Suppose $g$ is a field of cardinality $64$. Let $g^\times$ be a multiplicative group of invertible elements of order $63$ and let $c \geq 0$ be an integer. Show that $g^\times$ contains at most $c$ elements of order $c$.
I know that an element of order $c$ must satisfy an equation but I'm not entirely sure where to go from here... any help would be appreciated!
You have an unusual notation, lower case $g$ for a field. Let me switch to more familiar $F$ for a field. Let $f(x)$ be a polynomial, say of degree $n>1$, then for every root $a\in F$, by applying remainder theorem, $f(x) = (x-a) f_1(x)$ for a polynomial $f_1(x)$ of degree $n-1$ having coefficients in $F$ again.
This shows that the polynomial $X^n-1=0$ has at most $n$ roots. These roots among them contain all elements of multiplicative order $n$ (and some thing which are not), and this proves what you need.
EDIT: This statement is false for rings which are not fields. In the ring R=Z/8Z the polynomial $f(X) = X^2-1$ though of degree 2, has 4 solutions, namely $X=1,3,5,7$. So the multiplicative group of units of $R$ has 3 elements of order 2 (which along with 1 is a group isomorphic to Klein's four-group).
