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I need to prove that having group $G$ where for each $a,b\in G$ the statement $(ab)^{n}=a^{n}b^{n}$ is true for three consecutive and natural $n$ numbers then the group $G$ is abelian.

The prove in the textbook I am using reads like this:

If $m, m+1$ and $m+2$ are these three consecutive and natural numbers we know that $(ab)^{m}=a^{m}b^{m}$ and $(ab)^{m+1}=a^{m+1}b^{m+1}$.

Then the proving continues like this: $a^{m+1}b^{m+1} = (ab)^{m+1} = (ab)^{m}ab = a^{m}b^{m}ab$ and simplifying we have $ab^{m}=b^{m}a$.

How was this simplification done? How did it get from $a^{m+1}b^{m+1} = (ab)^{m+1} = (ab)^{m}ab = a^{m}b^{m}ab$ to $ab^{m}=b^{m}a$?

I can see that it removes $ab$ from both sides of $(ab)^{m}ab = a^{m}b^{m}ab$ but I can't figure out why it finally comes up with $ab^{m}=b^{m}a$.

Haritz
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  • you have $a^{m+1} b^{m+1} = a^m b^m ab$, so if you multiply with $a^{-m}$ "from the left" and $b^{-1}$ "from the right" you obtain $a b^m = b^m a$. – M.U. Sep 27 '15 at 13:44

1 Answers1

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$$a^{m+1}b^{m+1}=(ab)^{m+1}=(ab)^m\cdot ab = a^m b^m ab$$ Multiplying both sides from left by $a^{-m}$ we get $$ab^{m+1}=b^m.\cdot ab$$ Now multiplying both sides of this from right by $b^{-1}$ we get $$ab^m=b^m a$$

Similarly , $$a^{m+2}b^{m+2}=(ab)^{m+2}=(ab)^{m+1}\cdot ab=a^{m+1}b^{m+1}ab$$ And multiplying both sides by , $a^(-{m+1})$ from left and $b^{-1}$ from right , we get $$a\cdot b^{m+1}=b^{m+1}\cdot a$$ or, $$ab^{m+1}=b^{m+1} a=b\cdot (b^m\cdot a)=b\cdot (ab^m)$$ Now multiplying both sides by $b^{-m}$ from right we get $$a\cdot b=b\cdot a$$

Hence the group $G$ is commutative.

user118494
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