Elements with order 3 in group $F_{16}/\{0\}$

Question

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.

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. — Jyrki Lahtonen, Jun 15 '14 at 14:51

score 2 · Accepted Answer · answered Jun 15 '14 at 14:44

$\mathbb F_{16} - \{0\}$ 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.

score 1 · Answer 2 · answered Jun 15 '14 at 14:51

1

Hint:

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

answered Jun 15 '14 at 14:51

DonAntonio

211,718
17
136
287

Elements with order 3 in group $F_{16}/\{0\}$

2 Answers2