$Htrianglelefteq G$, $H$ intersects the commutator subgroup of $G$ trivially implies $H$ in center of $G$?
This question is related to this other post: I was wondering if proving that $ H trianglelefteq G$ and $Hcap G^{prime} ={e}$ (where $G^{prime}$ denotes the commutator subgroup of $G$) implies that the elements of $H$ commute with the elements of $G^{prime}$ is the same as proving that $H trianglelefteq G$ and $H cap G^{prime}={e}$ imply that $H subseteq C(G)$ (where $C(G)$ is the center of $G$)? If not, is there a way to modify/extend this proof to show that $H subseteq C(G)$? (It just needs to be a subset, not a subgroup).
If still not, how can I prove that$H trianglelefteq G$ and $H cap G^{prime}={e}$ imply that $Hsubseteq C(G)$?
Thank you.
abstract-algebra group-theory normal-subgroups
edited Apr 13 '17 at 12:21
Community♦
1
asked Jan 2 '17 at 5:25
ALannisterALannister
1,73731451
add a comment |
This question is related to this other post: I was wondering if proving that $ H trianglelefteq G$ and $Hcap G^{prime} ={e}$ (where $G^{prime}$ denotes the commutator subgroup of $G$) implies that the elements of $H$ commute with the elements of $G^{prime}$ is the same as proving that $H trianglelefteq G$ and $H cap G^{prime}={e}$ imply that $H subseteq C(G)$ (where $C(G)$ is the center of $G$)? If not, is there a way to modify/extend this proof to show that $H subseteq C(G)$? (It just needs to be a subset, not a subgroup).
If still not, how can I prove that$H trianglelefteq G$ and $H cap G^{prime}={e}$ imply that $Hsubseteq C(G)$?
Thank you.
abstract-algebra group-theory normal-subgroups
edited Apr 13 '17 at 12:21
Community♦
1
asked Jan 2 '17 at 5:25
ALannisterALannister
1,73731451
If $H$ is a subgroup of $G$ and a subset of the center $Z(G)$ which itself a subgroup of $G$, that automatically means $H$ would be a subgroup of $Z(G)$, so it's a bit jarring to see you say "it just needs to be a subset, not a subgroup." Also, "elements of $H$ commute with elements of $G',$" is equivalent to $Hsubseteq C(G')$ which is generally a weaker condition than $Hsubseteq Z(G)$, so more work will have to be done to show $H$ is central (if that does indeed follow from the hypotheses).
– arctic tern
Jan 2 '17 at 5:37
@arctictern When you say $C(G^{prime})$ you mean $Z(G^{prime})$ by your notation?
– ALannister
Jan 2 '17 at 5:46
No. $C(G')$ is short for $C_G(G')$ which means the set of all $gin G$ which commute with every element of $G'$. The notation $Z(G)$ means $C_G(G)$, the center, containing all $gin G$ which commute with every element of $G$. Thus, if someone wrote $Z(G')$, they would be talking about the elements of $G'$ which commute with every element of $G'$, and that is not a thing being discussed.
– arctic tern
Jan 2 '17 at 5:52
add a comment |
This question is related to this other post: I was wondering if proving that $ H trianglelefteq G$ and $Hcap G^{prime} ={e}$ (where $G^{prime}$ denotes the commutator subgroup of $G$) implies that the elements of $H$ commute with the elements of $G^{prime}$ is the same as proving that $H trianglelefteq G$ and $H cap G^{prime}={e}$ imply that $H subseteq C(G)$ (where $C(G)$ is the center of $G$)? If not, is there a way to modify/extend this proof to show that $H subseteq C(G)$? (It just needs to be a subset, not a subgroup).
If still not, how can I prove that$H trianglelefteq G$ and $H cap G^{prime}={e}$ imply that $Hsubseteq C(G)$?
