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
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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 |
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1 Answer
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1 Answer
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$begingroup$
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 |
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$begingroup$
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