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Let $G$ be a group. If the subgroup $H \leq G$ is generated by $\lbrace a_1,...,a_k \rbrace$, and $C_G(H)$ denotes the centralizer of $H$, then why do we have the following inequality $$[G : C_G(H)] \leq \prod_{i=1}^k [G:C_G(a_i)] $$

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By definition, $C_G(H)$ consists of all elements $x \in G$ such that $xh = hx$ for every $h \in H$. Therefore, $C_G(H)$ is equal to the intersection of the subgroups $C_G(a_i)$, $i = 1, \dots, k$.

It now remains to show that for two subgroups $A$ and $B$ (and $k$ subgroups by induction) of $G$ that

$$[G:A \cap B] \leq [G:A] [G:B] \qquad (*)$$

To prove $(*)$, show directly that the mapping $g A \cap B \mapsto (gA, gB)$ is injective.

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