Thank you.
abstract-algebra group-theory normal-subgroups
edited Apr 13 '17 at 12:21
Community♦
1
asked Jan 2 '17 at 5:25
ALannisterALannister
1,73731451
This question is related to this other post: I was wondering if proving that $ H trianglelefteq G$ and $Hcap G^{prime} ={e}$ (where $G^{prime}$ denotes the commutator subgroup of $G$) implies that the elements of $H$ commute with the elements of $G^{prime}$ is the same as proving that $H trianglelefteq G$ and $H cap G^{prime}={e}$ imply that $H subseteq C(G)$ (where $C(G)$ is the center of $G$)? If not, is there a way to modify/extend this proof to show that $H subseteq C(G)$? (It just needs to be a subset, not a subgroup).
If still not, how can I prove that$H trianglelefteq G$ and $H cap G^{prime}={e}$ imply that $Hsubseteq C(G)$?
Thank you.
abstract-algebra group-theory normal-subgroups
abstract-algebra group-theory normal-subgroups
edited Apr 13 '17 at 12:21
Community♦
1
asked Jan 2 '17 at 5:25
ALannisterALannister
1,73731451
edited Apr 13 '17 at 12:21
Community♦
1
asked Jan 2 '17 at 5:25
ALannisterALannister
1,73731451
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
asked Jan 2 '17 at 5:25
ALannisterALannister
1,73731451
asked Jan 2 '17 at 5:25
ALannisterALannister
1,73731451
asked Jan 2 '17 at 5:25
ALannisterALannister
1,73731451
1,73731451
If $H$ is a subgroup of $G$ and a subset of the center $Z(G)$ which itself a subgroup of $G$, that automatically means $H$ would be a subgroup of $Z(G)$, so it's a bit jarring to see you say "it just needs to be a subset, not a subgroup." Also, "elements of $H$ commute with elements of $G',$" is equivalent to $Hsubseteq C(G')$ which is generally a weaker condition than $Hsubseteq Z(G)$, so more work will have to be done to show $H$ is central (if that does indeed follow from the hypotheses).
– arctic tern
Jan 2 '17 at 5:37
@arctictern When you say $C(G^{prime})$ you mean $Z(G^{prime})$ by your notation?
– ALannister
Jan 2 '17 at 5:46
No. $C(G')$ is short for $C_G(G')$ which means the set of all $gin G$ which commute with every element of $G'$. The notation $Z(G)$ means $C_G(G)$, the center, containing all $gin G$ which commute with every element of $G$. Thus, if someone wrote $Z(G')$, they would be talking about the elements of $G'$ which commute with every element of $G'$, and that is not a thing being discussed.
– arctic tern
Jan 2 '17 at 5:52
add a comment |
If $H$ is a subgroup of $G$ and a subset of the center $Z(G)$ which itself a subgroup of $G$, that automatically means $H$ would be a subgroup of $Z(G)$, so it's a bit jarring to see you say "it just needs to be a subset, not a subgroup." Also, "elements of $H$ commute with elements of $G',$" is equivalent to $Hsubseteq C(G')$ which is generally a weaker condition than $Hsubseteq Z(G)$, so more work will have to be done to show $H$ is central (if that does indeed follow from the hypotheses).
– arctic tern
Jan 2 '17 at 5:37
@arctictern When you say $C(G^{prime})$ you mean $Z(G^{prime})$ by your notation?
– ALannister
Jan 2 '17 at 5:46
No. $C(G')$ is short for $C_G(G')$ which means the set of all $gin G$ which commute with every element of $G'$. The notation $Z(G)$ means $C_G(G)$, the center, containing all $gin G$ which commute with every element of $G$. Thus, if someone wrote $Z(G')$, they would be talking about the elements of $G'$ which commute with every element of $G'$, and that is not a thing being discussed.
– arctic tern
Jan 2 '17 at 5:52
If $H$ is a subgroup of $G$ and a subset of the center $Z(G)$ which itself a subgroup of $G$, that automatically means $H$ would be a subgroup of $Z(G)$, so it's a bit jarring to see you say "it just needs to be a subset, not a subgroup." Also, "elements of $H$ commute with elements of $G',$" is equivalent to $Hsubseteq C(G')$ which is generally a weaker condition than $Hsubseteq Z(G)$, so more work will have to be done to show $H$ is central (if that does indeed follow from the hypotheses).
