(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

suppose that H and K are subgroups of a group G such that K is a proper subgroup of H which is a proper subgroup of G and suppose (H : K) and (G : H) are both finite. Then (G : K) is finite, and (G : K) = (G : H)(H : K).

**that is to say that the proof must hold for infinite groups as well**

notation- (G : K) = |G|/|K| is the index of K in G

2. Relevant equations

Lagrange's Theorem - b/c G is finite implies that there is a finite subgroup in G (i.e. H) whose order divides that of G's.

**there is no mention that the group G in question is considered to be an abelian group.**

3. The attempt at a solution

if we say that {(a_i)H | i = 1, ... , r } is the collection of distinct left cosets of H in G and {(b_j)K | j = 1, ... , s } is the collection of distinct left cosets of K in H.

then in order to conclude the proof I have to show that:

{(a_i)(b_j)K | i = 1, ... , r; j = 1,...,s } is the collection of distinct left cosets of K in G.

i was not sure about a method of approach that came to me. so i was thinking of a few ways to solve it, but im not sure of the right one if any of them are correct.

*1-that is (a_i)H is the number of distinct left cosets of H in G. so b/c |G| is finite then |G| must either be prime or not prime. if |G| is prime then let |G| = p and let |H| = m where m is an element ofNso by Lagrange's Theorem we know that m divides p. and b/c p is prime then we know that m = p. but this is not true because H is a proper subset of G so p > m... then |G| must not be prime let |G| = y in that case then we can find an element "x" in theNwhere |H| = x such that x divides y and x < y.

im not sure if this is at all the correct way to approach it because i am having trouble relating this to the distinct cosets of H in G

i would really appreciate some help on the matter.

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# Lagrange, cosets, indicies

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