proving that various descriptions of the subgroup O^p(G) are equivalent

Question

Let $G$ be a group of order $p^{\alpha}m$, where $p$ is prime and $p$ does not divide $m$; my question pertains to the subgroup \begin{align*} O^p (G) \mathrel{\mathop:}= \min \left( \{ N \triangleleft G : |G/N| = p^{k} \text{ for some } k \in \mathbb{N}\cup \{0\} \} \right), \end{align*} i.e., the smallest normal subgroup of $N$ of $G$ such that $G/N$ is a $p$-group. I have already shown that $O^p (G) = \left< \{ g \in G : gcd(p,|g|) = 1 \}\right>$; given this description of $O^p(G)$ as well as the definition above, I am trying to show that $O^P (G) = \left<X\right>$, where \begin{align*} X = \{ Q : Q \in \bigcup_{q \in \mathbb{P} - \{p\}} {Syl}_q (G) \} \end{align*} and $\mathbb{P} \mathrel{\mathop:}= \{ p \in \mathbb{N} : p \text{ is prime}\}$. I know that this is the case when $m =1$ and thus have assumed that $m > 1$; I also know that $\left<X\right> \triangleleft G$ and $\left<X\right> \le O^p(G)$. Since $m > 1$, by the Fundamental Theorem of Arithmetic there exist distinct primes $q_1,q_2,...,q_n$ such that \begin{align*} m = {q_1}^{{\beta}_1} \cdots {q_n}^{{\beta}_n}; \end{align*} select representatives $Q_1,...,Q_n \in {Syl}_{q_1}(G),...,{Syl}_{q_n} (G)$, respectively, so each $Q_i \le G$ such that $|Q_i| = {q_i}^{{\beta}_i}$, so $X$ consists of $Q_1,...,Q_n$ and each of their conjugates in $G$. I understand that the idea is to show that $G/\left<X\right>$ is a $p$-group and then use the minimality of $O^p (G)$ to get the reverse containment, but this is precisely where I'm stuck. The partial proof given by the author says, "As $\left<X\right>$ contains $q$-Sylow groups of G for all primes $q \ne p$, no prime $q \ne p$ can divide $|G : \left<X\right>|$. Thus $|G : \left<X\right>|$ is a power of $p$ and thus $O^p(G) ⊆ \left<X\right>$." The author is correct that $\left<X\right>$ contains each of the $Q_i$'s, but it contains them as elements and not as subsets (i.e., not as subgroups, so we cannot use Lagrange's Theorem applied to each $Q_i$). I understand that this means by a corollary to Lagrange's Theorem that the order of each $Q_i$ as an element of $\left<X\right>$ divides $|\left<X\right>|$, but as I understand it, I cannot assume that the order of any element of $\left<X\right>$ is $q_i$ or some power of $q_i$ for some $i \in \{1,...,n\}$. I know by assumption that $|G/O^p(G)| = p^{k}$ for some $k \ge 0$, so $|O^p(G)| = p^{\alpha - k}m$, and by Lagrange's Theorem, as $\left<X\right> \le O^p(G)$, $p^{\alpha - k}m = |O^p(G)| = |\left<X\right>|\cdot j$ for some $j \in \mathbb{Z}$, which implies that $|G/\left<X\right>| = p^{k}j$. I have been unsuccessful in all attempts to prove that $j$ must be a power of $p$, and I am very confused as to the justification of the author's reasoning that $|G/\left<X\right>|$ must be a power of $p$. Any tips or clarification would be much appreciated.

Answer 1

Let H be the subgroup generated by all those Sylows. It is normal, because the conjugate of a Sylow is a Sylow. Since every element of G of order coprime to p has to be in O, every Sylow of order coprime to p is contained in O, and therefore H is contained in O. On the other hand, it is clear that G/H has order a power of p. So H coincides with O by the definition of the latter.