Show that $c(S) = ain G : ain gSg^-1 forall gin G$ is normal in $G$ The Next CEO of Stack OverflowIntersection of Normal Subgroups is Normal in Subgroup but Not Group - Fraleigh p. 143 14.35$H$ is a normal subgroup of order $2$.Prove that the commutator subgroups of $G$ are normal in $G$If $Sleq G$, prove that $Sunlhd G iff gamma (S) leq S$ for every conjugation $gamma$Show that $langle Srangle$ is a normal subgroup of $G$.Show that the intersection of two normal subgroup of $G$ is normal subgroup of $G$.If$G$ is a group and $S subset G$, then $langle S rangle$ is a normal subgroup iff $forall g in G$, $forall s in S$ $gsg^-1 in H$Show that in a nilpotent group a subgroup of index $p^2$ is normal.Minimal normal subgroup?Basic property of normal subgroup 2
Can Sri Krishna be called 'a person'?
Calculate the Mean mean of two numbers
Incomplete cube
Why can't we say "I have been having a dog"?
How can a day be of 24 hours?
Horror film about a man brought out of cryogenic suspension without a soul, around 1990
Is a distribution that is normal, but highly skewed, considered Gaussian?
Is it a bad idea to plug the other end of ESD strap to wall ground?
Why doesn't Shulchan Aruch include the laws of destroying fruit trees?
My boss doesn't want me to have a side project
Does Germany produce more waste than the US?
Avoiding the "not like other girls" trope?
How do I keep Mac Emacs from trapping M-`?
Find the majority element, which appears more than half the time
Simplify trigonometric expression using trigonometric identities
How dangerous is XSS
Planeswalker Ability and Death Timing
How to implement Comparable so it is consistent with identity-equality
Masking layers by a vector polygon layer in QGIS
Why does freezing point matter when picking cooler ice packs?
Is it possible to make a 9x9 table fit within the default margins?
Can a person "agarrar" something? ¿Puede una persona "agarrar" algo?
Gauss' Posthumous Publications?
What difference does it make matching a word with/without a trailing whitespace?
Show that $c(S) = ain G : ain gSg^-1 forall gin G$ is normal in $G$
The Next CEO of Stack OverflowIntersection of Normal Subgroups is Normal in Subgroup but Not Group - Fraleigh p. 143 14.35$H$ is a normal subgroup of order $2$.Prove that the commutator subgroups of $G$ are normal in $G$If $Sleq G$, prove that $Sunlhd G iff gamma (S) leq S$ for every conjugation $gamma$Show that $langle Srangle$ is a normal subgroup of $G$.Show that the intersection of two normal subgroup of $G$ is normal subgroup of $G$.If$G$ is a group and $S subset G$, then $langle S rangle$ is a normal subgroup iff $forall g in G$, $forall s in S$ $gsg^-1 in H$Show that in a nilpotent group a subgroup of index $p^2$ is normal.Minimal normal subgroup?Basic property of normal subgroup 2
$begingroup$
Question:
Let $G$ be a group, $S leq G$, and define
$$c(S) := ain G : a in gSg^-1 textrmfor all gin G$$
Show that $c(S)$ is a normal subgroup of $G$ contained in $S$
Attempt:
I have been able to show that $c(S)$ is a subgroup of $G$:
Let $a$, $b$ $in c(S)$. Then we have that
$$
ain gSg^-1 implies a = gs_1g^-1 textrmfor some s_1in S \
bin gSg^-1 implies b = gs_2g^-1 textrmfor some s_2in S
$$
Proceeding with the subgroup test:
beginalign
ab^-1 &= left(g s_1 g^-1right)left(g s_2 g^-1right)^-1 \
&= g s_1 g^-1 g s_2^-1 g^-1 \
&= gs_1s_2^-1g^-1 \
&in gSg^-1
endalign
So $c(S)leq G$.
I'm struggling in particular with showing that $c(S)$ is normal in $G$. I have so far attempted to show that $c(S)$ is closed under conjugation by elements from $G$:
Let $bin c(S)$. If $gbg^-1in c(S)$ for all $gin G$, then $c(S)triangleleft G$:
beginalign
bin c(S) &implies b = gsg^-1 textrmfor some sin Stextrm, for all gin G. \
&implies gbg^-1 = ggsg^-1g^-1 = (gg)s(gg)^-1
endalign
I'm not sure how to proceed from this point. I am also confused at the definition of $c(S)$ itself, I definitely do not have any intuition as to what the object $c(S)$ actually is.
