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Thread: Group Theory - Sylow Theory and simple groups

  1. #16
    Member Jason Bourne's Avatar
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    Quote Originally Posted by Gamma View Post
    Sorry I don't have much time to write it out myself for you, I have to get to a final exam.

    basically p divides |G| and each term of the sum, but if |Z(G)|=1 this would be impossible (move the sum over to the other side and p divides the left side, but not the right. Z(G) is a subgroup, so by lagrange its order has to divide $\displaystyle p^2$ and we see it is not 1.

    Here is a proof of the G/Z(G) cyclic thing.
    http://crypto.stanford.edu/pbc/notes...quotient.xhtml

    Sorry again for the brevity.
    Thanks, hope your exam went well.
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  2. #17
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by Jason Bourne View Post
    so the class equation is some thing like $\displaystyle |G| = |Z(G)| + \sum_{i=1}^{r} [G:N(x_i)]$

    I don't think I understand the Class Equation properly. So I don't have this proof anywhere, $\displaystyle G/Z(G)$ is cyclic so this implies G is abelian, how?
    Hi Jason Bourne.

    Here is a detailed proof of the class equation posted here not so long ago: http://www.mathhelpforum.com/math-he...al-center.html

    To show that $\displaystyle G/Z(G)$ cylic, say $\displaystyle G/Z(G)=\left<gZ(G)\right>$ for some $\displaystyle g\in G,$ implies $\displaystyle G$ Abelian, let $\displaystyle x,y\in G.$ Then $\displaystyle x,y$ belong to some left cosets of $\displaystyle Z(G)$ say $\displaystyle x=g^iz_1,\,y=g^jz^2$ for some $\displaystyle z_1,z_2\in Z(G)$ and $\displaystyle i,j\in\mathbb Z.$ It is then a straightforward matter to verify that $\displaystyle xy=yx.$
    Last edited by TheAbstractionist; May 17th 2009 at 05:09 AM.
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