I have a doubt in Dummit&Foote's Abstract Algebra on page142 :
Corollary 20. Let $P$ be a Sylow $p$-subgroup of $G$ . Then the following are equivalent:
(1) $P$ is the unique Sylow $p$-subgroup of $G$, i.e., $n_p = 1$
(2) $P$ is normal in $G$
(3) $P$ is characteristic in $G$
(4) All subgroups generated by elements of $p$-power order are $p$-groups, i.e., if $X$ is any subset of $G$ such that $|x|$ is a power of $p$ for all $x\in X$ , then $\langle X\rangle$ is a $p$-group.
And the latter proof by the authors shall show that $(4)\Rightarrow(1)$ :
Conversely, if (4) holds, let $X$ be the union of all Sylow $p$-subgroups of $G$ . If $P$ is any Sylow $p$-subgroup, $P$ is a subgroup of the $p$-group $\langle X\rangle$ . Since $P$ is a $p$-subgroup of $G$ of maximal order, we must have $P = \langle X \rangle$ , so (1) holds.
i.e., $X=\bigcup_\limits{P\in\mathrm{Syl}_p(G)}P$ , but I think it does not satisfy that " $X$ is any subset of $G$ such that $|x|$ is a power of $p$ for all $x\in X$ ", which is the definition of $X$ in Corollary20(4). For instance, $1_G\in P$ but $|1_G|=1\ne p$ .
So how does it go?
What we want is that every element $x\in X$ has $|x|=p^n$ for some $n\in\mathbb{N}$. In this case, $|1_G|=1=p^0$. In general, since every element $x\in X$ is also an element of some $p$-group P, it must be the case that $|x|$ divides $|P|=p^k$, and so $|x|=p^n$ for some $n\leq k$.
We then use 4 to conclude that $\langle X\rangle$ is a $p$-group. By construction, X contains all Sylow p-groups, and by their maximality they all in fact are equal to $\langle X\rangle$.