Mixing
Commutator subgroup of cyclic group
$begingroup$
Find commutator subgroup of cyclic group.
I know that every cyclic group is abelian group and commutator subgroup of any abelian group is trivial group.
I just need help to prove this.
abstract-algebra abelian-groups cyclic-groups
asked Jan 25 at 18:06
user638212user638212
302
$endgroup$
add a comment |
$begingroup$
Find commutator subgroup of cyclic group.
I know that every cyclic group is abelian group and commutator subgroup of any abelian group is trivial group.
I just need help to prove this.
abstract-algebra abelian-groups cyclic-groups
asked Jan 25 at 18:06
user638212user638212
302
$endgroup$
2
$begingroup$
Please include the definitions you've learned: What is the commutator subgroup of a group? What do you mean by the "the commutator subgroup of any abelian group is the trivial group"? Remember, a cyclic group happens to be an abelian group.
$endgroup$
– Namaste
Jan 25 at 18:11
add a comment |
$begingroup$
Find commutator subgroup of cyclic group.
I know that every cyclic group is abelian group and commutator subgroup of any abelian group is trivial group.
I just need help to prove this.
abstract-algebra abelian-groups cyclic-groups
asked Jan 25 at 18:06
user638212user638212
302
$endgroup$
Find commutator subgroup of cyclic group.
I know that every cyclic group is abelian group and commutator subgroup of any abelian group is trivial group.
I just need help to prove this.
abstract-algebra abelian-groups cyclic-groups
abstract-algebra abelian-groups cyclic-groups
asked Jan 25 at 18:06
user638212user638212
302
asked Jan 25 at 18:06
user638212user638212
302
asked Jan 25 at 18:06
user638212user638212
302
asked Jan 25 at 18:06
user638212user638212
302
asked Jan 25 at 18:06
user638212user638212
302
302
2
$begingroup$
Please include the definitions you've learned: What is the commutator subgroup of a group? What do you mean by the "the commutator subgroup of any abelian group is the trivial group"? Remember, a cyclic group happens to be an abelian group.
$endgroup$
– Namaste
Jan 25 at 18:11
add a comment |
2
$begingroup$
Please include the definitions you've learned: What is the commutator subgroup of a group? What do you mean by the "the commutator subgroup of any abelian group is the trivial group"? Remember, a cyclic group happens to be an abelian group.
$endgroup$
– Namaste
Jan 25 at 18:11
2
2
$begingroup$
Please include the definitions you've learned: What is the commutator subgroup of a group? What do you mean by the "the commutator subgroup of any abelian group is the trivial group"? Remember, a cyclic group happens to be an abelian group.
$endgroup$
– Namaste
Jan 25 at 18:11
$begingroup$
Please include the definitions you've learned: What is the commutator subgroup of a group? What do you mean by the "the commutator subgroup of any abelian group is the trivial group"? Remember, a cyclic group happens to be an abelian group.
$endgroup$
– Namaste
Jan 25 at 18:11
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Every cyclic group is abelian:
Indeed, if $G$ is cyclic, then there exists $a$ such that $G=left<aright>$.
Let $g,hin G$; then there exists $n,minmathbb Z$ such that $g=a^n$ and $h=a^m$. Therefore $$gh=(a^n)(a^m) = a^{n+m}=(a^m)(a^n) = hg$$
The commutator subgroup of every abelian group is trivial:
Indeed, the commutator subgroup is generated by elements of the form $[g,h]:=ghg^{-1}h^{-1}$ for $g,hin G$. If $g,h$ commutes, then $ghg^{-1}h^{-1}=gg^{-1}hh^{-1}= ecdot e = e$. Therefore, the commutator subgroup is trivial.
edited Jan 25 at 19:16
J. W. Tanner
3,4601320
answered Jan 25 at 18:15
YankoYanko
7,9551830
$endgroup$
add a comment |
$begingroup$
The commutator is defined as $[a,b]=aba^{-1}b^{-1}$. Since the group is cyclic, it's Abelian. So, we have that $forall a,b: ab=ba$. This gives that $aba^{-1}b^{-1}=e$ always. So, the commutator subgroup is $e$.
answered Jan 25 at 18:15
stressed outstressed out
6,5931939
$endgroup$
add a comment |
$begingroup$
Forgive me if perhaps I misunderstood, but is there anything more to this question than:
[$g,b$]=$g^{-1}b^{-1}gb$= $c^{-i}c^{-j}c^ic^j$=$e$ and thus the the commutator subgroup is the minimum subgroup on the set {$e$}, which is the trivial group.
