1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lagrange, cosets, indicies

  1. Dec 8, 2008 #1
    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. jcsd
  3. Dec 8, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    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.
  4. Dec 8, 2008 #3
    |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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Lagrange, cosets, indicies
  1. Stabilizers of cosets (Replies: 1)

  2. Cosets of Subgroups (Replies: 1)

  3. Right Cosets (Replies: 5)

  4. Coset Question (Replies: 12)