Mixing
understanding the commutator of dihedral group [duplicate]
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This question already has an answer here:
Commutator subgroup of a dihedral group.
2 answers
Let $G=D2n=⟨x,y|x^2=y^n=e, $ $yx=xy^{n-1}⟩$
i need to find G' [ the commutattor of G]
now i understand the G' is the subgroup that is generated from $ U=xyx^{-1}y^{-1} , $ $forall x,y in G$
so $<u> ={u^1,u^2...e}$
whats the strategy here ?
abstract-algebra group-theory dihedral-groups
asked Dec 12 '14 at 14:44
Nizar HallounNizar Halloun
739
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marked as duplicate by Dietrich Burde, Iuʇǝƃɹɐʇoɹ, Anastasiya-Romanova 秀, Stefan Hamcke, Cheerful Parsnip Dec 12 '14 at 15:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Commutator subgroup of a dihedral group.
2 answers
Let $G=D2n=⟨x,y|x^2=y^n=e, $ $yx=xy^{n-1}⟩$
i need to find G' [ the commutattor of G]
now i understand the G' is the subgroup that is generated from $ U=xyx^{-1}y^{-1} , $ $forall x,y in G$
so $<u> ={u^1,u^2...e}$
whats the strategy here ?
abstract-algebra group-theory dihedral-groups
asked Dec 12 '14 at 14:44
Nizar HallounNizar Halloun
739
$endgroup$
marked as duplicate by Dietrich Burde, Iuʇǝƃɹɐʇoɹ, Anastasiya-Romanova 秀, Stefan Hamcke, Cheerful Parsnip Dec 12 '14 at 15:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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I'd suggest trying with small $n$ first. Also you're missing generators, you need the commutators of all elements while here you've only written $[x,y]$.
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– GPerez
Dec 12 '14 at 14:49
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@GPerez how can i find the commutattor of group in general..i'm not getting it
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– Nizar Halloun
Dec 12 '14 at 14:57
$begingroup$
@NizarHalloun: Terminology issue: A "commutator" is an element of a group. You are talking about the "commutator subgroup," which is the subgroup generated by commutators.
$endgroup$
– Cheerful Parsnip
Dec 12 '14 at 15:40
add a comment |
$begingroup$
This question already has an answer here:
Commutator subgroup of a dihedral group.
2 answers
Let $G=D2n=⟨x,y|x^2=y^n=e, $ $yx=xy^{n-1}⟩$
i need to find G' [ the commutattor of G]
now i understand the G' is the subgroup that is generated from $ U=xyx^{-1}y^{-1} , $ $forall x,y in G$
so $<u> ={u^1,u^2...e}$
whats the strategy here ?
abstract-algebra group-theory dihedral-groups
asked Dec 12 '14 at 14:44
Nizar HallounNizar Halloun
739
$endgroup$
This question already has an answer here:
Commutator subgroup of a dihedral group.
2 answers
Let $G=D2n=⟨x,y|x^2=y^n=e, $ $yx=xy^{n-1}⟩$
i need to find G' [ the commutattor of G]
now i understand the G' is the subgroup that is generated from $ U=xyx^{-1}y^{-1} , $ $forall x,y in G$
so $<u> ={u^1,u^2...e}$
whats the strategy here ?
This question already has an answer here:
Commutator subgroup of a dihedral group.
2 answers
abstract-algebra group-theory dihedral-groups
abstract-algebra group-theory dihedral-groups
asked Dec 12 '14 at 14:44
Nizar HallounNizar Halloun
739
asked Dec 12 '14 at 14:44
Nizar HallounNizar Halloun
739
edited Dec 12 '14 at 14:53
Nizar Halloun
asked Dec 12 '14 at 14:44
Nizar HallounNizar Halloun
739
asked Dec 12 '14 at 14:44
Nizar HallounNizar Halloun
739
asked Dec 12 '14 at 14:44
Nizar HallounNizar Halloun
739
739
marked as duplicate by Dietrich Burde, Iuʇǝƃɹɐʇoɹ, Anastasiya-Romanova 秀, Stefan Hamcke, Cheerful Parsnip Dec 12 '14 at 15:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Dietrich Burde, Iuʇǝƃɹɐʇoɹ, Anastasiya-Romanova 秀, Stefan Hamcke, Cheerful Parsnip Dec 12 '14 at 15:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
I'd suggest trying with small $n$ first. Also you're missing generators, you need the commutators of all elements while here you've only written $[x,y]$.