Of course, it would depend on how exactly the commutator was defined to you( maybe your professor defined it through quotients, which would yield another simple solution)
answered Jan 25 at 18:17
Marat AlievMarat Aliev
1312
$endgroup$
$begingroup$
I don't understand your formula for $[gb]$.
$endgroup$
– Yanko
Jan 25 at 19:36
$begingroup$
Oh sorry i forgot the comma.
$endgroup$
– Marat Aliev
Jan 25 at 20:26
$begingroup$
That's not that. I don't understand where the $c$'s come from.
$endgroup$
– Yanko
Jan 26 at 13:28
1
$begingroup$
Oh, sorry forgot that as well. the $c$ is to denote the generator of the cyclic group. Needs another edit, sorry for the mistype
$endgroup$
– Marat Aliev
Jan 26 at 18:35
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Every cyclic group is abelian:
Indeed, if $G$ is cyclic, then there exists $a$ such that $G=left<aright>$.
Let $g,hin G$; then there exists $n,minmathbb Z$ such that $g=a^n$ and $h=a^m$. Therefore $$gh=(a^n)(a^m) = a^{n+m}=(a^m)(a^n) = hg$$
The commutator subgroup of every abelian group is trivial:
Indeed, the commutator subgroup is generated by elements of the form $[g,h]:=ghg^{-1}h^{-1}$ for $g,hin G$. If $g,h$ commutes, then $ghg^{-1}h^{-1}=gg^{-1}hh^{-1}= ecdot e = e$. Therefore, the commutator subgroup is trivial.
edited Jan 25 at 19:16
J. W. Tanner
3,4601320
answered Jan 25 at 18:15
YankoYanko
7,9551830
$endgroup$
add a comment |
$begingroup$
Every cyclic group is abelian:
Indeed, if $G$ is cyclic, then there exists $a$ such that $G=left<aright>$.
Let $g,hin G$; then there exists $n,minmathbb Z$ such that $g=a^n$ and $h=a^m$. Therefore $$gh=(a^n)(a^m) = a^{n+m}=(a^m)(a^n) = hg$$
The commutator subgroup of every abelian group is trivial:
Indeed, the commutator subgroup is generated by elements of the form $[g,h]:=ghg^{-1}h^{-1}$ for $g,hin G$. If $g,h$ commutes, then $ghg^{-1}h^{-1}=gg^{-1}hh^{-1}= ecdot e = e$. Therefore, the commutator subgroup is trivial.
edited Jan 25 at 19:16
J. W. Tanner
3,4601320
answered Jan 25 at 18:15
YankoYanko
7,9551830
$endgroup$
add a comment |
$begingroup$
Every cyclic group is abelian:
Indeed, if $G$ is cyclic, then there exists $a$ such that $G=left<aright>$.
Let $g,hin G$; then there exists $n,minmathbb Z$ such that $g=a^n$ and $h=a^m$. Therefore $$gh=(a^n)(a^m) = a^{n+m}=(a^m)(a^n) = hg$$
The commutator subgroup of every abelian group is trivial:
Indeed, the commutator subgroup is generated by elements of the form $[g,h]:=ghg^{-1}h^{-1}$ for $g,hin G$. If $g,h$ commutes, then $ghg^{-1}h^{-1}=gg^{-1}hh^{-1}= ecdot e = e$. Therefore, the commutator subgroup is trivial.
edited Jan 25 at 19:16
J. W. Tanner
3,4601320
answered Jan 25 at 18:15
YankoYanko
7,9551830
$endgroup$
Every cyclic group is abelian:
Indeed, if $G$ is cyclic, then there exists $a$ such that $G=left<aright>$.
