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Thread: 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 $\displaystyle K'\trianglelefteq G$ where $\displaystyle K'=[K,K] =\langle [a, b]| a, b \in K\rangle$.

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

    and

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


    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 $\displaystyle K'\trianglelefteq G$ where $\displaystyle K'=[K,K] =\langle [a, b]| a, b \in K\rangle$.

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

    and

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


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

    the reason is that if $\displaystyle h \in H,$ then $\displaystyle h=x_1^{\pm 1} x_2^{\pm 1} \cdots x_n^{\pm 1},$ for some $\displaystyle x_i \in X.$ but then: $\displaystyle ghg^{-1}=(gx_1g^{-1})^{\pm 1}(gx_2 g^{-1})^{\pm 1} \cdots (gx_ng^{-1})^{\pm 1}.$ i think you see my point now!
    Last edited by NonCommAlg; May 1st 2009 at 04:20 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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