Proving an Intersection of Subgroups is Normal
$begingroup$
Let $N$ be a subgroup of the group $G$.
Show that $$X =bigcap_{g in G} g^{-1}Ng $$
is normal in $G$.
It is easy to show $X$ is a subgroup in itself, by just showing $g^{-1}Ng $ is a subgroup for arbitrary $g$ then the intersection of subgroups is a subgroup in itself.
But I'm struggling to show that it is normal, I have tried showing that $$g^{-1}xg in X quad forall g in G, x in X $$
and also directly from the definition $$Xg=gX quad forall g in G$$
Yet have had no luck. It woud be greatly appreciated if someone could give me a helping hand to see where I'm going wrong.
group-theory normal-subgroups
edited Dec 2 '18 at 17:19
ervx
10.3k31338
asked Dec 2 '18 at 17:07
B. ActonB. Acton
376
$endgroup$
add a comment |
$begingroup$
Let $N$ be a subgroup of the group $G$.
Show that $$X =bigcap_{g in G} g^{-1}Ng $$
is normal in $G$.
It is easy to show $X$ is a subgroup in itself, by just showing $g^{-1}Ng $ is a subgroup for arbitrary $g$ then the intersection of subgroups is a subgroup in itself.
But I'm struggling to show that it is normal, I have tried showing that $$g^{-1}xg in X quad forall g in G, x in X $$
and also directly from the definition $$Xg=gX quad forall g in G$$
Yet have had no luck. It woud be greatly appreciated if someone could give me a helping hand to see where I'm going wrong.
group-theory normal-subgroups
edited Dec 2 '18 at 17:19
ervx
10.3k31338
asked Dec 2 '18 at 17:07
B. ActonB. Acton
376
$endgroup$
add a comment |
$begingroup$
Let $N$ be a subgroup of the group $G$.
Show that $$X =bigcap_{g in G} g^{-1}Ng $$
is normal in $G$.
It is easy to show $X$ is a subgroup in itself, by just showing $g^{-1}Ng $ is a subgroup for arbitrary $g$ then the intersection of subgroups is a subgroup in itself.
But I'm struggling to show that it is normal, I have tried showing that $$g^{-1}xg in X quad forall g in G, x in X $$
and also directly from the definition $$Xg=gX quad forall g in G$$
Yet have had no luck. It woud be greatly appreciated if someone could give me a helping hand to see where I'm going wrong.
group-theory normal-subgroups
edited Dec 2 '18 at 17:19
ervx
10.3k31338
asked Dec 2 '18 at 17:07
B. ActonB. Acton
376
$endgroup$
Let $N$ be a subgroup of the group $G$.
Show that $$X =bigcap_{g in G} g^{-1}Ng $$
is normal in $G$.
It is easy to show $X$ is a subgroup in itself, by just showing $g^{-1}Ng $ is a subgroup for arbitrary $g$ then the intersection of subgroups is a subgroup in itself.
But I'm struggling to show that it is normal, I have tried showing that $$g^{-1}xg in X quad forall g in G, x in X $$
and also directly from the definition $$Xg=gX quad forall g in G$$
Yet have had no luck. It woud be greatly appreciated if someone could give me a helping hand to see where I'm going wrong.
group-theory normal-subgroups
group-theory normal-subgroups
edited Dec 2 '18 at 17:19
ervx
10.3k31338
asked Dec 2 '18 at 17:07
B. ActonB. Acton
376
edited Dec 2 '18 at 17:19
ervx
10.3k31338
asked Dec 2 '18 at 17:07
B. ActonB. Acton
376
edited Dec 2 '18 at 17:19
ervx
10.3k31338
edited Dec 2 '18 at 17:19
ervx
10.3k31338
edited Dec 2 '18 at 17:19
ervx
10.3k31338
10.3k31338
asked Dec 2 '18 at 17:07
B. ActonB. Acton
376
asked Dec 2 '18 at 17:07
B. ActonB. Acton
376
asked Dec 2 '18 at 17:07
B. ActonB. Acton
376
376
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Fix $xin X$ and $gin G$. We'll show that $g^{-1}xgin X$. Since $xin X$, it follows that $xin h^{-1}Nh$ for all $hin G$. Now, fix a particular $tin G$. Our goal is to show that $g^{-1}xgin t^{-1}Nt$. (This will prove that $g^{-1}xgin X$, since $t$ is arbitrary.) Note that $tg^{-1}in G$. Hence, by assumption, $xin (tg^{-1})^{-1}N(tg^{-1})=gt^{-1}Ntg^{-1}$. Thus, $g^{-1}xgin t^{-1}Nt$, as desired.
answered Dec 2 '18 at 17:16
ervxervx
10.3k31338
$endgroup$
$begingroup$
Thank you very much, I think I was sniffing around the right result, but messed up what I was keeping arbitrary/ fixed.
