I'm studying right now Automorphisms in Dummit & Foote's Abstract Algebra (Section 4.4). In pages 135-136, the following example is given:
and here's Proposition 16 muntionned in the proof:
(Note that the authors proved the existence of an element of order $p$ without using Cauchy's Theorem, a result that we know and that is proved so far only for abelian groups. We can prove the existence of an element of order $q$ by the same reasoning)
Something that I didn't understand in the proof is: why do we have $C_G(H)=H$ if $Z(G)=1$ (underlined line)? The way the authors said it means that it's not hard to see, but after spending time trying to prove it I managed to prove it in a quite long way (In case I made a mistake please tell me):
Let $y$ be an element of $G$ of order $p$ ($y$ exists as we showed in the discussion below Proposition 16).
We can easly prove that the left cosets $H,\,yH,\,y^2H,\,\dots,\,y^{p-1}H$ are all distinct and so they are pairwise disjoint. Also they all have the order of $H$ which is $q$. Thus:$$\left|\bigcup_{i=0}^{p-1}y^iH\right|=\sum_{i=0}^{p-1}|y^iH|=\sum_{i=0}^{p-1}q=pq=|G|$$
Therefore $G=\bigcup_{i=0}^{p-1}y^iH$ and so $H,\,yH,\,y^2H,\,\dots,\,y^{p-1}H$ form a partition of $G$.
Now since $H$ is cyclic, we have $H\le C_G(H)$ and so $q$ divides $|C_G(H) |$ by Lagrange's Theorem. Since $C_G(H)\le G$ we have again by Lagrange's Theorem: $|C_G(H)|$ divides $pq$. Therefore $|C_G(H)|\in\{q,pq\}$.
Let's now prove that $C_G(H)=H$. Suppose that that's not the case, then $|C_G(H)|\neq q$ because $H\subset C_G(H)$. Thus $|C_G(H)|=pq=|G|$ and so $C_G(H)=G$.
Let $g\in G$ such as $g\neq 1$. Let's prove that $g\in Z(G)$.
Let $z\in G$. Thus $z\in y^iH$ for some $i\in\{0,\dots , p-1\}$ and so $\exists \alpha\in\mathbb{Z}^+,\,z=y^ix^\alpha$. We have then $z^{-1}=x^{-\alpha}y^{-i}$. We also have $gz\in G$ and so $\exists j\in\{0,\dots , p-1\},\,\exists \beta\in\mathbb{Z}^+,\,gz=y^jx^\beta$. Therefore $g=(gz)z^{-1}=y^jx^{\beta-\alpha}y^{-i}$. Since $C_G(H)=G$, all elements of $G$ commute with $H$'s elements. In particular: $x^{\beta-\alpha}y^{-i}=y^{-i}x^{\beta-\alpha}$. Therefore $g=y^{j-i}x^{\beta -\alpha}$.
Then we have:$$zg=y^ix^\alpha y^{j-i}x^{\beta -\alpha}=y^jx^\beta=gz$$ because $C_G(H)=G$ and so $y^{j-i}$ and $x^\alpha$ commute.
Therefore $g\in Z(G)$ and so $g=1$ ($Z(G)=1$) which contradicts the fact that $g\neq 1$.
Finally:$$C_G(H)=H$$
So could you please tell me if there's any simple reason for $C_G(H)=H$?
Since $H$ is cyclic hence abelian, we have $H\subseteq C_G(H) \subseteq G$. By order considerations, $C_G(H)$ must be either $H$ or $G$. Suppose $C_G(H) = G$. Then in particular $x$ commutes with every element of $G$, so that $x\in Z(G)$, contradiction.