Factor Groups: What am I thinking about wrongly here?

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Discussion Overview

The discussion revolves around the properties of factor groups, specifically focusing on the relationship between the order of an element in a group and the order of the corresponding coset in the factor group. Participants explore the implications of Lagrange's theorem in this context.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant asserts that if K is normal in G, then for an element g in G with order n, the coset Kg^n equals K, leading to the conclusion that the order of Kg divides n.
  • Another participant agrees with the initial reasoning and references Lagrange's theorem to support the claim that the order of any element in G/K must divide the order of G, which is n.
  • A third participant challenges the assumption that |G| = n, clarifying that |g| = n does not imply G is cyclic, thus questioning the direct application of the previous statements.
  • The second participant acknowledges the misinterpretation regarding the order of G and concedes that while the conclusion about |Kg| dividing n does not directly follow, it remains valid based on the initial reasoning presented.

Areas of Agreement / Disagreement

Participants exhibit some agreement on the implications of Lagrange's theorem but also highlight a misunderstanding regarding the relationship between the orders of G and g. The discussion remains unresolved regarding the implications of these orders on the properties of factor groups.

Contextual Notes

There are limitations in the assumptions made about the group G, particularly regarding its cyclic nature and the implications of element orders. The discussion does not resolve these assumptions.

PsychonautQQ
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say K is normal in G hence we have a factor group G/K.

let g be an element of G where |g| = n.

so Kg^n = K since g^n = 1.
and using the properties of factor groups, we know Kg^n = (Kg)^n
hence (Kg)^n = K
So we know that the order of Kg divides n.

Is this correct thinking? Factor groups are trippin me out
 
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Yes, this looks correct. In fact, this shouldn't be surprising, since by Lagrange's theorem, the order of the factor group G/K is |G|/|K| which divides |G| = n.
Thus the order of any element of G/K must divide |G/K| which divides n. You can easily show that a divisor of a divisor of n is also a divisor of n.
 
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you said |G| = n. I originally said it was an element of G, |g| = n. Doesn't |G| = |g| iff G = <g> (G is a cyclic group generated by g?) The question doesn't say G is cyclic.
 
Oh my bad I misread that the order of G was n. In that case the statement that |Kg| divides n does not directly follow from what I said, but it is still true by what you showed.
 

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