Edit (In response to Chrystomath's comment):
Let $bin c(S)$. Then $b = lhl^-1$ for fixed $lin G$, $hin S$.
Let $k$ be any element of $G$:
beginaligned
kbk^-1 &= k(lhl^-1)k^-1 \
&= klhl^-1k^-1 \
&= (kl)h(kl)^-1 \
&in gSg^-1
endaligned
Writing $g=klin G$.
My questions are:
- Was I correct in the way I showed that $c(S)leq G$?
- How can I interpret the definition of $c(S)$?
- How do I show that $c(S)$ is normal in $G$?
Thank you
group-theory normal-subgroups
asked Mar 20 at 8:00
AndrewAndrew
284
$endgroup$
add a comment |
$begingroup$
Question:
Let $G$ be a group, $S leq G$, and define
$$c(S) := ain G : a in gSg^-1 textrmfor all gin G$$
Show that $c(S)$ is a normal subgroup of $G$ contained in $S$
Attempt:
I have been able to show that $c(S)$ is a subgroup of $G$:
Let $a$, $b$ $in c(S)$. Then we have that
$$
ain gSg^-1 implies a = gs_1g^-1 textrmfor some s_1in S \
bin gSg^-1 implies b = gs_2g^-1 textrmfor some s_2in S
$$
Proceeding with the subgroup test:
beginalign
ab^-1 &= left(g s_1 g^-1right)left(g s_2 g^-1right)^-1 \
&= g s_1 g^-1 g s_2^-1 g^-1 \
&= gs_1s_2^-1g^-1 \
&in gSg^-1
endalign
So $c(S)leq G$.
I'm struggling in particular with showing that $c(S)$ is normal in $G$. I have so far attempted to show that $c(S)$ is closed under conjugation by elements from $G$:
Let $bin c(S)$. If $gbg^-1in c(S)$ for all $gin G$, then $c(S)triangleleft G$:
beginalign
bin c(S) &implies b = gsg^-1 textrmfor some sin Stextrm, for all gin G. \
&implies gbg^-1 = ggsg^-1g^-1 = (gg)s(gg)^-1
endalign
I'm not sure how to proceed from this point. I am also confused at the definition of $c(S)$ itself, I definitely do not have any intuition as to what the object $c(S)$ actually is.
Edit (In response to Chrystomath's comment):
Let $bin c(S)$. Then $b = lhl^-1$ for fixed $lin G$, $hin S$.
Let $k$ be any element of $G$:
beginaligned
kbk^-1 &= k(lhl^-1)k^-1 \
&= klhl^-1k^-1 \
&= (kl)h(kl)^-1 \
&in gSg^-1
endaligned
Writing $g=klin G$.
My questions are:
- Was I correct in the way I showed that $c(S)leq G$?
- How can I interpret the definition of $c(S)$?
- How do I show that $c(S)$ is normal in $G$?
Thank you
group-theory normal-subgroups
asked Mar 20 at 8:00
AndrewAndrew
284
$endgroup$
2
$begingroup$
When you state "$b=gsg^-1$ for some $s$ for all $g$", you have to be careful. It is not the same $s$ for all $g$ of course. For each $g$ there is an $s$ with that property. Secondly you have to prove $gbg^-1in c(S)$ for all $g$, so don't use the same letter $g$ as in $b=gsg^-1$. It can be any other element, so call it $h$ say.
$endgroup$
– Chrystomath
Mar 20 at 8:13
$begingroup$
@Chrystomath I have edited my post with these changes
$endgroup$
– Andrew
Mar 20 at 8:52
add a comment |
$begingroup$
Question:
Let $G$ be a group, $S leq G$, and define
$$c(S) := ain G : a in gSg^-1 textrmfor all gin G$$
Show that $c(S)$ is a normal subgroup of $G$ contained in $S$
Attempt:
I have been able to show that $c(S)$ is a subgroup of $G$:
Let $a$, $b$ $in c(S)$. Then we have that
$$
ain gSg^-1 implies a = gs_1g^-1 textrmfor some s_1in S \
bin gSg^-1 implies b = gs_2g^-1 textrmfor some s_2in S
$$
Proceeding with the subgroup test:
beginalign
ab^-1 &= left(g s_1 g^-1right)left(g s_2 g^-1right)^-1 \
&= g s_1 g^-1 g s_2^-1 g^-1 \
&= gs_1s_2^-1g^-1 \
&in gSg^-1
endalign
So $c(S)leq G$.
