Prove that a group of order 200 has a normal Sylow 5-subgroup.
$\textit{Proof}$ : Note that $200=2^3 \cdot 5^2$. By Sylow (iii), we know that
$$n_5 \equiv 1(mod5)$$
and
$$n_5 | 2^3$$
So the possible values of $n_5$ are $n_5=1,6$. But $6 \nmid 8$. So, we are forced that
$$n_5=1$$
By Sylow (ii), Sylow p-subgroups are conjugate to each other. So, this Sylow 5-subgroup, $P$, which has order $5^2=25$ is conjugate to itself i.e.
$$gPg^{-1}=P \quad \forall g\in{G} \Longrightarrow P \unlhd G$$
This is corollary 20, p. 142. So, for groups of order 200, they must always contain a normal Sylow 5-subgroup. $\blacksquare$
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