Theorem 3

Let G be a finite group. (a) The Sylow p-subgroups of G are conjugate to one another. Their number is congruent to 1 mod. p. (b) Every subgroup of G which is a p-group is contained in a Sylow p-subgroup.

Let P be a Sylow $p$-subgroup of G (Theorem 2) and let H be a $p$-subgroup of G. Let E = G/P and consider the operation of H on G/P. As Card(E) $\not\equiv 0$ mod. $p$, Proposition 11 of no. 5, shows that there exists $x \in G/P$ such that $hx = x$ for all $h \in H$. If g is a representative of x in G, this means that $H \subset gPg^{-1}$, whence assertion (b).

If H is a Sylow $p$-subgroup, then Card(H) = Card(P) = Card($gPg^{-1}$), whence $H = gPg^{-1}$, which proves the first assertion of (a).

We now prove the second assertion of (a). Let $\mathscr{S}$ be the set of Sylow $p$-subgroups of G and let P operate on $\mathscr{S}$ by inner automorphisms. The element $P \in \mathscr{S}$ is a fixed point under this operation, we show that it is the only one. Let $Q \in \mathscr{S}$ be a fixed point; Q is a Sylow subgroup of G normalized by P and hence P is contained in the normalizer N of Q. The groups P and Q are Sylow $p$-subgroups of N; hence there exists $n \in N$ such that $P = nQn^{-1} = Q$. By no. 5, Proposition 11, Card($\mathscr{S}^P$) $\equiv$ Card($\mathscr{S}^{P^P}$) = 1 (mod. $p$).