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I have the following question :

$G$ is a group ִ$a\in G$ while $O(a)=n$ then $O(a)=O(a^{-1})$

What I did :

We know that :

$(a^n)=e$

we know that

$(a^n)^{-1}=(a^{-n})^=(a^{-1})^n$

Well, I don't know that to do now. I'm not sure I approach this correctly.

Any ideas?

Any help will be appreciated.

JaVaPG
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3 Answers3

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You have already shown that $$(a^{-1})^n=e^{-1}=e$$ so you know that $O(a^{-1})\leq O(a)$.

Now, set $b=a^{-1}$. You know that $O(b^{-1})\leq O(b)$, so $O(a)\leq O(a^{-1})$. Together with the previous inequality, you have equality.

5xum
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  • Can you explain why we know that $O(a^{-1})\leq O(a)$? – JaVaPG Jan 03 '17 at 10:33
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    @JaVaPG Because $n$ is the order of $a$, and $(a^{-1})^n = e$. Now, you know that $O(a^{-1})$ is the smallest value of $k$ such that $(a^{-1})^k = e$, so $O(a^{-1})$ must be smaller or equal than $n$. – 5xum Jan 03 '17 at 10:35
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You've just shown $(a^{-1})^n=e$.

If you know $a^k=e \iff (a^{-1})^k=e$, what happens if $k<n$?

David P
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As you've shown, we have: $$(a^{-1})^n = (a^n)^{-1} = e^{-1} = e.$$