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Suppose that $H,C_G(H) \leq G$ where $G=HC_G(H)$. I want to show that $H\cap C_G(H) \leq Z(G).$(H being a subgroup, $C_G$ the centraliser of H and Z the centre of the group.)

My attempt :

$H,C_G(H) \leq G, \Rightarrow H \cap C_G(H) \leq G$ and that $|HC_G(H)|=\tfrac{|H||C_G(H)|}{|H\cap C_G(H)|}$

However

$|HC_G(H)|=|C_G(H)| \Rightarrow |H\cap C_G(H)|=|H|$.

Next consider Z(G)

$|Z(G)|=|G|-\sum|K(g)|$

But looking at K(g) and G

$K(g)=\{y \in G | y=x^{-1}gx , x \in G\}$

$G=\{hg|g=h^{-1}gh\}$ which means that for all g, |K(g)|=0.

and so we are left with $|Z|=|G|$.

Now we know that $|H|$ divides $|G|$, which means that $|H\cap C_G(H)|$ divides $|Z(G)|$ and so by lagrange's theorem $|H\cap C_G(H)|\leq |Z(G)|$

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  • $\begingroup$ Is it given to you that $G$ is finite? $\endgroup$ – Anurag A Oct 23 '18 at 23:16
  • $\begingroup$ @AnuragA No G is any group. $\endgroup$ – excalibirr Oct 23 '18 at 23:43
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    $\begingroup$ In which case you cannot use the cardinality of different subgroups or Lagrange's theorem. $\endgroup$ – Anurag A Oct 23 '18 at 23:45
  • $\begingroup$ damn, back to the drawing board $\endgroup$ – excalibirr Oct 24 '18 at 13:47

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