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The multiplicative group of any finite field is cyclic, a generator of this group is called a primitive element of the finite field. Is it true that the number of primitive elements in the finite field $GF(s^2)$ is s+1? If it holds, how can I prove it?

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up vote 2 down vote accepted

The multiplicative group of $GF(p^n)$ has order $p^n-1$. An element $a+m\mathbb Z$ of the cyclic group $\mathbb Z/m\mathbb Z$ generates the full group iff $a$ and $m$ are relatively prime. Hence there are $\phi(m)$ generators of $\mathbb Z/m\mathbb Z$. Finally, this shows tha there are $\phi(p^n-1)$ primitive elements in $GF(p^n)$. For the case $n=2$ we know that $p^2-1=(p+1)(p-1)$, which makes $\Phi(p^2-1)=p+1$ highly unlikely. In fact, this holds for $p=3$ only, I guess.

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