Ok, let's make it more interesting!

 

Let [math]G[/math] be a group, and suppose there exist three consecutive integers: [math]n-1, n, n+1[/math] such that: [math](ab)^{n-1} = a^{n-1} b^{n-1}, (ab)^n = a^n b^n, (ab)^{n+1} = a^{n+1} b^{n+1}, \forall a,b \in G[/math]. Show that [math]G[/math] is abelian. Also show that such a conclusion needs not hold if the condition is assumed for only two consecutive integers.

 

PM me if you want the answer!

Any reason you keep posting random problems? I know you say they aren't homework, but it certainly seems like they might be.

Ok, let's make it more interesting!

 

Let [math]G[/math] be a group, and suppose there exist three consecutive integers: [math]n-1, n, n+1[/math] such that: [math](ab)^{n-1} = a^{n-1} b^{n-1}, (ab)^n = a^n b^n, (ab)^{n+1} = a^{n+1} b^{n+1}, \forall a,b \in G[/math]. Show that [math]G[/math] is abelian. Also show that such a conclusion needs not hold if the condition is assumed for only two consecutive integers.

 

 

Making it interesting would be good. But this isn't interesting -- just parlor tricks.

 

We have [math](ab)^{n-1} = a^{n-1} b^{n-1}, (ab)^n = a^n b^n, (ab)^{n+1} = a^{n+1} b^{n+1}, \forall a,b \in G[/math] and taking inverses [math](ab)^{1-n} = b^{1-n} a^{1-n}, (ab)^{-n} = b^{-n}n a^{-n}, (ab)^{-n-1} = b^{-n-1} a^{-n-1}, \forall a,b \in G[/math]

 

So

 

[math] ab = (ab)^n(ab)^{1-n} = a^nb^nb^{1-n}a^{1-n} = a^nba^{1-n}[/math]

 

[math] ab = (ab)^{n+1}(ab)^{-n} = a^{n+1}b^{n+1}b^{-n}a^{-n} = a^{n+1}ba^{-n}[/math]

 

Thus [math] a^nba^{1-n} = a^{n+1}ba^{-n}[/math] multiplying on the left by [math]a^{-n}[/math] and on the right by [math]a^n[/math] we have [math] ba=ab[/math] . QED

 

 

As to the insufficiency of two of the three conditions, consider a non-abelian group [math]G[/math] of order n, for some n. The last two conditions are satisfied. since [math] a^n=1 \ \ \forall a \in G[/math]

 

I am about through with this sort of thing.

!

Moderator Note

moved to homework help. Please read our rules and post threads in their rightful place, Obelix; don't evade our rules by posting homework problems in other forums.

I am about through with this sort of thing.

 

There, there, DrRocket!