$endgroup$
– B. Acton
Dec 2 '18 at 17:30
$begingroup$
You are very welcome.
$endgroup$
– ervx
Dec 3 '18 at 0:51
add a comment |
$begingroup$
Taking your former approach: for any $x in X$ and any $gin G$, we need to show that $g^{-1}xg$ is in $h^{-1}Nh$ for all $h in H$. But, in particular, $x in gh^{-1}Nhg^{-1}$ (since $hg^{-1}in G$, and $(hg^{-1})^{-1} = gh^{-1}$), so $g^{-1}xgin g^{-1}gh^{-1}Nhg^{-1}g = h^{-1}Nh$, hence the result.
answered Dec 2 '18 at 17:19
user3482749user3482749
4,057918
$endgroup$
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%2f3022900%2fproving-an-intersection-of-subgroups-is-normal%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
$begingroup$
Fix $xin X$ and $gin G$. We'll show that $g^{-1}xgin X$. Since $xin X$, it follows that $xin h^{-1}Nh$ for all $hin G$. Now, fix a particular $tin G$. Our goal is to show that $g^{-1}xgin t^{-1}Nt$. (This will prove that $g^{-1}xgin X$, since $t$ is arbitrary.) Note that $tg^{-1}in G$. Hence, by assumption, $xin (tg^{-1})^{-1}N(tg^{-1})=gt^{-1}Ntg^{-1}$. Thus, $g^{-1}xgin t^{-1}Nt$, as desired.
answered Dec 2 '18 at 17:16
ervxervx
10.3k31338
$endgroup$
$begingroup$
Thank you very much, I think I was sniffing around the right result, but messed up what I was keeping arbitrary/ fixed.
$endgroup$
– B. Acton
Dec 2 '18 at 17:30
$begingroup$
You are very welcome.
$endgroup$
– ervx
Dec 3 '18 at 0:51
add a comment |
$begingroup$
Fix $xin X$ and $gin G$. We'll show that $g^{-1}xgin X$. Since $xin X$, it follows that $xin h^{-1}Nh$ for all $hin G$. Now, fix a particular $tin G$. Our goal is to show that $g^{-1}xgin t^{-1}Nt$. (This will prove that $g^{-1}xgin X$, since $t$ is arbitrary.) Note that $tg^{-1}in G$. Hence, by assumption, $xin (tg^{-1})^{-1}N(tg^{-1})=gt^{-1}Ntg^{-1}$. Thus, $g^{-1}xgin t^{-1}Nt$, as desired.
answered Dec 2 '18 at 17:16
ervxervx
10.3k31338
$endgroup$
$begingroup$
Thank you very much, I think I was sniffing around the right result, but messed up what I was keeping arbitrary/ fixed.
$endgroup$
– B. Acton
Dec 2 '18 at 17:30
$begingroup$
You are very welcome.
$endgroup$
– ervx
Dec 3 '18 at 0:51
add a comment |
$begingroup$
Fix $xin X$ and $gin G$. We'll show that $g^{-1}xgin X$. Since $xin X$, it follows that $xin h^{-1}Nh$ for all $hin G$. Now, fix a particular $tin G$. Our goal is to show that $g^{-1}xgin t^{-1}Nt$. (This will prove that $g^{-1}xgin X$, since $t$ is arbitrary.) Note that $tg^{-1}in G$. Hence, by assumption, $xin (tg^{-1})^{-1}N(tg^{-1})=gt^{-1}Ntg^{-1}$. Thus, $g^{-1}xgin t^{-1}Nt$, as desired.
answered Dec 2 '18 at 17:16
ervxervx
10.3k31338
$endgroup$
Fix $xin X$ and $gin G$. We'll show that $g^{-1}xgin X$. Since $xin X$, it follows that $xin h^{-1}Nh$ for all $hin G$. Now, fix a particular $tin G$. Our goal is to show that $g^{-1}xgin t^{-1}Nt$. (This will prove that $g^{-1}xgin X$, since $t$ is arbitrary.) Note that $tg^{-1}in G$. Hence, by assumption, $xin (tg^{-1})^{-1}N(tg^{-1})=gt^{-1}Ntg^{-1}$. Thus, $g^{-1}xgin t^{-1}Nt$, as desired.
