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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 $2^3-1$ ?

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    $\begingroup$ Yes, the order of $(\Bbb F^{\times},\cdot)$ is the order of the field $\Bbb F$ minus $1$, because we have to take out $0$. $\endgroup$ – Dietrich Burde May 17 at 19:36
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    $\begingroup$ No, a cyclic group $C_n$ has $1$ generator, but the order is $n$. $\endgroup$ – Dietrich Burde May 17 at 19:38
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    $\begingroup$ @DietrichBurde $C_n$ has $\varphi(n)$ generators. $\endgroup$ – Chris Custer May 17 at 19:47
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    $\begingroup$ 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). $\endgroup$ – Bernard May 17 at 23:36
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    $\begingroup$ $C_n$ 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. $\endgroup$ – Dietrich Burde May 18 at 10:55
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Yes. The order of a group is how many elements the group has. Since the multiplicative group of a field is every element except 0 the order of the group is exactly one less than the amount of elements in the field.

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