– arctic tern
Jan 2 '17 at 5:37
If $H$ is a subgroup of $G$ and a subset of the center $Z(G)$ which itself a subgroup of $G$, that automatically means $H$ would be a subgroup of $Z(G)$, so it's a bit jarring to see you say "it just needs to be a subset, not a subgroup." Also, "elements of $H$ commute with elements of $G',$" is equivalent to $Hsubseteq C(G')$ which is generally a weaker condition than $Hsubseteq Z(G)$, so more work will have to be done to show $H$ is central (if that does indeed follow from the hypotheses).
– arctic tern
Jan 2 '17 at 5:37
@arctictern When you say $C(G^{prime})$ you mean $Z(G^{prime})$ by your notation?
– ALannister
Jan 2 '17 at 5:46
@arctictern When you say $C(G^{prime})$ you mean $Z(G^{prime})$ by your notation?
– ALannister
Jan 2 '17 at 5:46
No. $C(G')$ is short for $C_G(G')$ which means the set of all $gin G$ which commute with every element of $G'$. The notation $Z(G)$ means $C_G(G)$, the center, containing all $gin G$ which commute with every element of $G$. Thus, if someone wrote $Z(G')$, they would be talking about the elements of $G'$ which commute with every element of $G'$, and that is not a thing being discussed.
– arctic tern
Jan 2 '17 at 5:52
No. $C(G')$ is short for $C_G(G')$ which means the set of all $gin G$ which commute with every element of $G'$. The notation $Z(G)$ means $C_G(G)$, the center, containing all $gin G$ which commute with every element of $G$. Thus, if someone wrote $Z(G')$, they would be talking about the elements of $G'$ which commute with every element of $G'$, and that is not a thing being discussed.
– arctic tern
Jan 2 '17 at 5:52
add a comment |
1 Answer
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Suppose $hin H$ and $gin G$ are arbitrary, and consider the commutator $[g,h]=(ghg^{-1})h^{-1}$.
Using the fact $H$ is normal, can you deduce anything? (Hint: look at how I parenthesized it.)
answered Jan 2 '17 at 5:45
arctic ternarctic tern
11.8k31535
1
what you parenthesized is contained in $H$ since $H$ is normal? And so $[g,h]$ is contained in $H$?
– ALannister
Jan 2 '17 at 5:47
I made an edit to that comment.
– ALannister
Jan 2 '17 at 5:51
Yes. As $H$ is normal, $ghg^{-1}in H$, so $[g,h]=ghg^{-1}h^{-1}in H$. Now what can you deduce from here?
– arctic tern
Jan 2 '17 at 5:52
1
You already deduced $[g,h]$ is in $H$. Go further. Note that $[g,h]$ is a commutator. Is there anything in our set of hypotheses that talks about commutators and $H$?
– arctic tern
Jan 2 '17 at 5:57
1
Yes. If $[g,h]$ is in $G'$ (which it is since it's a commutator) and $[g,h]$ is in $H$, and the only thing in both $G'$ and $H$ is the identity, then $[g,h]$ is the identity. What does that say about $g$ and $h$?
– arctic tern
Jan 2 '17 at 6:00
|
show 8 more comments
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Suppose $hin H$ and $gin G$ are arbitrary, and consider the commutator $[g,h]=(ghg^{-1})h^{-1}$.
Using the fact $H$ is normal, can you deduce anything? (Hint: look at how I parenthesized it.)
answered Jan 2 '17 at 5:45
arctic ternarctic tern
11.8k31535
1
what you parenthesized is contained in $H$ since $H$ is normal? And so $[g,h]$ is contained in $H$?
– ALannister
Jan 2 '17 at 5:47
I made an edit to that comment.
– ALannister
Jan 2 '17 at 5:51
Yes. As $H$ is normal, $ghg^{-1}in H$, so $[g,h]=ghg^{-1}h^{-1}in H$. Now what can you deduce from here?
