Can anyone tell my, why we have to mod $r$ ? Thank you so much.
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Since $a^{m+r}=a^m$ then all operations with exponents are performed on $\mod r$.
Boris Novikov
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Do I am right? Now, we consider elements in $K_{a}$ and know that if $a^{n}=a^{m}$, then $n=m$. Thus we have $m+u+m+x=m+v$. I struck here and I cannot connect it to $\mod r$. – Mr.Lilly Sep 10 '13 at 16:14
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1No. You can only say that if $a^s = a^t$ is an element of $K_a$, then $s \equiv t \bmod r$. For instance, you have $a^m = a^{m+r}$, but $m \not= m +r$. – J.-E. Pin Sep 10 '13 at 17:18
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Thanks. And I have a question. Are the elements in $K_{a}$ all distinct? – Mr.Lilly Sep 10 '13 at 17:31
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@J.-E. Pin: thank you for help :-) – Boris Novikov Sep 10 '13 at 17:33
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@nameless: Yes, they are distinct. – Boris Novikov Sep 10 '13 at 17:34
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If all elements in $K_{a}$ are all distinct, then for two elements in $K_{a}$, $a^{m+u}, a^{m+v}$ such that $a^{m+u}=a^{m+v}$. Why we don not have $m+u=m+v$? – Mr.Lilly Sep 10 '13 at 17:44
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1@nameless: All distinct elements of $K_a$ are: $a^m,\ldots,a^{m+r-1}$. So, for example, $a^{m+r}=a^m\in K_a$, but $m+r\ne m$! – Boris Novikov Sep 10 '13 at 18:08
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I got it!!! Thank you. – Mr.Lilly Sep 10 '13 at 18:21
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@nameless: You are welcome! – Boris Novikov Sep 10 '13 at 18:23