# Calculate order of multiplicative group of finite field

How can one calculate the order of a multiplicative group of a finite field such as:

$$(\mathbb{F}(2^3) \backslash \{0\}, \times)$$

Is it as simple as doing $$2^3-1$$ ?

• Yes, the order of $(\Bbb F^{\times},\cdot)$ is the order of the field $\Bbb F$ minus $1$, because we have to take out $0$. – Dietrich Burde May 17 at 19:36
• No, a cyclic group $C_n$ has $1$ generator, but the order is $n$. – Dietrich Burde May 17 at 19:38
• @DietrichBurde $C_n$ has $\varphi(n)$ generators. – Chris Custer May 17 at 19:47
• Yes. In your case, $\mathbf F_8^\times$ is a cyclic group with $7$ elements, and its generators are $\varphi(7)=6$. In other words any element $\ne 0,1$ is a generator (and it has order $7$ since it is a generator). – Bernard May 17 at 23:36
• $C_n$ is only generated by one single element, i.e., "it has one generator". This sentence is misleading. Of course we can chose $\phi(n)$ different elements for this one generator. – Dietrich Burde May 18 at 10:55