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This is part of practice midterm that i have been given ( our prof doesn't post any solutions to it) Id like to know whats right before i right the midterm on monday this was actually a 4 part question im post just 1 piece as a question in its own cause it was much too long.

Let $G$ be an abelian group.

Suppose that $a$ is in $G$ and has order $m$ (such that $m$ is finite) and that the positive integer $k$ divides $m$.

(iii) Suppose that $a$ is in $G$ and has order $m$. show that, for any positive integer $l$, the order $a^l$ divides $m$.

as show below if $l>m$ l mod m = k for some k s.t $0<k<m+1$ and we know that the order of $<a^{k}>$ and we know that the order of k is $m/k$ which is also the order of l

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2 Answers 2

up vote 2 down vote accepted

The point is, if you pick $l>m$, you can find $k\equiv l\mod m$ where $1\leq k\leq m$.

Then, $a^k=a^l$. Now, you only needs to deal with $a^k$.

The proof of the theorem is not hard. If the order of $a^k$ does not divides $m$, $(a^k)^m\neq e$. On the other hand, $(a^k)^m=a^{mk}=(a^m)^k=e^k=e$, contradiction.

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Thank you guys, i think my biggest problem is actually translating the question into something i can read, thank you very much for the help. –  Faust7 Mar 17 '13 at 5:41

Hint: Note that if $|a|=n<\infty,~~a\in G$ and $(m,n)=d$ then $$|a^m|=\frac{n}{d}$$

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