-2

Assume there exist $2$ natural numbers that are coprime ($m$ and $n$ such that $(m,n)=1$) such that for each $g$, $h \in G$ we have $g^m h^m = h^m g^m$ and $g^n h^n = h^n g^n$. Then $G$ is abelian group. (That means $ab=ba$.)

Hint: $(m,n)=1$ then there exist $c,d\in\mathbb{Z}$ s.t. $cm+dn=1$. thus $$gh = g^{cm+dn} h^{cm+dn} = g^{cm} g^{dn} h^{cm} h^{dn} = (g^m)^c (g^n)^d (h^m)^c (h^n)^d = (g^n)^d (g^m)^c (h^m)^c (h^n)^d$$ Use property in hypothesis to reach $gh=hg$.

Please help me to continue.

Community
  • 1
dibak chopra
  • 33

2 Answers2

4

HINT: you need to show that $g^{nd}h^{nd}=h^{nd}g^{nd}$ and $g^{cm}h^{cm}=h^{cm}g^{cm}$.

Bobby
  • 7,396
  • 1
    A good way to start is $g^dg^dh^dh^d=g^dh^dg^dh^d=h^dh^dg^dh^d=h^dh^dg^dg^d$. – Bobby Jun 06 '15 at 22:02
  • are you have simle way to proof?if we have same exponent or assumation in hypothesis we can commute – dibak chopra Jun 07 '15 at 19:21
  • Try this $gh = g^{cm+dn} h^{cm+dn} = g^{dn} g^{cm} h^{cm} h^{dn} = (g^d)^n (g^c)^m (h^m)^c (h^n)^d$. – Bobby Jun 07 '15 at 20:50
  • here problem is commutative ...what theorems help us to commute elements in group?same exponent? – dibak chopra Jun 07 '15 at 21:53
  • we use the same exponent property and show the hint. – Bobby Jun 08 '15 at 06:01
  • hint that you introduce when it is perform that we are same exponent but in your hint in comment not exist same exponent – dibak chopra Jun 08 '15 at 17:15
  • i confused...i read soloution about this problem in here:http://math.stackexchange.com/questions/326702/how-to-prove-that-a-group-with-some-properties-is-abelian/327766# is there anyone to give me best answer and solution of this problem – dibak chopra Jun 08 '15 at 19:58
0

Try starting with $$gh = g^{dn+cm}h^{cm+dn} = (g^d)^n(g^c)^m(h^c)^m(g^d)^n,$$ then apply the identity together with some regrouping.

rogerl
  • 22,399