# Lagrange, cosets, indicies

1. Dec 8, 2008

### proplaya201

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 of N so 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 the N where |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.

2. Dec 8, 2008

### HallsofIvy

So K is a subgroup of H which means that order(K) is an integer that divides order(H). That is If h= order(H) and k= order(K), h= nk for some integer n. But order(H) is an integer that divides order(G) so if g= order(G), g= mh for some integer m. Now write g in terms of k.

3. Dec 8, 2008

### proplaya201

|G| = g = mh = m(nk) = (mn)k. so that makes sense, but how do you relate that to the left cosets of K in G. or in other words, how do you relate that to the index of K in G. so does it suffice to state that a coset in H partitions H into |G| subsets? im just having difficulty seeing the connection of the order of the groups to the distinct left cosets.