Sea $a,b\in G$ entonces usando la hipótesis tenemos que \[a^2=(abb^{-1})^2=(b^{-1}ab)^2=b^{-1}a^2b\] entonces multiplicando por la izquierda por $b$ obtenemos que $ba^2=a^2b \forall a,b\in G$. Tomando $y=ba^{-1}$, $x=a$ y usando la ultima igualdad tenemos que $ba=yx^2=x^2y=a^2ba^{-1}$, entonces multiplicando por $a^{-1}$ por la izquierda obtenemos la siguiente identidad \[aba^{-1}=a^{-1}ba----(1)\]
Ahora veamos que $(aba^{-1}b^{-1})^2=aba^{-1}b^{-1}aba^{-1}b^{-1}=abab^{-1}a^{-1}ba^{-1}b^{-1}=ababa^{-1}b^{-1}a^{-1}b^{-1}$ donde las ultimas dos igualdades se deben a usar dos veces (1), sustituyendo $abab=baba$ en lo ultimo obtenemos que $(aba^{-1}b^{-1})^2=babaa^{-1}b^{-1}a^{-1}b^{-1}=e$ por lo tanto como no hay elementos de orden 2 se deduce que $aba^{-1}b^{-1}=e$ que multiplicando por $ba$ por la derecha se obtiene $ab=ba$ por lo tanto $G$ es abeliano.
Hi,this problem has a very simple solution:by Cauchy theorem the order of the group is odd,thus ab has odd order m,that is also the order of ba,since they are conjiugated,so call j the inverse of 2 mod m,you get (ab)^{2j}=ab=(ba)^{2j}=ba. Or said faster taking the 2-power is bjective if the order of the group is odd. This is true for every prime p that does not divide the order of the group. One proof is just the same as this posted here,taking inverses mod p. Another is quite funny,and i told you(Irving)while walking in Villa borghese:fix x,take the p-uple (a_1,...,a_p) in G^p,such that prod_{i=1}^{p}a_i lies in the coniugacy class of x,Cl(x),this p-uples are invariant under cyclic permutation. thus we have an action of Z_p on the set of this p-uples,but they are in cardinality C=(|G|^{p-1})(|Cl(x)|),indeed fix randomly the first p-1 and then just force the last one so that the product is in the coniugacy class. Now C is equivalent to |Cl(x)| mod p,that is different from 0(indeed |Cl(x)| divides |G|,that is coprime with p by ipothesis). So the action of Z-p,having just orbits of lenght p or 1 (by orbit stabilizer lemma),needs a number of fixed point that is equivalent(it will be equal at the end) to Cl(x) mod p. But a fixed point is a p-uple,of the form (a,a,...,a),so we have an a,such that a^p lies in Cl(x). but now conjugating you get that x is a p-power,so the map of the p power is surjective,and so bjective. With this idea,you can also prove Cauchy theorem. Bye Carlo p.s:this is actually off-topic,but Irving i've the doubt that you haven't received my mails,one with a solution of a problem and another with just talking. Did you received them?
