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Suppose k be a positive integer such that k divides p-1, where p is a prime. Prove that $\mathbb{Z}^*_p$ has an element of order k.

I think this is suppose to be the converse of Lagrange.

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Hint: If $k$ divides $p-1$ then we must have an element $m$ such that $km=p-1$. Now take an element in your group, say $a$. Then, $(a^m)^k=a^{(p-1)}=1$, by Fermat's Little theorem.

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  • $\begingroup$ This shows that $a^m$ has order at most $k$, maybe not actually $k$. $\endgroup$ – David Feb 10 '14 at 1:29
  • $\begingroup$ @David: That's why this is a hint. It's easy to get over this difficulty. It's not right to give the full solution for a HW problem. $\endgroup$ – voldemort Feb 10 '14 at 1:30
  • $\begingroup$ @David: In fact, unfortunately your and mine solution can be combined to give a full solution for this HW problem :( $\endgroup$ – voldemort Feb 10 '14 at 1:31
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Hint. Suppose that $p-1=km$ where $m$ is a positive integer. Let $g$ be a primitive root (or primitive element, or generator) modulo $p$. Then see what you can say about $$g^m\,,\ g^{2m}\,,\ \ldots\,,\ g^{(k-1)m}$$ and $$g^{km}\ .$$

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