$endgroup$
– GPerez
Dec 12 '14 at 14:49
$begingroup$
@GPerez how can i find the commutattor of group in general..i'm not getting it
$endgroup$
– Nizar Halloun
Dec 12 '14 at 14:57
$begingroup$
@NizarHalloun: Terminology issue: A "commutator" is an element of a group. You are talking about the "commutator subgroup," which is the subgroup generated by commutators.
$endgroup$
– Cheerful Parsnip
Dec 12 '14 at 15:40
add a comment |
1
$begingroup$
I'd suggest trying with small $n$ first. Also you're missing generators, you need the commutators of all elements while here you've only written $[x,y]$.
$endgroup$
– GPerez
Dec 12 '14 at 14:49
$begingroup$
@GPerez how can i find the commutattor of group in general..i'm not getting it
$endgroup$
– Nizar Halloun
Dec 12 '14 at 14:57
$begingroup$
@NizarHalloun: Terminology issue: A "commutator" is an element of a group. You are talking about the "commutator subgroup," which is the subgroup generated by commutators.
$endgroup$
– Cheerful Parsnip
Dec 12 '14 at 15:40
1
1
$begingroup$
I'd suggest trying with small $n$ first. Also you're missing generators, you need the commutators of all elements while here you've only written $[x,y]$.
$endgroup$
– GPerez
Dec 12 '14 at 14:49
$begingroup$
I'd suggest trying with small $n$ first. Also you're missing generators, you need the commutators of all elements while here you've only written $[x,y]$.
$endgroup$
– GPerez
Dec 12 '14 at 14:49
$begingroup$
@GPerez how can i find the commutattor of group in general..i'm not getting it
$endgroup$
– Nizar Halloun
Dec 12 '14 at 14:57
$begingroup$
@GPerez how can i find the commutattor of group in general..i'm not getting it
$endgroup$
– Nizar Halloun
Dec 12 '14 at 14:57
$begingroup$
@NizarHalloun: Terminology issue: A "commutator" is an element of a group. You are talking about the "commutator subgroup," which is the subgroup generated by commutators.
$endgroup$
– Cheerful Parsnip
Dec 12 '14 at 15:40
$begingroup$
@NizarHalloun: Terminology issue: A "commutator" is an element of a group. You are talking about the "commutator subgroup," which is the subgroup generated by commutators.
$endgroup$
– Cheerful Parsnip
Dec 12 '14 at 15:40
add a comment |
1 Answer
1
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Some ideas: if we write
$$D_{2n}=langle;x,y;:;;x^2=y^n=1;,;;xyx=y^{n-1};rangle$$
then we get that
$$forall,kinBbb N;,;;[x,y^k]=x^{-1}y^{-k}xy^k=left(xy^{-k}xright)y^k=left(y^{n-1}right)^{-k};y^k=y^{2k}$$
and this means every element in $;langle;y^2;rangle;$ is a commutator.
OTOH, we have that
$$xy(yx)^{-1}=xyxy^{-1}=y^{n-1}y^{-1}=y^{-2}inlangle;y^2;rangleimplies G/langle;y^2;rangle;;text{is abelian}iff G'lelangle;y^2;rangle$$
The above yields $;G'=[G:G]=langle;y^2;rangle;$ .
answered Dec 12 '14 at 15:26
TimbucTimbuc
31k22145
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Some ideas: if we write
$$D_{2n}=langle;x,y;:;;x^2=y^n=1;,;;xyx=y^{n-1};rangle$$
then we get that
$$forall,kinBbb N;,;;[x,y^k]=x^{-1}y^{-k}xy^k=left(xy^{-k}xright)y^k=left(y^{n-1}right)^{-k};y^k=y^{2k}$$
and this means every element in $;langle;y^2;rangle;$ is a commutator.
