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Math Help - abstract algebra

  1. #1
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    abstract algebra

    Let G be a group of order 3825. Prove that if H is a normal subgroup of order 17 in G, then H is a subgroup of Z(G).

    Z(G) = { g in G | xg = gx for every x in G}
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  2. #2
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    Quote Originally Posted by dori1123 View Post
    Let G be a group of order 3825. Prove that if H is a normal subgroup of order 17 in G, then H is a subgroup of Z(G).

    Z(G) = { g in G | xg = gx for every x in G}
    Lemma: both equations x^3 \equiv 1 \mod 17 and x^5 \equiv 1 \mod 17 have the unique solution x \equiv 1 \mod 17.

    Proof: if x^3 \equiv 1 \mod 17, then x^{15} \equiv 1 \mod 17. hence x^{16} \equiv x \mod 17. but x^{16} \equiv 1 \mod 17. thus: x \equiv 1 \mod 17. a similar argument works for x^5 \equiv 1 \mod 17. \ \ \ \Box


    now suppose |G|=3825=3^2 \times 5^2 \times 17 and let H=<h> be the normal subgroup of G with o(h)=17. in order to show that H \subseteq Z(G), we only need to prove that h \in Z(G).

    let g \in G. we want to show that g commutes with h. let o(g)=m. since H is normal, there exists an integer n such that ghg^{-1}=h^n. thus: h=g^mhg^{-m}=h^{n^m}, which gives us

    n^m \equiv 1 \mod 17. we have m=3^r \times 5^s \times 17^t, for some 0 \leq r \leq 2, \ 0 \leq s \leq 2, \ 0 \leq t \leq 1. since n^{17} \equiv n \mod 17, we may assume that t=0. now by the Lemma we must have

    n \equiv 1 \mod 17 and thus ghg^{-1}=h^n=h. therefore gh=hg, i.e. h \in Z(G). \ \ \Box


    Exercise: (very easy!) make the problem interesting by extending this result to a class of finite groups!
    Last edited by NonCommAlg; November 4th 2008 at 02:48 PM.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    Lemma: both equations x^3 \equiv 1 \mod 17 and x^5 \equiv 1 \mod 17 have the unique solutions x \equiv 1 \mod 17.

    Proof: if x^3 \equiv 1 \mod 17, then x^{15} \equiv 1 \mod 17. hence x^{16} \equiv x \mod 17. but x^{16} \equiv 1 \mod 17. thus: x \equiv 1 \mod 17. a similar argument works for x^5 \equiv 1 \mod 17. \ \ \ \Box
    Here is another way (not that there is anything bad about what you did) for dori.

    If you have the equation x^q \equiv 1(\bmod p) and q\not | (p-1) then x\equiv 1(\bmod p) where p,q are primes. This is because \mathbb{Z}_p^{\times} is a group. And if x^q \equiv 1 and x\not \equiv 1 then it means x has order q. By Lagrange's theorem it means q divides |\mathbb{Z}_p^{\times}| = p-1 a contradiction.

    thus ghg^{-1}=h^n=h. therefore xh=hx, i.e. h \in Z(G). \ \ \Box
    You mean gh=hg. But whatever it is not any significant.
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