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

Find a subgroup of order $6\in U\left(700\right)$.
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Let z be some complex number such that $z={e}^{i\frac{\pi }{3}}$.

Then ${z}^{6}=1,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{z}^{i}\ne 1$
for $0.

Define a $\left(700×700\right)$ matrix $A=\left[{a}_{ij}\right]$ by

Then ${A}^{\cdot }=\left[{b}_{ij}\right]$ is

which means that
$A{A}^{\cdot }={A}^{\cdot }A=I$,
so A is unitary.
Furthermore, ${A}^{k}=\left[{a}_{ij}^{k}\right]$ is given by

since A is a diagonal matrix.
Now assume that k is the smallest positive integer such that ${A}^{k}=I$. Then .

Furthermore, it is easy to see that ${A}^{6}=I$ (other elements remain 1 or 0 whatever k is).
Therefore, A is of order 6, so one subgroup of order $6\in U\left(700\right)$ is given by the cyclic group $⟨A⟩$.
Result:
Hint:Define a $\left(700×700\right)$ matrix $A=\left[{a}_{ij}\right]$ by
${a}_{ij}=\left\{\left({e}^{-i\frac{\pi }{3}},i=j\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}i=1,2,\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}3\right),\left(1,i=j\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}i>3\right),\left(i\ne j\right)$