G is a group and $\displaystyle x\in G$ and $\displaystyle a,b\in\mathbb{N}$.

Prove $\displaystyle x^{a+b}=x^ax^b$

I don't see how to show this. Multiplying by the inverse would just use the property I have to prove.

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- Sep 9th 2011, 05:44 PM #1

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- Sep 9th 2011, 05:53 PM #2
## Re: x^{a+b}=x^ax^b

Could you just write this:

$\displaystyle x^{a+b}=\underbrace{x\cdot x\cdot...\cdot x}_{a+b\text{ times}}=\underbrace{x\cdot x\cdot...\cdot x}_{a\text{ times}}\cdot\underbrace{x\cdot x\cdot...\cdot x}_{b\text{ times}}=x^{a}x^{b},$

and claim you're done? I would definitely claim that there isn't much to show here. The proof can't be that complex.

- Sep 10th 2011, 03:41 AM #3

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