OTOH, we have that
$$xy(yx)^{-1}=xyxy^{-1}=y^{n-1}y^{-1}=y^{-2}inlangle;y^2;rangleimplies G/langle;y^2;rangle;;text{is abelian}iff G'lelangle;y^2;rangle$$
The above yields $;G'=[G:G]=langle;y^2;rangle;$ .
answered Dec 12 '14 at 15:26
TimbucTimbuc
31k22145
$endgroup$
add a comment |
$begingroup$
Some ideas: if we write
$$D_{2n}=langle;x,y;:;;x^2=y^n=1;,;;xyx=y^{n-1};rangle$$
then we get that
$$forall,kinBbb N;,;;[x,y^k]=x^{-1}y^{-k}xy^k=left(xy^{-k}xright)y^k=left(y^{n-1}right)^{-k};y^k=y^{2k}$$
and this means every element in $;langle;y^2;rangle;$ is a commutator.
OTOH, we have that
$$xy(yx)^{-1}=xyxy^{-1}=y^{n-1}y^{-1}=y^{-2}inlangle;y^2;rangleimplies G/langle;y^2;rangle;;text{is abelian}iff G'lelangle;y^2;rangle$$
The above yields $;G'=[G:G]=langle;y^2;rangle;$ .
answered Dec 12 '14 at 15:26
TimbucTimbuc
31k22145
$endgroup$
add a comment |
$begingroup$
Some ideas: if we write
$$D_{2n}=langle;x,y;:;;x^2=y^n=1;,;;xyx=y^{n-1};rangle$$
then we get that
$$forall,kinBbb N;,;;[x,y^k]=x^{-1}y^{-k}xy^k=left(xy^{-k}xright)y^k=left(y^{n-1}right)^{-k};y^k=y^{2k}$$
and this means every element in $;langle;y^2;rangle;$ is a commutator.
OTOH, we have that
$$xy(yx)^{-1}=xyxy^{-1}=y^{n-1}y^{-1}=y^{-2}inlangle;y^2;rangleimplies G/langle;y^2;rangle;;text{is abelian}iff G'lelangle;y^2;rangle$$
The above yields $;G'=[G:G]=langle;y^2;rangle;$ .
answered Dec 12 '14 at 15:26
TimbucTimbuc
31k22145
$endgroup$
Some ideas: if we write
$$D_{2n}=langle;x,y;:;;x^2=y^n=1;,;;xyx=y^{n-1};rangle$$
then we get that
$$forall,kinBbb N;,;;[x,y^k]=x^{-1}y^{-k}xy^k=left(xy^{-k}xright)y^k=left(y^{n-1}right)^{-k};y^k=y^{2k}$$
and this means every element in $;langle;y^2;rangle;$ is a commutator.
OTOH, we have that
$$xy(yx)^{-1}=xyxy^{-1}=y^{n-1}y^{-1}=y^{-2}inlangle;y^2;rangleimplies G/langle;y^2;rangle;;text{is abelian}iff G'lelangle;y^2;rangle$$
The above yields $;G'=[G:G]=langle;y^2;rangle;$ .
answered Dec 12 '14 at 15:26
TimbucTimbuc
31k22145
answered Dec 12 '14 at 15:26
TimbucTimbuc
31k22145
answered Dec 12 '14 at 15:26
TimbucTimbuc
31k22145
answered Dec 12 '14 at 15:26
TimbucTimbuc
31k22145
31k22145
add a comment |
add a comment |
1
$begingroup$
I'd suggest trying with small $n$ first. Also you're missing generators, you need the commutators of all elements while here you've only written $[x,y]$.
$endgroup$
– GPerez
Dec 12 '14 at 14:49
$begingroup$
@GPerez how can i find the commutattor of group in general..i'm not getting it
$endgroup$
– Nizar Halloun
Dec 12 '14 at 14:57
$begingroup$
@NizarHalloun: Terminology issue: A "commutator" is an element of a group. You are talking about the "commutator subgroup," which is the subgroup generated by commutators.
$endgroup$
– Cheerful Parsnip
Dec 12 '14 at 15:40