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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: $$P_1=\begin{pmatrix} 1&0 &0 \\0 &0 &1 \\0 &1 &0 \end{pmatrix}\qquad P_2=\begin{pmatrix} 0&1 &0 \\1 &0 &0 \\0 &0 &1 \end{pmatrix}\qquad P_3=\begin{pmatrix} 0&1 &0 \\0 &0 &1 \\1 &0 &0 \end{pmatrix}$$ $$P_4=\begin{pmatrix} 0&0 & 1\\0 &1 &0 \\ 1& 0&0 \end{pmatrix}\qquad P_5=\begin{pmatrix} 0& 0& 1\\ 1& 0& 0\\0 & 1& 0\end{pmatrix}$$

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put on hold as off-topic by Did, Najib Idrissi, studiosus, Claude Leibovici, Hakim yesterday

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@Jessica: I don't know if you are posting homework or problems from a book; if the former, you may want to tag them with the "homework" tag. In either case, you should write a bit about what you've tried or why you are confused. That way you will get hints to point you in the right direction. In particular, try not to write your questions as if you were assigning homework to the readers of this site. –  Arturo Magidin Oct 22 '10 at 3:24
will do thanks for the advice –  Jessica Oct 22 '10 at 3:49

2 Answers 2

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).

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What I did was made a cayley table. –  Jessica Oct 27 '10 at 19:45

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...

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Well... the group G is going to be the group isomorphic to the symmetric group $S_3$ (since it consists of all of the 3 by 3 permutation matrices). You still need to work out if the group U(14) is isomorphic to G or not. –  Douglas S. Stones Oct 27 '10 at 21:57