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Prove that, If o(a)=n for some $a \in G$, then $a^m=e$ if n is a divisor of m.
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written 3.1 years ago by
prajapatijaimin
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Solution:
Suppose that n is a division of m then there exist an integer q, such that $q=\frac{m}{n}$ or, $n q=m$.
Now, $a^m=\left(a^n\right)^q=e^q=e$
If $O(a)=n$ and n is a divisor of m then $a^m=e$.
Now, $\quad a^m=e$
$ \Rightarrow \quad O(a)=m $
Hence, $m=n q$ i.e. n is a …
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