I'm struggling in particular with showing that $c(S)$ is normal in $G$. I have so far attempted to show that $c(S)$ is closed under conjugation by elements from $G$:
Let $bin c(S)$. If $gbg^-1in c(S)$ for all $gin G$, then $c(S)triangleleft G$:
beginalign
bin c(S) &implies b = gsg^-1 textrmfor some sin Stextrm, for all gin G. \
&implies gbg^-1 = ggsg^-1g^-1 = (gg)s(gg)^-1
endalign
I'm not sure how to proceed from this point. I am also confused at the definition of $c(S)$ itself, I definitely do not have any intuition as to what the object $c(S)$ actually is.
Edit (In response to Chrystomath's comment):
Let $bin c(S)$. Then $b = lhl^-1$ for fixed $lin G$, $hin S$.
Let $k$ be any element of $G$:
beginaligned
kbk^-1 &= k(lhl^-1)k^-1 \
&= klhl^-1k^-1 \
&= (kl)h(kl)^-1 \
&in gSg^-1
endaligned
Writing $g=klin G$.
My questions are:
- Was I correct in the way I showed that $c(S)leq G$?
- How can I interpret the definition of $c(S)$?
- How do I show that $c(S)$ is normal in $G$?
Thank you
group-theory normal-subgroups
asked Mar 20 at 8:00
AndrewAndrew
284
$endgroup$
Question:
Let $G$ be a group, $S leq G$, and define
$$c(S) := ain G : a in gSg^-1 textrmfor all gin G$$
Show that $c(S)$ is a normal subgroup of $G$ contained in $S$
Attempt:
I have been able to show that $c(S)$ is a subgroup of $G$:
Let $a$, $b$ $in c(S)$. Then we have that
$$
ain gSg^-1 implies a = gs_1g^-1 textrmfor some s_1in S \
bin gSg^-1 implies b = gs_2g^-1 textrmfor some s_2in S
$$
Proceeding with the subgroup test:
beginalign
ab^-1 &= left(g s_1 g^-1right)left(g s_2 g^-1right)^-1 \
&= g s_1 g^-1 g s_2^-1 g^-1 \
&= gs_1s_2^-1g^-1 \
&in gSg^-1
endalign
So $c(S)leq G$.
I'm struggling in particular with showing that $c(S)$ is normal in $G$. I have so far attempted to show that $c(S)$ is closed under conjugation by elements from $G$:
Let $bin c(S)$. If $gbg^-1in c(S)$ for all $gin G$, then $c(S)triangleleft G$:
beginalign
bin c(S) &implies b = gsg^-1 textrmfor some sin Stextrm, for all gin G. \
&implies gbg^-1 = ggsg^-1g^-1 = (gg)s(gg)^-1
endalign
I'm not sure how to proceed from this point. I am also confused at the definition of $c(S)$ itself, I definitely do not have any intuition as to what the object $c(S)$ actually is.
Edit (In response to Chrystomath's comment):
Let $bin c(S)$. Then $b = lhl^-1$ for fixed $lin G$, $hin S$.
Let $k$ be any element of $G$:
beginaligned
kbk^-1 &= k(lhl^-1)k^-1 \
&= klhl^-1k^-1 \
&= (kl)h(kl)^-1 \
&in gSg^-1
endaligned
Writing $g=klin G$.
My questions are:
- Was I correct in the way I showed that $c(S)leq G$?
- How can I interpret the definition of $c(S)$?
- How do I show that $c(S)$ is normal in $G$?
