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Math Help - commutator subgroup eep!

  1. #1
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    commutator subgroup eep!

    here's the prob

    Prove if K is a normal subgroup of a group G then K'\trianglelefteq  G where K'=[K,K] =\langle [a, b]| a, b \in K\rangle.

    I have a theorem i feel like i should be able to use but i'm not sure how...this is it
    Let H\leq G and let x, y\in G then H\trianglelefteq G \iff [H,G]=\leq H

    and

    if G'\leq H then H\trianglelefteq G.


    any ideas?
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  2. #2
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    Quote Originally Posted by delilahjam View Post
    here's the prob

    Prove if K is a normal subgroup of a group G then K'\trianglelefteq  G where K'=[K,K] =\langle [a, b]| a, b \in K\rangle.

    I have a theorem i feel like i should be able to use but i'm not sure how...this is it
    Let H\leq G and let x, y\in G then H\trianglelefteq G \iff [H,G]=\leq H

    and

    if G'\leq H then H\trianglelefteq G.


    any ideas?
    I think that g[a,b]g^{-1} = [gag^{-1},gbg^{-1}].
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    I think that g[a,b]g^{-1} = [gag^{-1},gbg^{-1}].
    I agree, but I can't just take a single commutator. The elements of K' are of the form [a_1,b_1]^{r_1}[a_2,b_2]^{r_2}\cdots[a_n,b_n]^{r_n} for some r_1, r_2\cdots \in \mathbb{Z}
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  4. #4
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    Quote Originally Posted by delilahjam View Post
    I agree, but I can't just take a single commutator. The elements of K' are of the form [a_1,b_1]^{r_1}[a_2,b_2]^{r_2}\cdots[a_n,b_n]^{r_n} for some r_1, r_2\cdots \in \mathbb{Z}
    it's a simple observation that if a subgroup H of a group G is generated by a set of elements, say X, then H is normal in G if and only if gxg^{-1} \in H, for all g \in G, \ x \in X.

    the reason is that if h \in H, then h=x_1^{\pm 1} x_2^{\pm 1} \cdots x_n^{\pm 1}, for some x_i \in X. but then: ghg^{-1}=(gx_1g^{-1})^{\pm 1}(gx_2 g^{-1})^{\pm 1} \cdots (gx_ng^{-1})^{\pm 1}. i think you see my point now!
    Last edited by NonCommAlg; May 1st 2009 at 05:20 AM.
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  5. #5
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    awesome thanks!
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  6. #6
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    actually thanks from me!

    i guess we love mathhelpforum here so much we never sign out!
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