Let $g,hin G$; then there exists $n,minmathbb Z$ such that $g=a^n$ and $h=a^m$. Therefore $$gh=(a^n)(a^m) = a^{n+m}=(a^m)(a^n) = hg$$
The commutator subgroup of every abelian group is trivial:
Indeed, the commutator subgroup is generated by elements of the form $[g,h]:=ghg^{-1}h^{-1}$ for $g,hin G$. If $g,h$ commutes, then $ghg^{-1}h^{-1}=gg^{-1}hh^{-1}= ecdot e = e$. Therefore, the commutator subgroup is trivial.
edited Jan 25 at 19:16
J. W. Tanner
3,4601320
answered Jan 25 at 18:15
YankoYanko
7,9551830
edited Jan 25 at 19:16
J. W. Tanner
3,4601320
edited Jan 25 at 19:16
J. W. Tanner
3,4601320
edited Jan 25 at 19:16
J. W. Tanner
3,4601320
3,4601320
answered Jan 25 at 18:15
YankoYanko
7,9551830
answered Jan 25 at 18:15
YankoYanko
7,9551830
answered Jan 25 at 18:15
YankoYanko
7,9551830
7,9551830
add a comment |
add a comment |
$begingroup$
The commutator is defined as $[a,b]=aba^{-1}b^{-1}$. Since the group is cyclic, it's Abelian. So, we have that $forall a,b: ab=ba$. This gives that $aba^{-1}b^{-1}=e$ always. So, the commutator subgroup is $e$.
answered Jan 25 at 18:15
stressed outstressed out
6,5931939
$endgroup$
add a comment |
$begingroup$
The commutator is defined as $[a,b]=aba^{-1}b^{-1}$. Since the group is cyclic, it's Abelian. So, we have that $forall a,b: ab=ba$. This gives that $aba^{-1}b^{-1}=e$ always. So, the commutator subgroup is $e$.
answered Jan 25 at 18:15
stressed outstressed out
6,5931939
$endgroup$
add a comment |
$begingroup$
The commutator is defined as $[a,b]=aba^{-1}b^{-1}$. Since the group is cyclic, it's Abelian. So, we have that $forall a,b: ab=ba$. This gives that $aba^{-1}b^{-1}=e$ always. So, the commutator subgroup is $e$.
answered Jan 25 at 18:15
stressed outstressed out
6,5931939
$endgroup$
The commutator is defined as $[a,b]=aba^{-1}b^{-1}$. Since the group is cyclic, it's Abelian. So, we have that $forall a,b: ab=ba$. This gives that $aba^{-1}b^{-1}=e$ always. So, the commutator subgroup is $e$.
answered Jan 25 at 18:15
stressed outstressed out
6,5931939
answered Jan 25 at 18:15
stressed outstressed out
6,5931939
answered Jan 25 at 18:15
stressed outstressed out
6,5931939
answered Jan 25 at 18:15
stressed outstressed out
6,5931939
6,5931939
add a comment |
add a comment |
$begingroup$
Forgive me if perhaps I misunderstood, but is there anything more to this question than:
[$g,b$]=$g^{-1}b^{-1}gb$= $c^{-i}c^{-j}c^ic^j$=$e$ and thus the the commutator subgroup is the minimum subgroup on the set {$e$}, which is the trivial group.
Of course, it would depend on how exactly the commutator was defined to you( maybe your professor defined it through quotients, which would yield another simple solution)
answered Jan 25 at 18:17
Marat AlievMarat Aliev
1312
$endgroup$
$begingroup$
I don't understand your formula for $[gb]$.
$endgroup$
– Yanko
Jan 25 at 19:36
$begingroup$
Oh sorry i forgot the comma.
$endgroup$
– Marat Aliev
Jan 25 at 20:26
$begingroup$
That's not that. I don't understand where the $c$'s come from.
$endgroup$
– Yanko
Jan 26 at 13:28
1
$begingroup$
Oh, sorry forgot that as well. the $c$ is to denote the generator of the cyclic group. Needs another edit, sorry for the mistype
$endgroup$
– Marat Aliev
Jan 26 at 18:35
add a comment |
$begingroup$
Forgive me if perhaps I misunderstood, but is there anything more to this question than:
[$g,b$]=$g^{-1}b^{-1}gb$= $c^{-i}c^{-j}c^ic^j$=$e$ and thus the the commutator subgroup is the minimum subgroup on the set {$e$}, which is the trivial group.
Of course, it would depend on how exactly the commutator was defined to you( maybe your professor defined it through quotients, which would yield another simple solution)
answered Jan 25 at 18:17
Marat AlievMarat Aliev
1312
$endgroup$
$begingroup$
I don't understand your formula for $[gb]$.
$endgroup$
– Yanko
Jan 25 at 19:36
$begingroup$
Oh sorry i forgot the comma.
$endgroup$
– Marat Aliev
Jan 25 at 20:26
$begingroup$
That's not that. I don't understand where the $c$'s come from.
