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From: mina_world on 21 Nov 2008 06:53
Hello teacher~

Let D_2n = <r, s | r^n = s^2 = e, s.r.s = r^-1>
be the usual presentation of the dihedral group
of order 2n
and let k be a positive integer dividing n.

(a) Prove that <r^k> is a normal subgroup of D_2n.

(b) Prove that D_2n / <r^k> ~ D_2k.

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pf)
(a)
D_2n = {e, r, r^2, ... , r^(n-1), s, sr, sr^2, ... , sr^(n-1)}

(1) r^i.<r^k> = <r^k>.r^i

(2) s.<r^k> = <r^k>.s [ <==> s.<r^k>.s = <r^k> ]

Because,
s.r.s = r^-1 ==> s.r^(km).s = r^(-km)

Since s.r^(km).s = r^(-km) in <r^k>
and |s.<r^k>.s| = |<r^k>|,
so, s.<r^k>.s = <r^k>
so, s.<r^k> = <r^k>.s

(3) s.r^i.<r^k> = <r^k>.s.r^i
[ <==> s.r^k.<r^k>.r^-i.s^-1 = <r^k> ]

Because,
s.r^i.r^(km).r^-i.s^-1 = s.r^(km).s^-1 = r^(-km) in <r^k>
and |s.r^k.<r^k>.r^-i.s^-1| = |<r^k>|,
so, s.r^k.<r^k>.r^-i.s^-1 = <r^k>
so, s.r^i.<r^k> = <r^k>.s.r^i

It means that <r^k> is a normal subgroup of D_2n.

(b)
Let k | n.

so, |<r>| = n, |<r^k>| = n / (k, n) = n/k

so, |D_2n / <r^k>| = (2n) / (n/k) = 2k

|r<r^k>| = k and |s<r^k>| = 2

[s.<r^k>][r.<r^k>][s.<r^k>] = s.r.s.<r^k> = r^-1.<r^k>

It means that D_2n / <r^k> ~ D_2k.


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