# Group order calculation

The number of elements in a group G is called group order. It specifies the number of elements in the group and is denoted |G|.

Example: $\mathbb{𝔽}_3={0,1,2},|\mathbb{𝔽}_3|=3$

There is a law which states that any group g element to the power of the group order is 1.

So let's try: g=2

$2^3 \equiv 2 \mod 3$. The result is 2 and not 1. Why?

On the other hand I now, that $|\mathbb{𝔽}_p|=p−1$ for $p$ prime. Does it mean that my upper definition of group order is not correct, though I read it everywhere?!

• Here the group operation is addition (modulo $3$), not field multiplication; indeed, the "$3$rd power" of $g = 2$ is $g + g + g = 2 + 2 + 2 = 0$. – Travis May 31 '15 at 9:50
• Whenever raising an element to some power occurs in a group, the group is assumed to be multiplicative. The field $\Bbb{F}_3$ is a group only under addition, so instead of power (repeated multiplication), you need to do repeated addition aka integer multiple. $2+2+2=0$ in your group, and the law holds. – Jyrki Lahtonen May 31 '15 at 9:51
• I understand. Thanks. But why is $|\mathbb{F}_p| = p-1$? There are p elements and not p-1. – null May 31 '15 at 9:57

As mentionned in the comments to check Lagrange's theorem in $\mathbb{F}_p$ you need to consider the addition and not multiplication ($0$ has no inverse for the multiplication, hence ($\mathbb{F}_p,\times$) is not even a group).
Now if you have been told that $|\mathbb{F}_p|=p-1$ it is false. The cardinal is $p$.
However $|\mathbb{F}_p^*|= p-1$ where $\mathbb{F}_p^*$ is the field without the zero element. For any element $x$ in this set, we have $x^p=x$ and hence $x^{p-1}=1$ which is coherent with Lagrange's theorem because $(\mathbb{F}_p^*,\times)$ is a group.