Thank you
group-theory normal-subgroups
group-theory normal-subgroups
asked Mar 20 at 8:00
AndrewAndrew
284
asked Mar 20 at 8:00
AndrewAndrew
284
edited Mar 20 at 8:56
Andrew
asked Mar 20 at 8:00
AndrewAndrew
284
asked Mar 20 at 8:00
AndrewAndrew
284
asked Mar 20 at 8:00
AndrewAndrew
284
284
2
$begingroup$
When you state "$b=gsg^-1$ for some $s$ for all $g$", you have to be careful. It is not the same $s$ for all $g$ of course. For each $g$ there is an $s$ with that property. Secondly you have to prove $gbg^-1in c(S)$ for all $g$, so don't use the same letter $g$ as in $b=gsg^-1$. It can be any other element, so call it $h$ say.
$endgroup$
– Chrystomath
Mar 20 at 8:13
$begingroup$
@Chrystomath I have edited my post with these changes
$endgroup$
– Andrew
Mar 20 at 8:52
add a comment |
2
$begingroup$
When you state "$b=gsg^-1$ for some $s$ for all $g$", you have to be careful. It is not the same $s$ for all $g$ of course. For each $g$ there is an $s$ with that property. Secondly you have to prove $gbg^-1in c(S)$ for all $g$, so don't use the same letter $g$ as in $b=gsg^-1$. It can be any other element, so call it $h$ say.
$endgroup$
– Chrystomath
Mar 20 at 8:13
$begingroup$
@Chrystomath I have edited my post with these changes
$endgroup$
– Andrew
Mar 20 at 8:52
2
2
$begingroup$
When you state "$b=gsg^-1$ for some $s$ for all $g$", you have to be careful. It is not the same $s$ for all $g$ of course. For each $g$ there is an $s$ with that property. Secondly you have to prove $gbg^-1in c(S)$ for all $g$, so don't use the same letter $g$ as in $b=gsg^-1$. It can be any other element, so call it $h$ say.
$endgroup$
– Chrystomath
Mar 20 at 8:13
$begingroup$
When you state "$b=gsg^-1$ for some $s$ for all $g$", you have to be careful. It is not the same $s$ for all $g$ of course. For each $g$ there is an $s$ with that property. Secondly you have to prove $gbg^-1in c(S)$ for all $g$, so don't use the same letter $g$ as in $b=gsg^-1$. It can be any other element, so call it $h$ say.
$endgroup$
– Chrystomath
Mar 20 at 8:13
$begingroup$
@Chrystomath I have edited my post with these changes
$endgroup$
– Andrew
Mar 20 at 8:52
$begingroup$
@Chrystomath I have edited my post with these changes
$endgroup$
– Andrew
Mar 20 at 8:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have done the subgroup test for $c(S)$.
Consider any element in $g^-1 c(S) g$, wlog let it be $h =g^-1 g' s g'^-1 g$, where $s in S$, according to the definition of $c(S)$.
But it follows by definition that $h in c(S)$, and $c(S) lhd G$.
However, it would seem that $S subset c(S)$, not the other way.
Have I missed something?
answered Mar 20 at 9:10
AminopterinAminopterin
1,089213
$endgroup$
1
$begingroup$
No, I don't think you missed anything.
$endgroup$
– Shaun
Mar 20 at 15:57
1
$begingroup$
In my original question I was confused at how to quantify each variable I was using to show that $c(S)triangleleft G$. After Chrystomath cleared up this confusion your answer makes sense to me, thank you.
$endgroup$
– Andrew
Mar 22 at 0:32
add a comment |
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%2f3155143%2fshow-that-cs-a-in-g-a-in-gsg-1-forall-g-in-g-is-normal-in-g%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
$begingroup$
You have done the subgroup test for $c(S)$.
Consider any element in $g^-1 c(S) g$, wlog let it be $h =g^-1 g' s g'^-1 g$, where $s in S$, according to the definition of $c(S)$.
But it follows by definition that $h in c(S)$, and $c(S) lhd G$.
However, it would seem that $S subset c(S)$, not the other way.
Have I missed something?
answered Mar 20 at 9:10
AminopterinAminopterin
1,089213
$endgroup$
1
$begingroup$
No, I don't think you missed anything.
$endgroup$
– Shaun
Mar 20 at 15:57
1
$begingroup$
In my original question I was confused at how to quantify each variable I was using to show that $c(S)triangleleft G$. After Chrystomath cleared up this confusion your answer makes sense to me, thank you.
$endgroup$
– Andrew
Mar 22 at 0:32
add a comment |
$begingroup$
You have done the subgroup test for $c(S)$.
