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Exercise:

Prove, that for every element $a$ of the group: $(a^m)^{-1}=(a^{-1})^m$, where $m$ is natural.

On the #23 was proven that $(a_1 \centerdot \ldots \centerdot a_n)^{-1} = a_n^{-1} \centerdot \ldots \centerdot a_1^{-1}$ and I believe that #26 directly follows from this, when $a_i=a$ for all $i$ from $1$ to $n$. Would you be so kind to tell me am I right? (Author gives another, not so simple answer.)

Thanks.

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    I think author tries to give a self-contained answer to 26 and do it in new approach (with induction), instead of just saying "obviously it follows from 23". Roughly quoting his solution: "we need to show that (a^m) (a^-1)^m = e, [by definition of powers of a] (a^m) * (a^-1)^m = (a^m-1) * a * a^-1 * (a^-1)^m-1 = ... = a * a^-1 = e, and the same for another order (a^-1)^m * a^m = e". So, no worries, using 23 is equivalent.2017-05-23

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You are correct, assuming that $m$ is a natural number (if $m=0$ it doesn't follow from #26 but its not hard to see why its true in this case).

You can also prove this easily from the definition, just recall that $a,a^{-1}$ commute beetwen them

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    @AlekS - I'm glad to hear that :)2012-12-03