Ineed help to solve the following problem:
Let $G$ be a finite group and $H<G$. suppose that $C_G(x)\subset H$ for all $x\in H\backslash\left\{e\right\}.$
Show that $|H|$ is relatively prime to $[G:H]$.
Thanks in advance.
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Do you have any insights of your own? – Edward Evans Sep 30 '17 at 11:25
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no i tried but i dont have any idea for the proof. I will be greatful if you could help me. Thanks. – Sep 30 '17 at 11:27
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I give up. I thought that drawing the sublattice of $G$ and applying second isomorphism theorem could help. D: – Sep 30 '17 at 12:16
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Let $p$ be a prime dividing $|H|$ and $P$ be a Sylow $p$ subgroup of $H$. Then $P$ is contained in some Sylow $p$ subgroup $Q$ of $G$. Note that if $q\in Z(Q)$ then $q\in C_G(x)$ for all $x\in P$ so $q\in H$. But $Z(Q)$ is non-trivial, so let $1\ne q\in Z(Q)$. $Q\le C_G(q)$ so $Q\le H$ and therefore $P=Q$.
Now we have for any $p$ dividing $|H|$ that the Sylow $p$ subgroups of $H$ are Sylow $p$ subgroups of $G$ so $p$ does not divide $[G:H]$. Hence $|H|$ and $[G:H]$ are relatively prime.
Robert Chamberlain
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Because $Q$ is a $p$-group. It follows directly from the class equation that $Z(Q)\neq 1$ – Sep 30 '17 at 13:37
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Every normal subgroup of a $p$ group intersects the center non-trivially - In particularly $p$ groups have non-trivial center. – Robert Chamberlain Sep 30 '17 at 13:39
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