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Finding G/H (H Subgroup of G)

  1. Apr 20, 2010 #1
    Does anybody know a general method to find the Group G/H (Where G is a Group and H is a subgroup of G)

    For example

    (1) What is the group S3/H ?????

    S3 = {e, a, a^2, b, ab, (a^2)b} (Permutation group of order 6)
    H =< a >= {e, a, a^2} is a cyclic subgroup of G

    (2) What is the group GL(n,R)/GL+(n;R)???

    GL(n,R) = { nxn Matrices with real entries whose determinants are not zero}
    GL+(n;R)= { nxn matrices with real entries whose determiants are positive}

    (3) Consider the dihedral group D4 =< a, b >= {e, a, a^2, a^3, b, ab, (a^2)b, (a^3)b}
    Find the groups D4/H where H is a normal proper subgroup of D4

    (4) Consider S to be the set of all transformations on R such that if x belongs to R
    s : x --> x' = ax + b
    with a, b real numbers and a not = 0.
    Let S1 be all transformations of the form x --> x' = x+b and S2 be all transformations
    of the form x --> x' = ax
    S/S1 is a group, which group is it?

    Help will be greatly appreciated
     
  2. jcsd
  3. Apr 20, 2010 #2
    All those examples are manageable with direct computation. The elements of G/H are all pairwise disjoint, and the identity is H. Furthermore, if G is finite, [G : H] = |G / H| = |G| / |H|. For example in S3 = {e, a, a^2, b, ab, (a^2)b}, H = {e, a, a^2}, since |G / H| = 2, [G : H] = 2, so there is only one other coset aside from H in G/H, and it has to be {b,ab, (a^2)b}. (2) and (3) are almost identically solved.


    (4): If F (x) = ax + b, what is F*S1?
     
  4. Apr 21, 2010 #3
    Just a few things.

    What is [G : H]?
    Why is it that you said the identity is H?
    As the only other coset aside from H is {b,ab, (a^2)b} how does this then determine what G/H is? you have only found two cosets...and not the group G/H?

    Last thing. How is this identical to (2) because GL (n,R) and GL +(n,R) have infinite order right? so you can't use the same [G : H] = |G / H| = |G| / |H|

    thanks
     
  5. Apr 21, 2010 #4
    1) Up to isomorphism, how many groups have order 2?
    2) How many groups satisfy the property that the product of any two non-identity elements is the identity?
     
  6. Apr 21, 2010 #5
    [G : H] is the index of the subgroup H in G. I.e., the number of cosets induced by H (this can be infinite).

    You are aware that if G is finite and H normal in G, |G / H| = |G| / |H| right? Each coset is of the same size and disjoint from the others, and each element of G falls into a coset of H, so it follows that there must be |G| / |H| of them. Also, in G/H, the identity element is H, this you should know.

    "As the only other coset aside from H is {b,ab, (a^2)b} how does this then determine what G/H is?" G/H is the group consisting of the cosets of H in G, with the group operation defined as aH*bH = abH and identity H.

    Although in (2) the two groups are infinite, [G : H] = 2, which is what you really need to know.
     
  7. Apr 22, 2010 #6
    in (2) the groups are infinite, how does that mean that [G:H] = 2?
     
  8. Apr 22, 2010 #7
    [G : H] is the size of G/H. It can be finite, even if G and H are both infinite. Can you see why G/H consists of precisely 2 elements in (2)?
     
  9. Apr 22, 2010 #8
    No i am not sure why G/H only consists of only 2 precise elements in (2)? Could you please explain why?
     
  10. Apr 22, 2010 #9
    I can't give you full out explanations, it's against the forum's policies. Rather than that, I can point you in the right direction:

    GL(n,R)/GL+(n;R) has identity GL+(n;R), so we know of one of its elements. Now let GL-(n;R) be the set of matrices with negative determinant. Can you prove this is the only other element in GL(n,R)/GL+(n;R)? First prove that it actually is a coset, and next prove that an element of GL(n,R) is in either GL+(n;R) or GL-(n;R) (this is immediate by the definition of those sets).
     
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