Subgroups:
Let
(2)
\begin{align} A=\begin{pmatrix} 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0\\ \end{pmatrix} \end{align}
therefore
(3)
\begin{align} A^{-1}=\begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 0 & 0\\ \end{pmatrix} \end{align}
We know by matrix multiplication that the following applies:
1. $A\cdot A=A^{-1}$
2. $A^2=A^{-1}$
3. $A^3=I_3$
4. $A^4=A^{-1}$
5. $(A^{-1})^2=A$
This subgroup, then, is $\{I_3, A, A^2\} \cong \mathbb{Z}_3.$
When
(4)
\begin{align} \[ B=\begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1\\ \end{pmatrix}, C=\begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0\\ \end{pmatrix}, D=\begin{pmatrix} 0 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 0\\ \end{pmatrix} \] \end{align}
then $B^{-1}=B, C^{-1}=C,$ and $D^{-1}=D$ so that $B^2 = I_3, C^2=I_3,$ and $D^2=I_3.$
And so the following are subgroups:
$\{I_3, B\}, \{I_3, C\},$ and $\{I_3, D\} \cong \mathbb{Z}_2$
Let
(5)
\begin{align} $$E=\begin{pmatrix} 1 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1\\ \end{pmatrix}$$ \end{align}
We see that
(6)
\begin{align} E^{-1}=\begin{pmatrix} 0 & 1 & 0\\ 1 & -1 & 0\\ 0 & 0 & 1\\ \end{pmatrix} \end{align}
which in $\mathbb{Z}{_2}$ is equal to
(7)
\begin{pmatrix} 0 & 1 & 0\\ 1 & 1 & 0\\ 0 & 0 & 1\\ \end{pmatrix}
We also see that
(8)
\begin{align} E^{2}=\begin{pmatrix} 2 & 1 & 0\\ 1 & 1 & 0\\ 0 & 0 & 1\\ \end{pmatrix} \end{align}
which in $\mathbb{Z}{_2}$ is equal to
(9)
\begin{pmatrix} 0 & 1 & 0\\ 1 & 1 & 0\\ 0 & 0 & 1\\ \end{pmatrix}
We can also calculate
(10)
\begin{align} E^{3}=\begin{pmatrix} 3 & 2 & 0\\ 2 & 1 & 0\\ 0 & 0 & 1\\ \end{pmatrix} \end{align}
which in $\mathbb{Z}{_2}$ is equal to $I_{3}.$
Then
(11)
\begin{align} (E^{-1})^{2}=\begin{pmatrix} 0 & 1 & 0\\ 1 & -1 & 0\\ 0 & 0 & 1\\ \end{pmatrix} \end{align}
which in $\mathbb{Z}{_2}$ is equal to E.
We now know the following based on the above information:
1. $E^{2}\cdot E^{-1} = E$
2. $(E^{-1})^{2} = E$
3. $(E^{-1})^{2} \cdot E = E^{2}$
4. $E^{2} = E^{-1}$
5. $E^{-1} \cdot E^{2} = E$
Therefore we have the subgroup $\{I_{3}, E, E^{2}\}.$