How Does xH Equal yH Imply x⁻¹y Belongs to H in Group Theory?

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SUMMARY

The discussion centers on proving the equivalence xH = yH ⇔ x⁻¹y ∈ H within group theory, where H is a subgroup of G. Participants clarify that if xH = yH, then both x and y must belong to H, leveraging the closure property of subgroups. The reverse implication is established by demonstrating that if x⁻¹y ∈ H, then x can be expressed as y multiplied by some element h in H, confirming that xH = yH. This proof utilizes the properties of subgroups and the definition of cosets.

PREREQUISITES
  • Understanding of group theory concepts, specifically subgroups and cosets.
  • Familiarity with the closure property of groups and subgroups.
  • Knowledge of the identity element and inverse elements in group theory.
  • Ability to manipulate algebraic expressions involving group elements.
NEXT STEPS
  • Study the properties of cosets in group theory, focusing on left and right cosets.
  • Learn about the Lagrange's theorem and its implications for subgroup orders.
  • Explore the concept of normal subgroups and their significance in group theory.
  • Investigate examples of groups and subgroups, such as integers under addition and modular arithmetic.
USEFUL FOR

This discussion is beneficial for students of abstract algebra, particularly those studying group theory, as well as educators and researchers looking to deepen their understanding of subgroup properties and coset relations.

gottfried
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Homework Statement


Let H be a subgroup of G
Prove xH=yH ⇔ x-1.y\inH


Homework Equations





The Attempt at a Solution


If x.H = y.H then x,y\inH
since H is a subgroup x-1,y-1\inH
and the closure of H means x-1.y\inH

Proving the reverse is my problem despite the fact that I'm sure is very easy but i just can't see it.

What I want to do is show that x-1.y\inH implies x,y\inH
In which case x.H=H=y.H.

How is the best way to show this?
 
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If xH = yH then x,y \in H. This is not true. Consider the integers. And conside the sub group 5Z. Now, 3+5Z = 8+5Z but neither 8 nor 3 is in 5Z. So, if xH=yH then for every h\in H there is a h' \in H with xh=yh'. Now, use that fact to prove this direction.

Try to do something similar for the other direction. You know that there is some h \in H with y = xh, see what you can make of that.
 
Thanks.

So we know that x=y.h for some h\inH
therefore e=x-1.y.h
e.h-1= x-1.y
Therefore x-1.y\inH since h-1\inH

So the reverse could be

x-1.y=h\inH
therefore y=x.h and this tells us xH=yH by the theorem that xH=yH ⇔ x=y.h for some h\inH
 
yes, that seems correct to me.
 
Thanks for the help
 

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