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When I'm proving this Lemma, I use the fact (recommended by my advisor) that

$\begin{array}{cl} \sum_{\alpha\in \Omega}|G_{\alpha}| & =\sum_{\alpha\in \Omega}\frac{|G|}{|G\cdot\alpha|} \\ & =|G|\sum_{\alpha\in \Omega}\frac{1}{|G\cdot\alpha|} \\ & =|G|\sum_{A\in X/G}\sum_{\alpha\in A}\frac{1}{|A|} \\ & =|G|\sum_{A\in X/G}1 \\ &=|G||X/G|.\end{array}$

But I don't understand why can I do $|G|\sum_{\alpha\in \Omega}\frac{1}{|G\cdot\alpha|}=|G|\sum_{A\in X/G}\sum_{\alpha\in A}\frac{1}{|A|}~?$

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    ty ever so much!!2012-10-20

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