# Is this the correct way to prove that this intersection is a subgroup of Z(G)

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 $$|H\cap C_G(H)|\leq |Z(G)|$$

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