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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.)


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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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Thank you. It looks obvious, but I tend to see obvious answers, where they aren't, so need to check. Many thanks. Answer in textbook was exactly you purpose. – AlekS Dec 3 '12 at 11:03
@AlekS - I'm glad to hear that :) – Belgi Dec 3 '12 at 11:17

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