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Subgroups-commutator, normal, Abelian

  1. Oct 6, 2008 #1
    Subgroups--commutator, normal, Abelian

    1. The problem statement, all variables and given/known data
    Let G be a group and g,h in G. Define the commutator of g and h as [g,h]= gh(hg)^-1. Then define the commutator subgroup, denoted [G,G], of G as the subgroup generated by all the commutators of elements of G, i.e. [G,G]=<{[g,h]: g,h in G}>.

    (a) What does an element of [G,G] look like?
    Hint: it is not enough to only consider elements of the form [g,h].

    (b) Prove that [G,G] is a normal subgroup of G.

    (c) Prove that G/[G,G] is an Abelian group.

    3. The attempt at a solution

    (a) Since [G,G] is cyclic, I thought an element would be of some form of [gh(hg)^-1]^n for some n in N. But the hint kind of threw me off...

    (b) Initially I wanted to show that gh(hg)^-1 is mm^-1 but I have to show commutativity there and I don't know how to...so there must be another way? I can't seem to get started on this one...

    (c) Is this just a consequence of part (b)? Since to show a subgroup is normal you need to use their cosets as an example.
  2. jcsd
  3. Oct 6, 2008 #2


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    Re: Subgroups--commutator, normal, Abelian

    (a) [G,G] is not necessarily cyclic - it's generated by a set. I suggest you look up what this means.

    (b) I don't understand what you're saying here. What's the definition of "normal subgroup"?

    (c) It's a consequence of (b) that G/[G,G] is a group. Why is it an abelian one?
  4. Oct 6, 2008 #3
    Re: Subgroups--commutator, normal, Abelian

    A normal subgroup is some H such that xH=Hx for x in G.
    Equivalently, xhx^-1 is in H.
  5. Oct 7, 2008 #4


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    Re: Subgroups--commutator, normal, Abelian

    Right. Now use the definition to show that [G,G] is normal in G.
  6. Oct 7, 2008 #5
    Re: Subgroups--commutator, normal, Abelian

    Well...that's my question here.

    This is what I attempted:
    Let H be a subgroup of G,
    g,g-inverse in G
    Then gh is in gH, and (g-inverse)(h-inverse) is in (g-inverse)H.
    So then [g,h]=gh(g-inverse)(h-inverse)=gH(g-inverse)H=g(g-inverse)H=eH=H
    (since the product of left cosets equals a left coset).

    Is that correct? If not...any hints?

    If it is right...then I have shown that an element of [G,G] looks like the subgroup H, and then all I need to do now is show xH=Hx?
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