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Let $q$ prime power and $\alpha \in GF(q^m)$ be primitive element. When is $c\alpha$ still primitive in $GF(q^m)$?

EDIT: More generally, if $\alpha \in GF(q^m)$ is any non-zero element, I'm also wondering if we can say something about the order of $c\alpha, c\in GF(q)^*$ with respect to the order of $\alpha$..

geo909
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    The units of a finite field are a cyclic group: it may help to convert the question into a question about cyclic groups without reference to fields. –  Jun 27 '14 at 19:41

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Let $$N=(q^m-1)/(q-1)=1+q+q^2+\cdots+q^{m-1}.$$ Then $\alpha^N$ is primitive in $GF(q)$, so $c=\alpha^{kN}$ for some $k, 0\le k<q-1$. Thus $c\alpha=\alpha^{1+kN}$. By the properties of cyclic groups this is a primitive element of $GF(q^m)$ if and only if $$d=\gcd(1+kN,q^m-1)=1.\qquad(*)$$ Here $$ (q-1)(1+kN)-k(q^m-1)=(q-1), $$ so any common divisor $d$ of $1+kN$ and $q^m-1$ is also a divisor of $q-1$. As $q-1\mid q^m-1$ the condition $(*)$ is satisfied, iff $\gcd(1+kN,q-1)=1$. Clearly $N\equiv m\pmod{q-1}$, so we get that $(*)$ is equivalent to $$ \gcd(1+km,q-1)=1.\qquad(**) $$ I can't say more in general. At least not right away. A fun special case is when all the prime factors of $q-1$ are also factors of $m$. Then $(**)$ is automatic for all $k$, and we can say that $c\alpha$ is primitive for all non-zero $c$.

Jyrki Lahtonen
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