– arctic tern
Jan 2 '17 at 5:52
1
You already deduced $[g,h]$ is in $H$. Go further. Note that $[g,h]$ is a commutator. Is there anything in our set of hypotheses that talks about commutators and $H$?
– arctic tern
Jan 2 '17 at 5:57
1
Yes. If $[g,h]$ is in $G'$ (which it is since it's a commutator) and $[g,h]$ is in $H$, and the only thing in both $G'$ and $H$ is the identity, then $[g,h]$ is the identity. What does that say about $g$ and $h$?
– arctic tern
Jan 2 '17 at 6:00
|
show 8 more comments
Suppose $hin H$ and $gin G$ are arbitrary, and consider the commutator $[g,h]=(ghg^{-1})h^{-1}$.
Using the fact $H$ is normal, can you deduce anything? (Hint: look at how I parenthesized it.)
answered Jan 2 '17 at 5:45
arctic ternarctic tern
11.8k31535
1
what you parenthesized is contained in $H$ since $H$ is normal? And so $[g,h]$ is contained in $H$?
– ALannister
Jan 2 '17 at 5:47
I made an edit to that comment.
– ALannister
Jan 2 '17 at 5:51
Yes. As $H$ is normal, $ghg^{-1}in H$, so $[g,h]=ghg^{-1}h^{-1}in H$. Now what can you deduce from here?
– arctic tern
Jan 2 '17 at 5:52
1
You already deduced $[g,h]$ is in $H$. Go further. Note that $[g,h]$ is a commutator. Is there anything in our set of hypotheses that talks about commutators and $H$?
– arctic tern
Jan 2 '17 at 5:57
1
Yes. If $[g,h]$ is in $G'$ (which it is since it's a commutator) and $[g,h]$ is in $H$, and the only thing in both $G'$ and $H$ is the identity, then $[g,h]$ is the identity. What does that say about $g$ and $h$?
– arctic tern
Jan 2 '17 at 6:00
|
show 8 more comments
Suppose $hin H$ and $gin G$ are arbitrary, and consider the commutator $[g,h]=(ghg^{-1})h^{-1}$.
Using the fact $H$ is normal, can you deduce anything? (Hint: look at how I parenthesized it.)
answered Jan 2 '17 at 5:45
arctic ternarctic tern
11.8k31535
Suppose $hin H$ and $gin G$ are arbitrary, and consider the commutator $[g,h]=(ghg^{-1})h^{-1}$.
Using the fact $H$ is normal, can you deduce anything? (Hint: look at how I parenthesized it.)
answered Jan 2 '17 at 5:45
arctic ternarctic tern
11.8k31535
edited Jan 2 '17 at 5:48
answered Jan 2 '17 at 5:45
arctic ternarctic tern
11.8k31535
answered Jan 2 '17 at 5:45
arctic ternarctic tern
11.8k31535
answered Jan 2 '17 at 5:45
arctic ternarctic tern
11.8k31535
11.8k31535
1
what you parenthesized is contained in $H$ since $H$ is normal? And so $[g,h]$ is contained in $H$?
– ALannister
Jan 2 '17 at 5:47
I made an edit to that comment.
– ALannister
Jan 2 '17 at 5:51
Yes. As $H$ is normal, $ghg^{-1}in H$, so $[g,h]=ghg^{-1}h^{-1}in H$. Now what can you deduce from here?
– arctic tern
Jan 2 '17 at 5:52
1
You already deduced $[g,h]$ is in $H$. Go further. Note that $[g,h]$ is a commutator. Is there anything in our set of hypotheses that talks about commutators and $H$?
– arctic tern
Jan 2 '17 at 5:57
1
Yes. If $[g,h]$ is in $G'$ (which it is since it's a commutator) and $[g,h]$ is in $H$, and the only thing in both $G'$ and $H$ is the identity, then $[g,h]$ is the identity. What does that say about $g$ and $h$?
– arctic tern
Jan 2 '17 at 6:00
|
show 8 more comments
1
what you parenthesized is contained in $H$ since $H$ is normal? And so $[g,h]$ is contained in $H$?
