Modern Algebra Factor Groups question

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SUMMARY

The discussion focuses on the proof that if K is a normal subgroup of G and the order of an element g in G is n, then the order of the coset Kg in the quotient group G/K divides n. The user attempts to demonstrate this by showing that (Kg)^n equals the identity element in G/K, concluding that the order of Kg must divide n. A crucial point raised is the necessity of K being normal in G to validate the equality (Kg)^n = Kg^n, which is essential for establishing that G/K forms a group.

PREREQUISITES
  • Understanding of group theory concepts, particularly normal subgroups.
  • Familiarity with quotient groups and their properties.
  • Knowledge of the order of elements in group theory.
  • Basic algebraic manipulation involving group elements and cosets.
NEXT STEPS
  • Study the properties of normal subgroups in group theory.
  • Learn about the structure and properties of quotient groups.
  • Explore the concept of the order of elements in finite groups.
  • Review proofs involving cosets and their implications in group theory.
USEFUL FOR

Students of abstract algebra, particularly those studying group theory, as well as educators seeking to clarify concepts related to normal subgroups and quotient groups.

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Homework Statement


If K is normal in G and |g| = n for some g in G, show that the order of Kg in G/K divides n.

Homework Equations


None

The Attempt at a Solution


Okay so I feel like I have a solution but I don't use all the information given so I'm trying to find holes in it...

g^n = 1

K = Kg^n = (Kg)^n. So it seems If we take Kg to the nth power we get it's identity. This means that the order of Kg must divide n.

What am I missing here?
 
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PsychonautQQ said:

Homework Statement


If K is normal in G and |g| = n for some g in G, show that the order of Kg in G/K divides n.

Homework Equations


None

The Attempt at a Solution


Okay so I feel like I have a solution but I don't use all the information given so I'm trying to find holes in it...

g^n = 1

K = Kg^n = (Kg)^n. So it seems If we take Kg to the nth power we get it's identity. This means that the order of Kg must divide n.

What am I missing here?

If you mean you aren't using that K is normal in G, you need that fact to justify saying (Kg)^n=Kg^n, or even that G/K is a group. Better review why.
 

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