Question

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

Complex numbers
Find a subgroup of order $$\displaystyle{6}\in{U}{\left({700}\right)}$$.

2021-01-09

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