– ALannister
Jan 2 '17 at 5:47
I made an edit to that comment.
– ALannister
Jan 2 '17 at 5:51
Yes. As $H$ is normal, $ghg^{-1}in H$, so $[g,h]=ghg^{-1}h^{-1}in H$. Now what can you deduce from here?
– arctic tern
Jan 2 '17 at 5:52
1
You already deduced $[g,h]$ is in $H$. Go further. Note that $[g,h]$ is a commutator. Is there anything in our set of hypotheses that talks about commutators and $H$?
– arctic tern
Jan 2 '17 at 5:57
1
Yes. If $[g,h]$ is in $G'$ (which it is since it's a commutator) and $[g,h]$ is in $H$, and the only thing in both $G'$ and $H$ is the identity, then $[g,h]$ is the identity. What does that say about $g$ and $h$?
– arctic tern
Jan 2 '17 at 6:00
1
1
what you parenthesized is contained in $H$ since $H$ is normal? And so $[g,h]$ is contained in $H$?
– ALannister
Jan 2 '17 at 5:47
what you parenthesized is contained in $H$ since $H$ is normal? And so $[g,h]$ is contained in $H$?
– ALannister
Jan 2 '17 at 5:47
I made an edit to that comment.
– ALannister
Jan 2 '17 at 5:51
I made an edit to that comment.
– ALannister
Jan 2 '17 at 5:51
Yes. As $H$ is normal, $ghg^{-1}in H$, so $[g,h]=ghg^{-1}h^{-1}in H$. Now what can you deduce from here?
– arctic tern
Jan 2 '17 at 5:52
Yes. As $H$ is normal, $ghg^{-1}in H$, so $[g,h]=ghg^{-1}h^{-1}in H$. Now what can you deduce from here?
– arctic tern
Jan 2 '17 at 5:52
1
1
You already deduced $[g,h]$ is in $H$. Go further. Note that $[g,h]$ is a commutator. Is there anything in our set of hypotheses that talks about commutators and $H$?
– arctic tern
Jan 2 '17 at 5:57
You already deduced $[g,h]$ is in $H$. Go further. Note that $[g,h]$ is a commutator. Is there anything in our set of hypotheses that talks about commutators and $H$?
– arctic tern
Jan 2 '17 at 5:57
1
1
Yes. If $[g,h]$ is in $G'$ (which it is since it's a commutator) and $[g,h]$ is in $H$, and the only thing in both $G'$ and $H$ is the identity, then $[g,h]$ is the identity. What does that say about $g$ and $h$?
– arctic tern
Jan 2 '17 at 6:00
Yes. If $[g,h]$ is in $G'$ (which it is since it's a commutator) and $[g,h]$ is in $H$, and the only thing in both $G'$ and $H$ is the identity, then $[g,h]$ is the identity. What does that say about $g$ and $h$?
– arctic tern
Jan 2 '17 at 6:00
|
show 8 more comments
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If $H$ is a subgroup of $G$ and a subset of the center $Z(G)$ which itself a subgroup of $G$, that automatically means $H$ would be a subgroup of $Z(G)$, so it's a bit jarring to see you say "it just needs to be a subset, not a subgroup." Also, "elements of $H$ commute with elements of $G',$" is equivalent to $Hsubseteq C(G')$ which is generally a weaker condition than $Hsubseteq Z(G)$, so more work will have to be done to show $H$ is central (if that does indeed follow from the hypotheses).
– arctic tern
Jan 2 '17 at 5:37
@arctictern When you say $C(G^{prime})$ you mean $Z(G^{prime})$ by your notation?
– ALannister
Jan 2 '17 at 5:46
No. $C(G')$ is short for $C_G(G')$ which means the set of all $gin G$ which commute with every element of $G'$. The notation $Z(G)$ means $C_G(G)$, the center, containing all $gin G$ which commute with every element of $G$. Thus, if someone wrote $Z(G')$, they would be talking about the elements of $G'$ which commute with every element of $G'$, and that is not a thing being discussed.
– arctic tern
Jan 2 '17 at 5:52