How is the multiplicative structure of $F_4$ same as that of `Z mod 3`

In this video - https://youtu.be/BbxsiGjbYD4?t=47

he talks about the Finite Field $F_4 = \{0, e, a, b\}$

A little ahead at this point (https://youtu.be/BbxsiGjbYD4?t=165), he says that the multiplicative structure of the field is same as the multiplicative structure of Z mod 3 (He writes it as $Z_3$)

I don't understand how this is true. The multiplicative structure of Z mod 3 has only 2 elements 1 & 2 in the multiplicative structure (zero does not have an inverse) while the multiplicative structure of this group has 3 elements {e, a, b}

EDIT: If he means the whole of $Z_3$, even then how is the same?

For $F_4$, the Cayley Table is

\begin{array} {|r|r|}\hline * & e & a & b \\ \hline e & e & a & b \\ \hline a & a & b & e \\ \hline b & b & e & a \\ \hline \end{array}

For "whole of $Z_3$" it is

\begin{array} {|r|r|}\hline * & 0 & 1 & 2 \\ \hline 0 & 0 & 0 & 0 \\ \hline 1 & 0 & 1 & 2 \\ \hline 2 & 0 & 2 & 1 \\ \hline \end{array}

These don't look like similar structures.

Or does he mean that the multiplicative structure of $F_4^\times$ is same as the additive structure of $Z_3$?

Solution 1:

Or does he mean that the multiplicative structure of $F_4^\times$ is same as the additive structure of $Z_3$?

Yes, that's exactly what he means. He even wrote "$F_4^\times\simeq\mathbb{Z}_3$" in the video.

First of all, because that is true (other variants are simply not).

But also because he is talking about groups. It is standard notation that $R^\times$ denotes the multiplicative group of a ring $R$ (the group of all units, in case of field $F^\times=F\backslash\{0\}$) with multiplication as group operation. On the other hand when someone treats a ring $R$ as a group, then typically what he means is "with addition".

Solution 2:

Or does he mean that the multiplicative structure of $F^\times_4$ is same as the additive structure of $Z_3$?

Without watching the video: this is almost certainly what he meant, since it is true, and more or less obvious once you assume that $F^\times_4$ are the units of a field with four elements. If he explicitly said "the multiplicative structure of $\mathbb Z\text{ mod }3$" then he probably misspoke.