On a characterization of abelian groups $G$ based on special commutator relations ($exists ninBbb N$ s.t....
3
$begingroup$
Let $G$ be a group. If $exists nin mathbb N$ such that $x^n=yx^n(y^{n+1}x)^{-1}xy^n,forall x,yin G$, then how to prove that $G$ is abelian?
Thoughts: the condition is same as saying $[x^n,y]=[x,y^{n+1}],forall x,y in G$, where $[a,b]:=a^{-1}b^{-1}ab$.
group-theory abelian-groups group-homomorphism
edited Dec 29 '18 at 8:47
Shaun
9,690113684
asked Dec 29 '18 at 5:14
user521337user521337
1,1981417
$endgroup$
add a comment |
3
$begingroup$
Let $G$ be a group. If $exists nin mathbb N$ such that $x^n=yx^n(y^{n+1}x)^{-1}xy^n,forall x,yin G$, then how to prove that $G$ is abelian?
Thoughts: the condition is same as saying $[x^n,y]=[x,y^{n+1}],forall x,y in G$, where $[a,b]:=a^{-1}b^{-1}ab$.
group-theory abelian-groups group-homomorphism
edited Dec 29 '18 at 8:47
Shaun
9,690113684
asked Dec 29 '18 at 5:14
user521337user521337
1,1981417
$endgroup$
$begingroup$
Potentially helpful observations: (1) The equation holds for the generators of the group in particular. (2) The LHS is a subword of the RHS, so $g^n=h^kg^n((h^{n+1}g)^{-1}gh^n)^k$ for all generators $g,h$ of $G$ and all $kinBbb N$.
$endgroup$
– Shaun
Dec 29 '18 at 7:44
1
$begingroup$
@Shaun: the condition also relates to commutators ... see my latest edit ...
$endgroup$
– user521337
Dec 29 '18 at 7:46
$begingroup$
Why use the group-homomorphism tag?
$endgroup$
– Shaun
Dec 29 '18 at 8:45
$begingroup$
@Shaun: no particular reason ... could be related ...
$endgroup$
– user521337
Dec 29 '18 at 8:48
1
$begingroup$
If you could prove that $x[x,y] = [x,y]x$ and $y[x,y]=[x,y]y, forall x, y in G$, then you are done, because then you have that $[x^n,y] = [x,y]^n = [x,y^n], forall x,y in G$.
$endgroup$
– Cosmin
Dec 29 '18 at 9:09
add a comment |
3
3
3
4
$begingroup$
Let $G$ be a group. If $exists nin mathbb N$ such that $x^n=yx^n(y^{n+1}x)^{-1}xy^n,forall x,yin G$, then how to prove that $G$ is abelian?
Thoughts: the condition is same as saying $[x^n,y]=[x,y^{n+1}],forall x,y in G$, where $[a,b]:=a^{-1}b^{-1}ab$.
group-theory abelian-groups group-homomorphism
edited Dec 29 '18 at 8:47
Shaun
9,690113684
asked Dec 29 '18 at 5:14
user521337user521337
1,1981417
$endgroup$
Let $G$ be a group. If $exists nin mathbb N$ such that $x^n=yx^n(y^{n+1}x)^{-1}xy^n,forall x,yin G$, then how to prove that $G$ is abelian?
Thoughts: the condition is same as saying $[x^n,y]=[x,y^{n+1}],forall x,y in G$, where $[a,b]:=a^{-1}b^{-1}ab$.
group-theory abelian-groups group-homomorphism
group-theory abelian-groups group-homomorphism
edited Dec 29 '18 at 8:47
Shaun
9,690113684
asked Dec 29 '18 at 5:14
user521337user521337
1,1981417
edited Dec 29 '18 at 8:47
Shaun
9,690113684
asked Dec 29 '18 at 5:14
user521337user521337
1,1981417
edited Dec 29 '18 at 8:47
Shaun
9,690113684
edited Dec 29 '18 at 8:47
Shaun
9,690113684
edited Dec 29 '18 at 8:47
Shaun
9,690113684
9,690113684
asked Dec 29 '18 at 5:14
user521337user521337
1,1981417
asked Dec 29 '18 at 5:14
user521337user521337
1,1981417
asked Dec 29 '18 at 5:14
user521337user521337
1,1981417
1,1981417
$begingroup$
Potentially helpful observations: (1) The equation holds for the generators of the group in particular. (2) The LHS is a subword of the RHS, so $g^n=h^kg^n((h^{n+1}g)^{-1}gh^n)^k$ for all generators $g,h$ of $G$ and all $kinBbb N$.
$endgroup$
– Shaun
Dec 29 '18 at 7:44
1
$begingroup$
@Shaun: the condition also relates to commutators ... see my latest edit ...
$endgroup$
– user521337
Dec 29 '18 at 7:46
$begingroup$
Why use the group-homomorphism tag?
