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

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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