Consider any element in $g^-1 c(S) g$, wlog let it be $h =g^-1 g' s g'^-1 g$, where $s in S$, according to the definition of $c(S)$.
But it follows by definition that $h in c(S)$, and $c(S) lhd G$.
However, it would seem that $S subset c(S)$, not the other way.
Have I missed something?
answered Mar 20 at 9:10
AminopterinAminopterin
1,089213
$endgroup$
1
$begingroup$
No, I don't think you missed anything.
$endgroup$
– Shaun
Mar 20 at 15:57
1
$begingroup$
In my original question I was confused at how to quantify each variable I was using to show that $c(S)triangleleft G$. After Chrystomath cleared up this confusion your answer makes sense to me, thank you.
$endgroup$
– Andrew
Mar 22 at 0:32
add a comment |
$begingroup$
You have done the subgroup test for $c(S)$.
Consider any element in $g^-1 c(S) g$, wlog let it be $h =g^-1 g' s g'^-1 g$, where $s in S$, according to the definition of $c(S)$.
But it follows by definition that $h in c(S)$, and $c(S) lhd G$.
However, it would seem that $S subset c(S)$, not the other way.
Have I missed something?
answered Mar 20 at 9:10
AminopterinAminopterin
1,089213
$endgroup$
You have done the subgroup test for $c(S)$.
Consider any element in $g^-1 c(S) g$, wlog let it be $h =g^-1 g' s g'^-1 g$, where $s in S$, according to the definition of $c(S)$.
But it follows by definition that $h in c(S)$, and $c(S) lhd G$.
However, it would seem that $S subset c(S)$, not the other way.
Have I missed something?
answered Mar 20 at 9:10
AminopterinAminopterin
1,089213
edited Mar 20 at 9:16
answered Mar 20 at 9:10
AminopterinAminopterin
1,089213
answered Mar 20 at 9:10
AminopterinAminopterin
1,089213
answered Mar 20 at 9:10
AminopterinAminopterin
1,089213
1,089213
1
$begingroup$
No, I don't think you missed anything.
$endgroup$
– Shaun
Mar 20 at 15:57
1
$begingroup$
In my original question I was confused at how to quantify each variable I was using to show that $c(S)triangleleft G$. After Chrystomath cleared up this confusion your answer makes sense to me, thank you.
$endgroup$
– Andrew
Mar 22 at 0:32
add a comment |
1
$begingroup$
No, I don't think you missed anything.
$endgroup$
– Shaun
Mar 20 at 15:57
1
$begingroup$
In my original question I was confused at how to quantify each variable I was using to show that $c(S)triangleleft G$. After Chrystomath cleared up this confusion your answer makes sense to me, thank you.
$endgroup$
– Andrew
Mar 22 at 0:32
1
1
$begingroup$
No, I don't think you missed anything.
$endgroup$
– Shaun
Mar 20 at 15:57
$begingroup$
No, I don't think you missed anything.
$endgroup$
– Shaun
Mar 20 at 15:57
1
1
$begingroup$
In my original question I was confused at how to quantify each variable I was using to show that $c(S)triangleleft G$. After Chrystomath cleared up this confusion your answer makes sense to me, thank you.
$endgroup$
– Andrew
Mar 22 at 0:32
$begingroup$
In my original question I was confused at how to quantify each variable I was using to show that $c(S)triangleleft G$. After Chrystomath cleared up this confusion your answer makes sense to me, thank you.
$endgroup$
– Andrew
Mar 22 at 0:32
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%2f3155143%2fshow-that-cs-a-in-g-a-in-gsg-1-forall-g-in-g-is-normal-in-g%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
2
$begingroup$
When you state "$b=gsg^-1$ for some $s$ for all $g$", you have to be careful. It is not the same $s$ for all $g$ of course. For each $g$ there is an $s$ with that property. Secondly you have to prove $gbg^-1in c(S)$ for all $g$, so don't use the same letter $g$ as in $b=gsg^-1$. It can be any other element, so call it $h$ say.
$endgroup$
– Chrystomath
Mar 20 at 8:13
$begingroup$
@Chrystomath I have edited my post with these changes
$endgroup$
– Andrew
Mar 20 at 8:52