Ok,i felt a little bit bad to having left just a solution in a silly case with an off topic and a strange proof of a trivial fact :-). So now i try to give a natural proof of the real problem: in a group ab and ba are the tipical two element in the same coniugacy class,indeed saying ab is as to say b^{-1}(ba)b,and saying b^{-1}ab and a,is like to say b^{-1}(ab) and (ab)b^{-1}. Given this,is completely natural to reinterpeter the question as,any two elements in the same coniugacy class have the same square,there is no 2-torsion,then G is commutative. So we have x^2=y^{-1}x^2y whenever x and y are in G,that is to say x^2 is in the Center whenever x is in G. That is G/Z(G) is a group where evryone have order 2,it is easy to show that then the group is abelian,so we get yxy^{-1}=xz with z in the center, but then being y^2 in the center we get x=y^{-2}xy^2=xz^2,thus,z^2=1,thuen z=1,by ipothesis. Bye
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ResponderEliminarA ver según yo sale así:
ResponderEliminarSea $a,b\in G$ entonces usando la hipótesis tenemos que
\[a^2=(abb^{-1})^2=(b^{-1}ab)^2=b^{-1}a^2b\]
entonces multiplicando por la izquierda por $b$ obtenemos que $ba^2=a^2b \forall a,b\in G$. Tomando $y=ba^{-1}$, $x=a$ y usando la ultima igualdad tenemos que $ba=yx^2=x^2y=a^2ba^{-1}$, entonces multiplicando por $a^{-1}$ por la izquierda obtenemos la siguiente identidad \[aba^{-1}=a^{-1}ba----(1)\]
Ahora veamos que $(aba^{-1}b^{-1})^2=aba^{-1}b^{-1}aba^{-1}b^{-1}=abab^{-1}a^{-1}ba^{-1}b^{-1}=ababa^{-1}b^{-1}a^{-1}b^{-1}$ donde las ultimas dos igualdades se deben a usar dos veces (1), sustituyendo $abab=baba$ en lo ultimo obtenemos que $(aba^{-1}b^{-1})^2=babaa^{-1}b^{-1}a^{-1}b^{-1}=e$ por lo tanto como no hay elementos de orden 2 se deduce que $aba^{-1}b^{-1}=e$ que multiplicando por $ba$ por la derecha se obtiene $ab=ba$ por lo tanto $G$ es abeliano.
Hi,this problem has a very simple solution:by Cauchy theorem the order of the group is odd,thus ab has odd order m,that is also the order of ba,since they are conjiugated,so call j the inverse of 2 mod m,you get (ab)^{2j}=ab=(ba)^{2j}=ba. Or said faster taking the 2-power is bjective if the order of the group is odd. This is true for every prime p that does not divide the order of the group. One proof is just the same as this posted here,taking inverses mod p. Another is quite funny,and i told you(Irving)while walking in Villa borghese:fix x,take the p-uple (a_1,...,a_p) in G^p,such that prod_{i=1}^{p}a_i lies in the coniugacy class of x,Cl(x),this p-uples are invariant under cyclic permutation. thus we have an action of Z_p on the set of this p-uples,but they are in cardinality C=(|G|^{p-1})(|Cl(x)|),indeed fix randomly the first p-1 and then just force the last one so that the product is in the coniugacy class. Now C is equivalent to |Cl(x)| mod p,that is different from 0(indeed |Cl(x)| divides |G|,that is coprime with p by ipothesis). So the action of Z-p,having just orbits of lenght p or 1 (by orbit stabilizer lemma),needs a number of fixed point that is equivalent(it will be equal at the end) to Cl(x) mod p. But a fixed point is a p-uple,of the form (a,a,...,a),so we have an a,such that a^p lies in Cl(x). but now conjugating you get that x is a p-power,so the map of the p power is surjective,and so bjective. With this idea,you can also prove Cauchy theorem.
ResponderEliminarBye
Carlo
p.s:this is actually off-topic,but Irving i've the doubt that you haven't received my mails,one with a solution of a problem and another with just talking. Did you received them?
Ops i read again,and i was assuming the group is finite. Pardon
ResponderEliminarOk,i felt a little bit bad to having left just a solution in a silly case with an off topic and a strange proof of a trivial fact :-). So now i try to give a natural proof of the real problem: in a group ab and ba are the tipical two element in the same coniugacy class,indeed saying ab is as to say b^{-1}(ba)b,and saying b^{-1}ab and a,is like to say b^{-1}(ab) and (ab)b^{-1}. Given this,is completely natural to reinterpeter the question as,any two elements in the same coniugacy class have the same square,there is no 2-torsion,then G is commutative. So we have x^2=y^{-1}x^2y whenever x and y are in G,that is to say x^2 is in the Center whenever x is in G. That is G/Z(G) is a group where evryone have order 2,it is easy to show that then the group is abelian,so we get yxy^{-1}=xz with z in the center, but then being y^2 in the center we get x=y^{-2}xy^2=xz^2,thus,z^2=1,thuen z=1,by ipothesis.
ResponderEliminarBye