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  1. #1
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    p-group

    Prove that if G is a finite p-group, then its center is non-trivial.
    Use this fact to prove if |G|=p^2, then G is abelian.

    For the first part, I have:
    So |G|=p^{\alpha}m. By the class equation, |G|=|Z(G)|+\sum_{i=0}^r|G:C_G(g_i)|.

    Since p| \ |G|, p|\left(|Z(G)|+\sum_{i=0}^r|G:C_G(g_i)|\right)

    So p| \ |Z(G)| and p| \ \sum_{i=0}^r|G:C_G(g_i)|.

    Since p is prime, |Z(G)|\neq 1. Therefore, Z(G)\neq\{e\}

    So how do I use that to prove the next part?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: p-group

    Quote Originally Posted by dwsmith View Post
    Prove that if G is a finite p-group, then its center is non-trivial.
    Use this fact to prove if |G|=p^2, then G is abelian.

    For the first part, I have:
    So |G|=p^{\alpha}m. By the class equation, |G|=|Z(G)|+\sum_{i=0}^r|G:C_G(g_i)|.

    Since p| \ |G|, p|\left(|Z(G)|+\sum_{i=0}^r|G:C_G(g_i)|\right)

    So p| \ |Z(G)| and p| \ \sum_{i=0}^r|G:C_G(g_i)|.

    Since p is prime, |Z(G)|\neq 1. Therefore, Z(G)\neq\{e\}

    So how do I use that to prove the next part?
    The first part is fine, but there is a principle you should remember. If P is a finite p-group acting on a finite set X, and if X^P denotes the fix set of the action (i.e. those elements that are fixed by every then |X|\equiv |X^P|\text{ mod }p. So, if, in your example, you let X=G act on itself by conjugation you get that |X^G|\equiv |G|=p^2\equiv 0\text{ mod }p but evidently X^G=Z(G) and so the result follows. But, as I said, your solution is fine. This technique just generalizes and is nice to know.


    For the second part you need to be aware of the common fact that if N\leqslant Z(G)\leqslant G and G/N is cyclic then G is abelian (exercise). So, you know that |Z(G)|=p,p^2, suppose that |Z(G)|=p then |G/Z(G)|=p and so G/Z(G) is cyclic, and so by the previous sentence G is abelian, which contradicts that |Z(G)|=p. Thus, |Z(G)|=p^2 and so G=Z(G) as desired.
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  3. #3
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    Re: p-group

    Why is |Z(G)|=p a contradiction? We want G to be abelian.
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    MHF Contributor Drexel28's Avatar
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    Re: p-group

    Quote Originally Posted by dwsmith View Post
    Why is |Z(G)|=p a contradiction? We want G to be abelian.
    It's a weird argument. You can just conclude by saying that G is abelian, but it feels wrong. The contradiction is that if G is abelian than |Z(G)|=p^.
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  5. #5
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    Re: p-group

    i find it instructive to see why if N ⊆ Z(G), and G/N is cyclic, G is abelian.

    let xN be the (a) generator of G/N. so every element of G can be written as: x^kz, where z is in Z(G) (because N is contained within Z(G)).

    so for any g,h in G: gh = (x^kz_1)(x^mz_2) = x^{k+m}z_1z_2 (since z_1 is in Z(G))

    =(x^m)(x^k)z_2z_1 = (x^mz_2)(x^kz_1) = hg (since z_2 is in Z(G)), so G is abelian.

    (the contradiction is: assuming |Z(G)| = p, and Z(G) ≠ G, leads to Z(G) = G. therefore, |Z(G)| ≠ p).
    Last edited by Deveno; December 3rd 2011 at 01:02 AM.
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