Group theory problem

[Obelix]

Ok, let's make it more interesting!

 

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

[DJBruce]

 

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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.

[DrRocket]

  On 5/23/2011 at 12:03 AM, Obelix said:

Ok, let's make it more interesting!

 

Let $G$ be a group, and suppose there exist three consecutive integers: $n-1, n, n+1$ such that: $(ab)^{n-1} = a^{n-1} b^{n-1}, (ab)^n = a^n b^n, (ab)^{n+1} = a^{n+1} b^{n+1}, \forall a,b \in G$. Show that $G$ 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 $(ab)^{n-1} = a^{n-1} b^{n-1}, (ab)^n = a^n b^n, (ab)^{n+1} = a^{n+1} b^{n+1}, \forall a,b \in G$ and taking inverses $(ab)^{1-n} = b^{1-n} a^{1-n}, (ab)^{-n} = b^{-n}n a^{-n}, (ab)^{-n-1} = b^{-n-1} a^{-n-1}, \forall a,b \in G$

 

So

 

$ab = (ab)^n(ab)^{1-n} = a^nb^nb^{1-n}a^{1-n} = a^nba^{1-n}$

 

$ab = (ab)^{n+1}(ab)^{-n} = a^{n+1}b^{n+1}b^{-n}a^{-n} = a^{n+1}ba^{-n}$

 

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

 

 

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

 

I am about through with this sort of thing.

[mooeypoo]

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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.

[baxtrom]

  On 5/23/2011 at 5:57 AM, DrRocket said:

I am about through with this sort of thing.

 

There, there, DrRocket!