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I wonder if you explain that why Chermak and Delgado named it in their article by measuring lemma? Is the measuring lemma caused the measure on finite groups?

A. Chermak, A. L. Delgado, A measuring argument for finite groups, Proc. Amer. Math. Soc. 107 (1989), 907-914.

measuring lemma:let $ ‎A,B‎ ‎\in‎ ‎\alpha‎\mathcal{M}‎ $ and assume that either$ ‎A‎\cap B‎‎ ‎\neq ‎1‎ $‎‏ ‎ or that $ ‎m‎_{‎\alpha‎}‎‎‎ ‎\geq‎ ‎\vert H ‎\vert‎ $‎ .

Then ‎‎‎$ ‎‎AB=BA‎ \in‎ ‎\alpha‎\mathcal{M} $‎ and $ C_{H} ( A \cap B ) = C_{H} (A) C_{H}(B) $ Further, if $ A \cap B \neq 1$ then $ A \cap B \in \alpha‎\mathcal{M} $ . we have the following notation: Let $G$ be finite group acting on a finite group $H$, and let $ S^{*} (G) = $ $\brace all\ non-trivial\ subgroups\ of\ G $

$ ‎m‎_{‎\alpha‎} $

$= m‎_{‎\alpha‎} ( G,H ) $

$ = ‎‎ \max ‎\lbrace $ ‎‎‎$ ‎\vert ‎A‎ ‎ \vert‎^{‎\alpha‎}‎$ $ \vert C‎_{H}(A)‎‎ ‎\vert‎^{‎\alpha‎} $ $ \ \ \ ‎‎\vert‎ \ $ $ \ \ \ A ‎‎\in S‎^{‎\ast‎‎}(G)$ $‎\rbrace ‎‎‎‎‎‎‎‎$‎ ;

$ ‎\alpha‎\mathcal{M}$

$=‎\alpha‎\mathcal{M}(G,H)$

$=‎\lbrace $ $ A ‎‎\in$ $ S‎^{‎\*‎‎}(G) $ $ ‎‎\vert $ $ ‎‎ ‎\vert ‎A‎ ‎\vert‎^{‎\alpha‎}‎ $ $ \vert C‎_{H}(A)‎‎ ‎\vert‎^{‎\alpha‎} $

$ =m‎_{‎\alpha‎} ‎\rbrace ‎.‎$‎‎

$ ‎C‎_{H}(A)$

$=fix‎_{H}(A)$

$ =‎\lbrace $ $ h‎ \in ‎H $ ‎ $ ‎\vert $ $ h‎^{a}=h $ ‎ ‎‎ $\forall ‎‎a ‎\in A$ $‎‎‎‎‎‎‎‎‎‎‎\rbrace‎ $‎

Where $ h‎^{a} $ denotes the action of $a$ on $ h $.

Best Regards

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1  
Pls rephrase the lemma. –  Marc Palm Feb 13 '13 at 9:19
1  
Yes, please give the statement of the lemma in your question. You're asking for help, so it's a good idea to make it easy for someone to help you. –  HJRW Feb 13 '13 at 10:51
    
okay now the notation has to specified;-) –  Marc Palm Feb 22 '13 at 19:08

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