$endgroup$
– Yanko
Jan 26 at 13:28
1
$begingroup$
Oh, sorry forgot that as well. the $c$ is to denote the generator of the cyclic group. Needs another edit, sorry for the mistype
$endgroup$
– Marat Aliev
Jan 26 at 18:35
add a comment |
$begingroup$
Forgive me if perhaps I misunderstood, but is there anything more to this question than:
[$g,b$]=$g^{-1}b^{-1}gb$= $c^{-i}c^{-j}c^ic^j$=$e$ and thus the the commutator subgroup is the minimum subgroup on the set {$e$}, which is the trivial group.
Of course, it would depend on how exactly the commutator was defined to you( maybe your professor defined it through quotients, which would yield another simple solution)
answered Jan 25 at 18:17
Marat AlievMarat Aliev
1312
$endgroup$
Forgive me if perhaps I misunderstood, but is there anything more to this question than:
[$g,b$]=$g^{-1}b^{-1}gb$= $c^{-i}c^{-j}c^ic^j$=$e$ and thus the the commutator subgroup is the minimum subgroup on the set {$e$}, which is the trivial group.
Of course, it would depend on how exactly the commutator was defined to you( maybe your professor defined it through quotients, which would yield another simple solution)
answered Jan 25 at 18:17
Marat AlievMarat Aliev
1312
edited Jan 26 at 18:35
answered Jan 25 at 18:17
Marat AlievMarat Aliev
1312
answered Jan 25 at 18:17
Marat AlievMarat Aliev
1312
answered Jan 25 at 18:17
Marat AlievMarat Aliev
1312
1312
$begingroup$
I don't understand your formula for $[gb]$.
$endgroup$
– Yanko
Jan 25 at 19:36
$begingroup$
Oh sorry i forgot the comma.
$endgroup$
– Marat Aliev
Jan 25 at 20:26
$begingroup$
That's not that. I don't understand where the $c$'s come from.
$endgroup$
– Yanko
Jan 26 at 13:28
1
$begingroup$
Oh, sorry forgot that as well. the $c$ is to denote the generator of the cyclic group. Needs another edit, sorry for the mistype
$endgroup$
– Marat Aliev
Jan 26 at 18:35
add a comment |
$begingroup$
I don't understand your formula for $[gb]$.
$endgroup$
– Yanko
Jan 25 at 19:36
$begingroup$
Oh sorry i forgot the comma.
$endgroup$
– Marat Aliev
Jan 25 at 20:26
$begingroup$
That's not that. I don't understand where the $c$'s come from.
$endgroup$
– Yanko
Jan 26 at 13:28
1
$begingroup$
Oh, sorry forgot that as well. the $c$ is to denote the generator of the cyclic group. Needs another edit, sorry for the mistype
$endgroup$
– Marat Aliev
Jan 26 at 18:35
$begingroup$
I don't understand your formula for $[gb]$.
$endgroup$
– Yanko
Jan 25 at 19:36
$begingroup$
I don't understand your formula for $[gb]$.
$endgroup$
– Yanko
Jan 25 at 19:36
$begingroup$
Oh sorry i forgot the comma.
$endgroup$
– Marat Aliev
Jan 25 at 20:26
$begingroup$
Oh sorry i forgot the comma.
$endgroup$
– Marat Aliev
Jan 25 at 20:26
$begingroup$
That's not that. I don't understand where the $c$'s come from.
$endgroup$
– Yanko
Jan 26 at 13:28
$begingroup$
That's not that. I don't understand where the $c$'s come from.
$endgroup$
– Yanko
Jan 26 at 13:28
1
1
$begingroup$
Oh, sorry forgot that as well. the $c$ is to denote the generator of the cyclic group. Needs another edit, sorry for the mistype
$endgroup$
– Marat Aliev
Jan 26 at 18:35
$begingroup$
Oh, sorry forgot that as well. the $c$ is to denote the generator of the cyclic group. Needs another edit, sorry for the mistype
$endgroup$
– Marat Aliev
Jan 26 at 18:35
add a comment |
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2
$begingroup$
Please include the definitions you've learned: What is the commutator subgroup of a group? What do you mean by the "the commutator subgroup of any abelian group is the trivial group"? Remember, a cyclic group happens to be an abelian group.
$endgroup$
– Namaste
Jan 25 at 18:11