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I have a proof for the following proposition:

Suppose $G$ is a group, $g\in G$ and $m,n \in \mathbb{Z}$. Then $(g^m)^n=g^{mn}$

Proof

If $m$ and $n$ are positive then it is clear that $(g^m)^n=g^{mn}$ since $$(g^m)^n=\underset{\text{m times}}{\underbrace{(gg\cdot\cdot\cdot g)}}^n=\underset{\text{mn times}}{\underbrace{(gg\cdot\cdot\cdot g)}}$$ *Not sure on this part, might be better with induction.

Now if $m$ and $n$ are both negative, then $g^m=(g^{-m})^{-1}$ by definition and we also know $$(g^{-m})^{-n}=g^{mn}$$ since all exponents are positive in this equation. Therefore $$(g^m)^n=((g^{-m})^{-1})^{n}=((g^{-m})^{-1})^{-1(-n)}=((g^{-m})^{-1(-n)})^{-1}=((g^{mn})^{-1})^{-1}=g^{mn}$$ Last part of the equation from the fact of unique inverses in groups. Also not sure on this step.

Now if $m<0$ and $n$ positive then we know $$(g^{-m})^{n}=g^{-mn}$$ since all the exponents are positive. Therefore $$(g^{m})^n=((g^{-m})^{-1})^n=((g^{-m})^n)^{-1}=(g^{-mn})^{-1}=g^{mn}$$ Now if $m$ positive and $n<0$ we have $$(g^m)^{-n}=g^{-mn}$$ $$\therefore (g^m)^n=(g^m)^{-1(-n)}=(g^{-mn})^{-1}=g^{mn}$$

Now does this seem right? Firefox crashed twice whilst creating this question so I hope so. Any thoughts on how I could improve it?

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  • $\begingroup$ You are making this way more complicated than it really is. For $m, n$ both positive it is as obvious as that because you are simply counting how many $g$ there is. $\endgroup$ – Jack Yoon Apr 24 '14 at 15:07
  • $\begingroup$ I thought so, my tutor is a stickler for stupid amounts of rigor in proofs though and just wanted to make sure $\endgroup$ – George1811 Apr 24 '14 at 15:09
  • $\begingroup$ It might be easiest to induct on $n$ (the outer exponent) $\endgroup$ – Prahlad Vaidyanathan Apr 24 '14 at 15:11
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Your proof seems fine; however, for a more rigorous approach, follow the hint by @PrahladVaidyanathan. This question has been asked on MSE before and the inductive argument seems to be favourable.

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We start by accepting that for any $g$ and $m,n \in \Bbb Z$ we can unambiguously define $g^n$, writing as true

$\tag 1 g^0 = \text{ the identity element } \text{id}_G = 1_G \text{ in the group } G$

$\tag 2 g^1 = g$

$\tag 3 g^{-1} = \text{ satisfies } g \, g^{-1} = 1_G$

$\tag 4 g^{-n} = {(g^n)}^{-1} = {(g^{-1})}^{n} $

$\tag 5 g^{m} g^{m} = g^{m+n}$

Now prove the folloowing

Proposition 1: For every $m, n \in \Bbb Z$, if $m \ge 0$ and $n \ge 0$

$\tag 6 {(g^{m})}^{n} = g^{mn}$

Theorem 2: For every $m, n \in \Bbb Z$,

$\tag 7 {(g^{m})}^{n} = g^{mn}$

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