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If you have the finite field $GF(16)$ and you define the group $GF(16)/\{0\},*$ this group is cyclic. I need to determine how many elements in this group have order 3. Of course you could just try out all elements but I wonder if there isn't a more efficient technique. I don't need to know which elements have order 3, just how many.

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    $\begingroup$ Martini's answer (or DonAntonio's) is perfect for your purposes. If you actually want to find those elements, then you can take a peek here. $\endgroup$ – Jyrki Lahtonen Jun 15 '14 at 14:51
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$\mathbb F_{16} - \{0\}$ has 15 elements. A cyclic group with $15$ elements is isomorphic to $\def\Z{\mathbb Z}\Z/(15) \cong \Z/(3) \times \Z/(5)$. An element $(a,b)$ of the latter group has order $3$ if $b = 0$ and $a \ne 0$. Hence there are two of them.

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Hint:

Any cyclic group of order $\;n\;$ has one unique subgroup of order $\;k\;$ for any $\;k\mid n\;$ .

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