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 $.
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