# Group Theory - Sylow Theory and simple groups

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• May 12th 2009, 11:53 AM
Jason Bourne
Quote:

Originally Posted by Gamma
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.
• May 16th 2009, 11:10 AM
TheAbstractionist
Quote:

Originally Posted by Jason Bourne
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 $\displaystyle xy=yx.$
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