Exercise 4.5.38 (Sylow $p$-subgroups $P$ and $Q$ of $G$ such that ${|P : P \cap Q| = |Q : P \cap Q| = p}$)

Proof. We suppose that $n_p\not\equiv 1(\mathit{𝑚𝑜𝑑}{p}^2)$ (thus $n_p>1$). As in the proof of Sylow’s Theorem p. 140, let

be the set of the $r=n_p>1$ distinct Sylow $p$-subgroups of $G$. Then $P_1$ acts by conjugacy on $𝒮$. Write $𝒮$ as a disjoint union of orbits under this action by $P_1$:

Then

Renumber the elements of $𝒮$ is necessary so that the first $s$ elements of $𝒮$ are representatives of the $P_1$-orbits: $P_i\in {𝒪}_i,1\le i\le s$. In particular, the orbit of $P_1$ is ${𝒪}_1=\{P_1\}$, so

By (4.1) (p. 141), where $Q=P_1$,

For every $i\in [[2,s]]$, $P_i\ne P_1$, thus $|P_1:P_i\cap P_1|>1$, so

Assume for the sake of contradiction that $p^2$ divides $|P_1:P_i\cap P_1|$ for all $i\in [[2,s]]$. Then the relations (1), (2) and (3) show that $n_p\equiv 1(\mathit{𝑚𝑜𝑑}{p}^2)$, in contradiction with the hypothesis. This contradiction proves that there is some $i\in [[2,s]]$ such that

where $|P_1:P_i\cap P_1|$ is a power of $p$, therefore

If we put $P=P_1$ and $Q=P_i$, then $P,Q$ are distinct Sylow $p$-subgroups of $G$, and

Finally, since $P$ and $Q$ are both Sylow $p$-subgroups of $G$, then $|P|=|Q|$, thus

In conclusion, if $n_p\not\equiv 1(\mathit{𝑚𝑜𝑑}{p}^2)$, there are distinct Sylow $p$-subgroups $P$ and $Q$ of $G$ such that

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