I have a proof for the following proposition:

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


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?

  • $\begingroup$ 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

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.


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 $m, n \in \Bbb Z$,

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.