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Suppose that $a$ is an element of order $n$ in a group $G$. Prove:

i) $a^i = a^j$ if and only if $i \equiv j \pmod n$;

ii) if $d = (m,n)$, then the order of $a^m$ is $n/d$;

I was trying to self teach myself this and came to this question. How would you solve this? Can someone please show how to?

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3  
What's up with these irrelevant tags? You left out [calculus] and [category-theory] while you're at it. –  Asaf Karagila Apr 3 '13 at 17:47
    
@AsafKaragila: Like typing L' Hospital instead of L' Hopital. :D –  Babak S. Apr 3 '13 at 17:48
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@Babak: Actually writing L'Hospital is correct if you don't add an accent to the "o". Like writing Goedel if you don't have an umlaut key for the "o". –  Asaf Karagila Apr 3 '13 at 17:51
    
I see. Thanks Asaf. –  Babak S. Apr 3 '13 at 17:53

2 Answers 2

If $i \equiv j \pmod n$, then $i = j + kn$ for some $k \in \Bbb Z$. It follows that: $$ a^i = a^{j + kn} = a^ja^{kn} = a^j (a^n)^k = a^j $$

If you haven't proved the power properties $a^{p+q} = a^pa^q$ and $a^{pq} = (a^p)^q$ for $p, q \in \Bbb Z^+$, this is a good exercise to do now. Try using induction.

Now, if $a^i = a^j$, then: $$ a^i a^{-j} = a^{i - j} = 1 $$

And this is only possible if $i - j$ is a multiple of $n$. (i) follows.

The property $(a^n)^{-1} = a^{-n}$ was used here. Again, try proving it via induction.


For (ii), assume the order of $a^m$ is a number smaller than $n/d$, and try to use (i) to reach a contradiction.

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Hint $\rm\ (i)\ \ \ a^i = a^j\iff a^{i-j} = 1\iff n\mid i-j,\:$ since $\rm\ n = ord(a)$

$\rm(ii)\ \ \ a^{mk}\! = 1\iff n\mid mk\iff n\mid mk,nk\iff n\mid (mk,nk)=(m,n)k=dk \iff n/d\mid k$

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