Let z be some complex number such that \(\displaystyle{z}={e}^{{{i}\frac{\pi}{{3}}}}\).

Then \(\displaystyle{z}^{{6}}={1},{\quad\text{and}\quad}{z}^{{i}}\ne{1}\)

for \(0 < i < 6\).

Define a \(\displaystyle{\left({700}\times{700}\right)}\) matrix \(\displaystyle{A}={\left[{a}_{{{i}{j}}}\right]}\) by

\(a_{ij}=\begin{cases}e^{\frac{i\pi}{3}}, & i=j\ and\ i=1,2,\ or\ 3 \\ 1, & i =j\ and\ i>3 \\ 0 & i\neq j\end{cases}\)

Then \(\displaystyle{A}^{\cdot}={\left[{b}_{{{i}{j}}}\right]}\) is

\(b_{ij}=\begin{cases}e^{-\frac{i\pi}{3}}, & i=j\ and\ i=1,2,\ or\ 3 \\ 1, & i =j\ and\ i>3 \\ 0 & i\neq j\end{cases}\)

which means that

\(\displaystyle AA^{\cdot}={A}^{\cdot}{A}={I}\),

so A is unitary.

Furthermore, \(\displaystyle{A}^{{k}}={\left[{{a}_{{{i}{j}}}^{{{k}}}}\right]}\) is given by

\(a_{ij}^{k}=\begin{cases}e^{\frac{i\pi}{3}}, & i=j\ and\ i=1,2,\ or\ 3 \\ 1, & i =j\ and\ i>3 \\ 0 & i\neq j\end{cases}\)

since A is a diagonal matrix.

Now assume that k is the smallest positive integer such that \(\displaystyle{A}^{{k}}={I}\). Then \((e^{\frac{i\pi}{3}})^{k}=1,\ so\ k\geq6(\text{because, for}\ z=e^{\frac{i\pi}{3}},\ z^{6}=1,\ and\ z^{i}\neq1\ for\ 0<i<6).\).

Furthermore, it is easy to see that \(\displaystyle{A}^{{6}}={I}\) (other elements remain 1 or 0 whatever k is).

Therefore, A is of order 6, so one subgroup of order \(\displaystyle{6}\in{U}{\left({700}\right)}\) is given by the cyclic group \(\displaystyle{\left\langle{A}\right\rangle}\).

Result:

Hint:Define a \(\displaystyle{\left({700}\times{700}\right)}\) matrix \(\displaystyle{A}={\left[{a}_{{{i}{j}}}\right]}\) by

\(\displaystyle{a}_{{{i}{j}}}={\left\lbrace{\left({e}^{{-{i}\frac{\pi}{{3}}}},{i}={j}{\quad\text{and}\quad}{i}={1},{2},{\quad\text{or}\quad}{3}\right)},{\left({1},{i}={j}{\quad\text{and}\quad}{i}{>}{3}\right)},{\left({i}\ne{j}\right)}\right.}\)