Quotient Groups and their Index

  • Thread starter Pjennings
  • Start date
  • #1
16
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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[tex]\leq[/tex]K[tex]\leq[/tex]G |G:H|=|G:K||K:H|?
 

Answers and Replies

  • #2
200
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Have they not gone through Lagrange's theorem yet?
 
  • #3
Have they not gone through Lagrange's theorem yet?
Who said any of the groups were finite?
 
  • #5
TMM
92
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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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