Find a subgroup of order 6 in U(700).

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

Then ${\displaystyle z^6=1,\text{and}{z}^i\ne 1}$
for $0<i<6$.

Define a ${\displaystyle (700\times 700)}$ matrix ${\displaystyle A=\left[a_{ij}\right]}$ by
$a_{ij}=\{\begin{array}{ll}{e}^{\frac{i\pi }{3}},& i=jandi=1,2,or3\\ 1,& i=jandi>3\\ 0& i\ne j\end{array}$
Then ${\displaystyle A^{\cdot }=\left[b_{ij}\right]}$ is
$b_{ij}=\{\begin{array}{ll}{e}^{-\frac{i\pi }{3}},& i=jandi=1,2,or3\\ 1,& i=jandi>3\\ 0& i\ne j\end{array}$
which means that
${\displaystyle A{A}^{\cdot }=A^{\cdot }A=I}$,
so A is unitary.
Furthermore, ${\displaystyle A^k=\left[a_{ij}^k\right]}$ is given by
$a_{ij}^k=\{\begin{array}{ll}{e}^{\frac{i\pi }{3}},& i=jandi=1,2,or3\\ 1,& i=jandi>3\\ 0& i\ne j\end{array}$
since A is a diagonal matrix.
Now assume that k is the smallest positive integer such that ${\displaystyle A^k=I}$. Then $(e^{\frac{i\pi }{3}}{)}^k=1,sok\ge 6(\text{because, for}z=e^{\frac{i\pi }{3}},z^6=1,and{z}^i\ne 1for0<i<6).$.

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 ${\displaystyle ⟨A⟩}$.
Result:
Hint:Define a ${\displaystyle (700\times 700)}$ matrix ${\displaystyle A=\left[a_{ij}\right]}$ by
${\displaystyle a_{ij}=\{(e^{-i\frac{\pi }{3}},i=j\text{and}i=1,2,\text{or}3),(1,i=j\text{and}i>3),(i\ne j)}$