Question
Asked Oct 17, 2019
Let H and K be subgroups of a finite group G. Show that
|HK
|HK=
|HОКI
where HK (hk hE H, k E K}. (HK is not assumed to be a subgroup of G)
:
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Let H and K be subgroups of a finite group G. Show that |HK |HK= |HОКI where HK (hk hE H, k E K}. (HK is not assumed to be a subgroup of G) :

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check_circleExpert Solution
Step 1

let D = HK then D is a subgroup of k and there exist a decomposition of k into disjoint right cosets of D in k and

k Dk, U Dk, U..Dk
(1)
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k Dk, U Dk, U..Dk (1)

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Step 2

 

This impl...

|k=|Dk; U Dk,U..Dk,
= Dk+|Dk+..D
ID+|D|+..D (t times)
..(2)
t =
Again
Hk H Dk;
i1
i1
=UHk, DCH)
i1
=Hk UHk,U...U Hk
Now no two of Hk,,Hk,.. .Hk, can be equal as if Hk, = Hk, for some i,j then,
kkHkkEHNk = D
kk,ED
Dk, Dk,
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|k=|Dk; U Dk,U..Dk, = Dk+|Dk+..D ID+|D|+..D (t times) ..(2) t = Again Hk H Dk; i1 i1 =UHk, DCH) i1 =Hk UHk,U...U Hk Now no two of Hk,,Hk,.. .Hk, can be equal as if Hk, = Hk, for some i,j then, kkHkkEHNk = D kk,ED Dk, Dk,

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