Prove HK=KH iff HK is a Subgroup of G

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Homework Help Overview

The problem involves proving that the product of two subgroups HK equals KH if and only if HK is a subgroup of a group G. The discussion centers around subgroup properties and manipulations of group elements.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of subgroup properties, particularly focusing on the relationships between elements in HK and KH. There are attempts to manipulate expressions involving group elements to show subset relations.

Discussion Status

Some participants have offered insights into the relationships between elements in HK and KH, with one noting a successful manipulation that suggests a direction towards proving the subset relationship. However, there is still some uncertainty about the correctness of the reasoning presented.

Contextual Notes

Participants are working under the constraints of proving subgroup properties without using LaTeX, which may affect the clarity of their expressions. There is an ongoing exploration of the inverses of elements in the context of subgroup definitions.

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Homework Statement



if H and K are arbitrary subgroup of G, prove that HK=KH iff HK is a subgroup of G

Homework Equations



n/a

The Attempt at a Solution



no problem to prove => direction

for <= i can prove KH is a subset of HK

only i got troubled to show HK ia subset of KH

x in HK

x=hk for some h in H ,k in K

i manipulate it many ways and always got the form x=khk for some h in H ,k in K

HELP, and sorry no latex ,i'm very buzy now ;P
 
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Take b in HK so that b-1 = hk is in HK. How do you get b back?
 


so k^{-1}h^{-1} \in KH[/tex] <br /> <br /> and hk \in HK, what&#039;s its inverse?
 


aahhhh i see,

b in HK

b=hk for some h in H k in K

b^{-1} also is in HK

imply
<br /> b^{-1}=(hk)^{-1}=k^{-1}h^{-1} \in KH <br />

right ??
 


wait wrong,

for any x in HK, x^-1 in HK so x^-1=hk for some h and k

then <br /> <br /> x=(x^{-1})^{-1}=(hk)^{-1}=k^{-1}h^{-1} \in KH<br /> <br />

now this is correct right?
 


yeah that looks good
 

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