Find elements of group permutation that is isomorphic to $U(14)$

Question

Write out the elements of group permutation that is isomorphic to $U(14)$.

$G =\{I_3, P_1, P_2, P_3, P_4, P_5\}$ For $n = 3$ the permutation matrices are $I_3$ and the five matrices are: $$P_1=\begin{pmatrix} 1&0 &0 \\0 &0 &1 \\0 &1 &0 \end{pmatrix}\qquad P_2=\begin{pmatrix} 0&1 &0 \\1 &0 &0 \\0 &0 &1 \end{pmatrix}\qquad P_3=\begin{pmatrix} 0&1 &0 \\0 &0 &1 \\1 &0 &0 \end{pmatrix}$$ $$P_4=\begin{pmatrix} 0&0 & 1\\0 &1 &0 \\ 1& 0&0 \end{pmatrix}\qquad P_5=\begin{pmatrix} 0& 0& 1\\ 1& 0& 0\\0 & 1& 0\end{pmatrix}$$

Answer 1

If $U(14)$ means the group of units modulo $14$ under multiplication, then the units are $\{1, 3, 5, 9, 11, 13 \}$. (as on page 2 of this document: http://math.bu.edu/people/rpollack/Teach/541fall09/HW3_Solutions.pdf)

Hint: $3$ generates $U(14)$, and therefore has order $|U(14)| = 6$. Since the group you're seeking is isomorphic, it is also generated by an element of order $6$.

PS: there are no elements of order $6$ in the symmetric group of order $3$ (so $3$ by $3$ permutation matrices won't work).

Answer 2

I made a table using the labels $\{I_3, p_1, p_2, p_3, p_4, p_5\}$ instead of filling the table with the matrices.

$$\begin{array}{c|c|c|c|c|c|} & \mathbf{I_3} & \mathbf{p_1} & \mathbf{p_2} & \mathbf{p_3} & \mathbf{p_4} & \mathbf{p_5}\\ \hline \mathbf{I_3} & I_3 & p_1 & p_2 & p_3 & p_4 & p_5 \\ \hline \mathbf{p_1} & p_1 & I_3 & p_3 & p_2 & p_5 & p_4 \\ \hline \mathbf{p_2} & p_2 & p_5 & I_3 & p_4 & p_3 & p_1 \\ \hline \mathbf{p_3} & p_3 & p_4 & p_1 & p_5 & p_2 & I_3 \\ \hline \mathbf{p_4} & p_4 & p_3 & p_5 & p_1 & I_3 & p_2 \\ \hline \mathbf{p_5} & p_5 & p_2 & p_4 & I_3 & p_1 & p_3 \\ \hline \end{array}$$

$f_{I_3} (x) = I_3$

then looking at a matrix mapping looking at $p_1$ line:

$$G = \binom{I_3\quad p_1\quad p_2\quad p_3\quad p_4\quad p_5}{p_1\quad I_3 \quad p_3\quad p_2\quad p_5\quad p_4}$$

$p_1*I_3 = p_1 ,\quad p_1*p_1 = I_3$ this brings it back to start so $(I_3, p_1)$
$p_1*p_2 = p_3,\quad p_1*p_3 = p_2$ brings it back to start so $(I_3, p_1),(p_2, p_3)$
$p_1*p_4 = p_5,\quad p_1*p_5 = p_4$ brings it back to start so $(I_3, p_1),(p_2, p_3),(p_4, p_5)$

$f_{p_1} (x) = (I_3, p_1),(p_2, p_3),(p_4, p_5)$

then looking at a matrix mapping looking at $p_2$ line:

$$G = \binom{I_3\quad p_1\quad p_2\quad p_3\quad p_4\quad p_5}{p_2\quad p_5 \quad I_3\quad p_4\quad p_3\quad p_1}$$

$f_{p_2} (x) = (I_3, p_2), (p_1, p_5), (p_3, p_4)$

and so on...