$endgroup$
– Shaun
Dec 29 '18 at 8:45
$begingroup$
@Shaun: no particular reason ... could be related ...
$endgroup$
– user521337
Dec 29 '18 at 8:48
1
$begingroup$
If you could prove that $x[x,y] = [x,y]x$ and $y[x,y]=[x,y]y, forall x, y in G$, then you are done, because then you have that $[x^n,y] = [x,y]^n = [x,y^n], forall x,y in G$.
$endgroup$
– Cosmin
Dec 29 '18 at 9:09
add a comment |
$begingroup$
Potentially helpful observations: (1) The equation holds for the generators of the group in particular. (2) The LHS is a subword of the RHS, so $g^n=h^kg^n((h^{n+1}g)^{-1}gh^n)^k$ for all generators $g,h$ of $G$ and all $kinBbb N$.
$endgroup$
– Shaun
Dec 29 '18 at 7:44
1
$begingroup$
@Shaun: the condition also relates to commutators ... see my latest edit ...
$endgroup$
– user521337
Dec 29 '18 at 7:46
$begingroup$
Why use the group-homomorphism tag?
$endgroup$
– Shaun
Dec 29 '18 at 8:45
$begingroup$
@Shaun: no particular reason ... could be related ...
$endgroup$
– user521337
Dec 29 '18 at 8:48
1
$begingroup$
If you could prove that $x[x,y] = [x,y]x$ and $y[x,y]=[x,y]y, forall x, y in G$, then you are done, because then you have that $[x^n,y] = [x,y]^n = [x,y^n], forall x,y in G$.
$endgroup$
– Cosmin
Dec 29 '18 at 9:09
$begingroup$
Potentially helpful observations: (1) The equation holds for the generators of the group in particular. (2) The LHS is a subword of the RHS, so $g^n=h^kg^n((h^{n+1}g)^{-1}gh^n)^k$ for all generators $g,h$ of $G$ and all $kinBbb N$.
$endgroup$
– Shaun
Dec 29 '18 at 7:44
$begingroup$
Potentially helpful observations: (1) The equation holds for the generators of the group in particular. (2) The LHS is a subword of the RHS, so $g^n=h^kg^n((h^{n+1}g)^{-1}gh^n)^k$ for all generators $g,h$ of $G$ and all $kinBbb N$.
$endgroup$
– Shaun
Dec 29 '18 at 7:44
1
1
$begingroup$
@Shaun: the condition also relates to commutators ... see my latest edit ...
$endgroup$
– user521337
Dec 29 '18 at 7:46
$begingroup$
@Shaun: the condition also relates to commutators ... see my latest edit ...
$endgroup$
– user521337
Dec 29 '18 at 7:46
$begingroup$
Why use the group-homomorphism tag?
$endgroup$
– Shaun
Dec 29 '18 at 8:45
$begingroup$
Why use the group-homomorphism tag?
$endgroup$
– Shaun
Dec 29 '18 at 8:45
$begingroup$
@Shaun: no particular reason ... could be related ...
$endgroup$
– user521337
Dec 29 '18 at 8:48
$begingroup$
@Shaun: no particular reason ... could be related ...
$endgroup$
– user521337
Dec 29 '18 at 8:48
1
1
$begingroup$
If you could prove that $x[x,y] = [x,y]x$ and $y[x,y]=[x,y]y, forall x, y in G$, then you are done, because then you have that $[x^n,y] = [x,y]^n = [x,y^n], forall x,y in G$.
$endgroup$
– Cosmin
Dec 29 '18 at 9:09
$begingroup$
If you could prove that $x[x,y] = [x,y]x$ and $y[x,y]=[x,y]y, forall x, y in G$, then you are done, because then you have that $[x^n,y] = [x,y]^n = [x,y^n], forall x,y in G$.
$endgroup$
– Cosmin
Dec 29 '18 at 9:09
add a comment |
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$begingroup$
Potentially helpful observations: (1) The equation holds for the generators of the group in particular. (2) The LHS is a subword of the RHS, so $g^n=h^kg^n((h^{n+1}g)^{-1}gh^n)^k$ for all generators $g,h$ of $G$ and all $kinBbb N$.
$endgroup$
– Shaun
Dec 29 '18 at 7:44
1
$begingroup$
@Shaun: the condition also relates to commutators ... see my latest edit ...
$endgroup$
– user521337
Dec 29 '18 at 7:46
$begingroup$
Why use the group-homomorphism tag?
$endgroup$
– Shaun
Dec 29 '18 at 8:45
$begingroup$
@Shaun: no particular reason ... could be related ...
$endgroup$
– user521337
Dec 29 '18 at 8:48
1
$begingroup$
If you could prove that $x[x,y] = [x,y]x$ and $y[x,y]=[x,y]y, forall x, y in G$, then you are done, because then you have that $[x^n,y] = [x,y]^n = [x,y^n], forall x,y in G$.
$endgroup$
– Cosmin
Dec 29 '18 at 9:09