# Find generator of multiplicative group of $\mathbb{F}_{27}$

I have a finite field $\mathbb{F}_{27}$. I need to find an element of order 13. I know that multiplicative group of this field is cyclic with order 26. So I want to find a generator $g$ of this cyclic group, then $g^2$ will have order 13. Also I know that all elements of this field can be represented by polynomials from $\mathbb{F}_3[t]$, where t is root of irreducible polynom $x^3-x-1$.

Probably we can find generator of $\mathbb{Z}_{26}$ and than build an isomorphism to $\mathbb{F}_{27}\setminus 0$.

Anyway I don't know were to start.

Thanks!

Start with finding the order of, say, $t$. It can't be $2$ (since $t^2-1\neq0$), so it's either $13$ or $26$. So either $t$ or $t^2$ is the one you're looking for.

• And how to find order of t? Just raise it to a 13th power? – Kirill Losev May 6 '18 at 15:41
• Calculate (simplify) $t^{13}$ using $t^3=t+1$, and see whether it's equal to $1$. If it's not equal to $1$, then $t$ must have order $26$, so $t^2$ has order $13$. (Side note: If $t$ does have order $13$, then $-t$ has order $26$, and vice versa.) – Arthur May 6 '18 at 15:49
• Well, $t$ has order 13. Thanks! – Kirill Losev May 6 '18 at 15:52
• btw, why $t^2\neq 1$ ? – Kirill Losev May 6 '18 at 15:59
• @Kiril I meant that $x^3-x-1$ is the minimal polynomial of $t$ over $\Bbb F_3$, so it can't be a root of $x^2-1$, by definition of minimal. That was a bit unclear, sorry. Over $\Bbb F_{27}$, the polynomial $x^2-1$ can only have two roots, and $t$ is not one of them. – Arthur May 6 '18 at 17:19

Nothing wrong with Arthur's answer. But the Frobenius automorphism (or Freshman's dream) makes the calculation very easy:

Let $t$ be a zero of $x^3-x-1$. Then $t^3=t+1$. Consequently $$t^9=(t^3)^3=(t+1)^3=t^3+1=t+2=t-1.$$ Implying that $$t^{13}=t^{1+3+9}=t(t+1)(t-1)=t^3-t=1.$$ As $13$ is a prime and $t\neq1$, we can conclude that $t$ has order $13$.

It may be worth noting that $-t$ is then of order $26$, just in case you need a generator of $\Bbb{F}_{27}^*$.