answered Dec 2 '18 at 17:16
ervxervx
10.3k31338
answered Dec 2 '18 at 17:16
ervxervx
10.3k31338
answered Dec 2 '18 at 17:16
ervxervx
10.3k31338
answered Dec 2 '18 at 17:16
ervxervx
10.3k31338
10.3k31338
$begingroup$
Thank you very much, I think I was sniffing around the right result, but messed up what I was keeping arbitrary/ fixed.
$endgroup$
– B. Acton
Dec 2 '18 at 17:30
$begingroup$
You are very welcome.
$endgroup$
– ervx
Dec 3 '18 at 0:51
add a comment |
$begingroup$
Thank you very much, I think I was sniffing around the right result, but messed up what I was keeping arbitrary/ fixed.
$endgroup$
– B. Acton
Dec 2 '18 at 17:30
$begingroup$
You are very welcome.
$endgroup$
– ervx
Dec 3 '18 at 0:51
$begingroup$
Thank you very much, I think I was sniffing around the right result, but messed up what I was keeping arbitrary/ fixed.
$endgroup$
– B. Acton
Dec 2 '18 at 17:30
$begingroup$
Thank you very much, I think I was sniffing around the right result, but messed up what I was keeping arbitrary/ fixed.
$endgroup$
– B. Acton
Dec 2 '18 at 17:30
$begingroup$
You are very welcome.
$endgroup$
– ervx
Dec 3 '18 at 0:51
$begingroup$
You are very welcome.
$endgroup$
– ervx
Dec 3 '18 at 0:51
add a comment |
$begingroup$
Taking your former approach: for any $x in X$ and any $gin G$, we need to show that $g^{-1}xg$ is in $h^{-1}Nh$ for all $h in H$. But, in particular, $x in gh^{-1}Nhg^{-1}$ (since $hg^{-1}in G$, and $(hg^{-1})^{-1} = gh^{-1}$), so $g^{-1}xgin g^{-1}gh^{-1}Nhg^{-1}g = h^{-1}Nh$, hence the result.
answered Dec 2 '18 at 17:19
user3482749user3482749
4,057918
$endgroup$
add a comment |
$begingroup$
Taking your former approach: for any $x in X$ and any $gin G$, we need to show that $g^{-1}xg$ is in $h^{-1}Nh$ for all $h in H$. But, in particular, $x in gh^{-1}Nhg^{-1}$ (since $hg^{-1}in G$, and $(hg^{-1})^{-1} = gh^{-1}$), so $g^{-1}xgin g^{-1}gh^{-1}Nhg^{-1}g = h^{-1}Nh$, hence the result.
answered Dec 2 '18 at 17:19
user3482749user3482749
4,057918
$endgroup$
add a comment |
$begingroup$
Taking your former approach: for any $x in X$ and any $gin G$, we need to show that $g^{-1}xg$ is in $h^{-1}Nh$ for all $h in H$. But, in particular, $x in gh^{-1}Nhg^{-1}$ (since $hg^{-1}in G$, and $(hg^{-1})^{-1} = gh^{-1}$), so $g^{-1}xgin g^{-1}gh^{-1}Nhg^{-1}g = h^{-1}Nh$, hence the result.
answered Dec 2 '18 at 17:19
user3482749user3482749
4,057918
$endgroup$
Taking your former approach: for any $x in X$ and any $gin G$, we need to show that $g^{-1}xg$ is in $h^{-1}Nh$ for all $h in H$. But, in particular, $x in gh^{-1}Nhg^{-1}$ (since $hg^{-1}in G$, and $(hg^{-1})^{-1} = gh^{-1}$), so $g^{-1}xgin g^{-1}gh^{-1}Nhg^{-1}g = h^{-1}Nh$, hence the result.
answered Dec 2 '18 at 17:19
user3482749user3482749
4,057918
answered Dec 2 '18 at 17:19
user3482749user3482749
4,057918
answered Dec 2 '18 at 17:19
user3482749user3482749
4,057918
answered Dec 2 '18 at 17:19
user3482749user3482749
4,057918
4,057918
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%2f3022900%2fproving-an-intersection-of-subgroups-is-normal%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
