When is the centralizer of a subgroup equal to the subgroup itself

abstract-algebragroup-theory

I am going through a proof of the proposition

Assume $G$ is a group of order $pq$, where $p$ and $q$ are primes with $p\le q$ and $p$ does not divide $q-1$. Then $G$ is abelian.

in Abstract Algbra by Dummit and Foote, and I am stuck at one step. The relevant assumptions are: $Z(G)=1$ (the center of $G$ is the trivial subgroup), $G$ has an element $x$ of order $q$, and $H=\langle x\rangle$. At this point the author concludes that $C_G(H)=H$, ($C_G(H)$ represents the centralizer of $H$ in $G$) but I cannot see why.

I can see $H\subset C_G(H)$. Also $C_G(H)=C_G(x)$. But how to proceed? In particular, I do not know how the condition $Z(G)=1$ is relevant here.

Any help is appreciated. If you believe more information is needed in the context, I can provide it.

Best Answer

Argue using orders:

The centraliser $C_G(H)$ can only have order $1$, $p$, $q$, or $pq$, as it must divide the order of $G$.

  • As $H\leq C_G(H)$ and $|H|=q$, we have that $C_G(H)$ has order either $q$ or $pq$.
  • As $Z(G)=1$, we see that $C_G(H)\lneq G$ (as if $G=C_G(H)$ then the generator $x$ of $H$ would centralise every element of $G$, and so $x\in Z(G)$). So the order cannot be $pq$.

Hence, $|C_H(G)|=q$, and so $C_G(H)=H$.

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