to be a counterexample?
$endgroup$
– The Footprint
Mar 14 at 12:36
1
1
$begingroup$
In the answer of Brian M. Scott the set ${gin Gmid gHg^{-1} subset H}$ is not stable under inversion, hence not a group. It is only closed under composition.
$endgroup$
– Claudius
Mar 14 at 12:40
$begingroup$
In the answer of Brian M. Scott the set ${gin Gmid gHg^{-1} subset H}$ is not stable under inversion, hence not a group. It is only closed under composition.
$endgroup$
– Claudius
Mar 14 at 12:40
add a comment |
$begingroup$
You don't need any additional hypotheses. Multiplying both sides by $g^{-1}$ on the left and $g$ on the right yields
$$gHg^{-1} subset H implies H subset g^{-1}Hg$$
Since this is true for all $g in G$, we can substitute $g$ for $g^{-1}$, concluding
$$gHg^{-1} subset H subset gHg^{-1}$$
$$H = gHg^{-1}$$
This is equivalent to
$$gH = Hg$$
answered Mar 14 at 12:21
Alex G.Alex G.
6,2581128
$endgroup$
add a comment |
$begingroup$
You don't need any additional hypotheses. Multiplying both sides by $g^{-1}$ on the left and $g$ on the right yields
$$gHg^{-1} subset H implies H subset g^{-1}Hg$$
Since this is true for all $g in G$, we can substitute $g$ for $g^{-1}$, concluding
$$gHg^{-1} subset H subset gHg^{-1}$$
$$H = gHg^{-1}$$
This is equivalent to
$$gH = Hg$$
answered Mar 14 at 12:21
Alex G.Alex G.
6,2581128
$endgroup$
add a comment |
$begingroup$
You don't need any additional hypotheses. Multiplying both sides by $g^{-1}$ on the left and $g$ on the right yields
$$gHg^{-1} subset H implies H subset g^{-1}Hg$$
Since this is true for all $g in G$, we can substitute $g$ for $g^{-1}$, concluding
$$gHg^{-1} subset H subset gHg^{-1}$$
$$H = gHg^{-1}$$
This is equivalent to
$$gH = Hg$$
answered Mar 14 at 12:21
Alex G.Alex G.
6,2581128
$endgroup$
You don't need any additional hypotheses. Multiplying both sides by $g^{-1}$ on the left and $g$ on the right yields
$$gHg^{-1} subset H implies H subset g^{-1}Hg$$
Since this is true for all $g in G$, we can substitute $g$ for $g^{-1}$, concluding
$$gHg^{-1} subset H subset gHg^{-1}$$
$$H = gHg^{-1}$$
This is equivalent to
$$gH = Hg$$
answered Mar 14 at 12:21
Alex G.Alex G.
6,2581128
answered Mar 14 at 12:21
Alex G.Alex G.
6,2581128
answered Mar 14 at 12:21
Alex G.Alex G.
6,2581128
answered Mar 14 at 12:21
Alex G.Alex G.
6,2581128
6,2581128
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%2f3147904%2fif-ah-bh-implies-ha-hb-for-a-subgroup-h-having-finite-index-then-gh-hg-f%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
$begingroup$
Compare with this duplicate and its answers.
$endgroup$
– Dietrich Burde
Mar 14 at 12:00
2
$begingroup$
@DietrichBurde The answers in your linked question only prove $gHg^{-1} subset H$ and not equality. So I don't think they address the OP’s problem.
$endgroup$
– Claudius
Mar 14 at 12:15
$begingroup$
@Claudius This is not true. They prove equality, e.g., see Deven's answer.
$endgroup$
– Dietrich Burde
Mar 14 at 12:18
2
$begingroup$
@DietrichBurde Yes, Deven claims to have shown $gHg^{-1} =H$. But what he really did was to prove $gHg^{-1} subset H$. At least, I don't see how he proved the other inclusion.
$endgroup$
– Claudius
Mar 14 at 12:20