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I want to prove:

Let $G$ be a group and let $g \in G$. Given $m,n \in \mathbb{Z},$ then $g^{n+m} = g^ng^m$.

The problem follows by generalized associativity when $n,m > 0$. I also managed to prove it when $n,m < 0$. I have troubles proving the case where $m < 0, n >0$ (or the other way). I tried direct computation but this didn't lead me anywhere. I then tried to deduce that $g^{n+m}(g^ng^m)^{-1} = e$ and $(g^ng^m)^{-1}g^{n+m} = e$ from which the solution would follow but I get always stuck.

Does anyone have a hint or solution?

Thanks in advance.

2 Answers 2

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I suppose you have already proved that $(ab)^{-1} = b^{-1}a^{-1}$ for any members $a, b$ of the group, from which it follows that $(g^n g^m)^{-1} = (g^m)^{-1}(g^n)^{-1}$ for any integers $m$ and $n.$ I also supppose you have proved that $(g^k)^{-1} = g^{-k}$ for any integer $k.$

You might consider four cases separately: one case is $n>0,$ $m<0,$ and $n+m>0,$ and in each of the other cases we reverse the sign of $n,$ $n+m,$ or both, while making the sign of $m$ opposite to that of $n.$ In the first case, $g^{-m}$ and $g^{n+m}$ both have positive exponents, so if you have a product of those two elements (in either order) you can apply the fact you have already proved about multiplying two positive powers of $g.$ In other cases you might want to apply the fact you have already proved about multiplying two negative powers.

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    So I should take the way using the inverses (my second approach)? – 2017-02-18
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    I think you can make it work by considering the appropriate cases one by one. There are other ways to do the proof, but I'm not sure if any of them is particularly easier than your second approach. – 2017-02-18
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Hint: if you have proved it for positive exponents, the note that $g^ {-n}=(g^ {-1})^{n}$, so you can reduce this to things you have already proved.

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    I tried this, but got nowhere :/ – 2017-02-18