This question considers a proof of Sylow's first theorem (for every prime factor $p$ of $|G|$, a Sylow $p$-subgroup of $G$ exists).
My book proves this in the following way: prove the theorem for Abelian groups, then consider the centre $Z(G)$ in the case $G$ is not Abelian.
The part about Abelian groups in the proof seems very sloppy. Here is an English translation of the proof:
Let $S_p$ be the set of all elements with order $p^k$ for some $k$. In Abelian groups, the product of such elements also satisfy this property, so $S_p$ is a $p$-group. (Since there are no more $p$-elements, $S_p$ is maximal.) Due to Lagrange theorem, $|S_p|=p^n$ for some $n$. Again, since there are no more $p$-elements, $p^n\|G$, i.e., $p^{n+1}\nmid |G|$.
The problem here is that Lagrange theorem does not guarantee that $|S_p|=p^n$. It only guarantees that $mp^n=|S_p|$ for some $m$, $p\nmid m$. I worked quite hard to fill this gap. Let's do this by induction on $|S_p|$. Let $x\in G$. The order of $x$ is a power of $p$. $\langle x \rangle$ is a normal subgroup, and $|G/\langle x \rangle|$ is divisible by $m$. $G/\langle x \rangle$ is also a $p$-group of order $mp^{n'}$ (verification omitted). By induction assumption, $m=1$.
The last sentence above is also very imprecise. To prove that $p^{n+1}\nmid |G|$, let $Q=G/S_p$. If $p^{n+1}| \space |G|$, then $p| \space |Q|$. We now prove that $Q$ has an order $p^k$ element. We do this again by induction on the order of the group. Suppose $Q$ does not have an order-$p$ element. Then for any $y\in Q$, the order of $y$ is not divisible by $p$. Therefore, the order of the quotient group $Q/\langle y\rangle$ is divisible by $p$. By induction assumption, $Q/\langle y\rangle$ has an order-$p$ element. Let this element be $q\langle y\rangle, q\in Q$. Then $q^p=z\in \langle y\rangle$. Let the order of $z$ be $r$, then $q^r$ is an order-$p$ element, contradiction. So $G/S_p$ has an order $p$ element, and by the same argument, so does $G\backslash S_p$, contradicting the definition of $S_p$.
I have completed the proof, but it seems pretty strange that the author omits so many lines in his/her proof. Am I over complicating things? Is my proof correct? Is that a simpler proof?
PS: I admit that my proof should better be made into two lemmas.
PS: I have essentially proven Cauchy's theorem; it seems that the author hasn't proved it by this point.