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

EunoR 2021-01-08 Answered
Find a subgroup of order 6U(700).
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Expert Answer

Daphne Broadhurst
Answered 2021-01-09 Author has 109 answers

Let z be some complex number such that z=eiπ3.

Then z6=1,andzi1
for 0<i<6.

Define a (700×700) matrix A=[aij] by
aij={eiπ3,i=j and i=1,2, or 31,i=j and i>30ij
Then A=[bij] is
bij={eiπ3,i=j and i=1,2, or 31,i=j and i>30ij
which means that
AA=AA=I,
so A is unitary.
Furthermore, Ak=[aijk] is given by
aijk={eiπ3,i=j and i=1,2, or 31,i=j and i>30ij
since A is a diagonal matrix.
Now assume that k is the smallest positive integer such that Ak=I. Then (eiπ3)k=1, so k6(because, for z=eiπ3, z6=1, and zi1 for 0<i<6)..

Furthermore, it is easy to see that A6=I (other elements remain 1 or 0 whatever k is).
Therefore, A is of order 6, so one subgroup of order 6U(700) is given by the cyclic group A.
Result:
Hint:Define a (700×700) matrix A=[aij] by
aij={(eiπ3,i=jandi=1,2,or3),(1,i=jandi>3),(ij)

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