Prove that the following set of matrices under $\mathbb{Z}_2$ form a field. I'm trying to prove that the following set of matrices over $\mathbb{Z}_2$ forms a field:
$
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix},
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix},
\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 1 & 0
\end{pmatrix},
\begin{pmatrix}
1 & 1 & 0 \\
0 & 1 & 1 \\
1 & 1 & 1
\end{pmatrix},\\
\begin{pmatrix}
0 & 0 & 1 \\
1 & 1 & 0 \\
0 & 1 & 1
\end{pmatrix},
\begin{pmatrix}
1 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{pmatrix},
\begin{pmatrix}
0 & 1 & 1 \\
1 & 1 & 1 \\
1 & 0 & 1
\end{pmatrix},
\begin{pmatrix}
1 & 1 & 1 \\
1 & 0 & 1 \\
1 & 0 & 0
\end{pmatrix}.
$
The only way I can think of proving this is by adding and multiplying every single matrix to every other matrix to show that it's closed, that is going through every single case to see if each such case satisfies the field laws. Is there a more elegant/efficient way of showing the statement is true?
 A: HINT:
Since distributivity is obvious, we only have to show that this thing forms an abelian group under addition, and that this thing with the zero element excluded forms a abelian group under multiplication.
So why don't we invest some thought into figuring out what these groups could potentially be?
Additive group. We know it's abelian, and it's obvious from what the elements are that each non-zero element has order two. The only abelian group where all elements are of order two is $\mathbf Z_2 \oplus \mathbf Z_2 \oplus \mathbf Z_2$.
So I suggest you find three elements $v_1, v_2, v_3$ such that every one of the eight matrices can be written as $c_1 v_1 + c_2 v_2 + c_3 v_3$ for $c_1, c_2, c_3 \in \mathbf Z_2$.
Once you've done this, it will be obvious that the set forms an abelian group under addition.
Multiplicative group. This contains seven elements, and seven is prime, so it must be cyclic, and every non-identity element should be a generator.
So why not pick a non-zero, non-identity matrix $g$ from the set, and show that the powers of $g$ are all of the non-zero elements of the set?
Once you've done this, it will be obvious that the non-zero elements of the set form an abelian group under multiplication.
