How to Prove |G:H| = |G:K||K:H| for Quotient Groups?

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SUMMARY

The discussion centers on proving the equation |G:H| = |G:K||K:H| for quotient groups, where H is a subgroup of K, and K is a subgroup of G. Participants reference Lagrange's theorem and the lattice isomorphism theorem, emphasizing the importance of normal subgroups in this context. The proof involves counting cosets and establishing isomorphisms between the quotient groups. The conclusion is that the relationship holds true under the given conditions.

PREREQUISITES
  • Understanding of quotient groups and their properties
  • Familiarity with Lagrange's theorem in group theory
  • Knowledge of the lattice isomorphism theorem
  • Ability to work with cosets and subgroup relationships
NEXT STEPS
  • Study the proof of Lagrange's theorem in detail
  • Explore the lattice isomorphism theorem and its applications
  • Learn about normal subgroups and their significance in group theory
  • Practice counting cosets in various group structures
USEFUL FOR

Mathematics students, particularly those studying abstract algebra, group theorists, and anyone looking to deepen their understanding of quotient groups and their properties.

Pjennings
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As a way to keep busy in between semesters I decided to work my way through Algebra by Dummit and Foote in order to prepare for the fall. Working my way through quotient groups is proving to be quite difficult and as a result I'm stuck on an exercise that looks simple, but I just don't know where to start. Any ideas how to prove that given H\leqK\leqG |G:H|=|G:K||K:H|?
 
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Have they not gone through Lagrange's theorem yet?
 
eok20 said:
Have they not gone through Lagrange's theorem yet?
Who said any of the groups were finite?
 
Martin Rattigan said:
Who said any of the groups were finite?

good point.
 
Lattice isomorphism theorem. Assuming those are normal subgroups, (G/K)/(K/H) is isomorphic to G/H.

Of course, it's not hard to just count the number of cosets. G = g_1K + ... + g_nK where n = [G:K]. K = k_1H + ... + k_mH, where m = [H:K]. Then G = g_1(k_1 + ... + k_m)H + ... + g_n(k_1+...+k_m)H, where I have abused notation a little. This shows [G:H]<=[G:K][K:H]. Now think about the other direction.
 

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