= Solution
The <quaternion group> of order eight has a unique <involution>, which generates its <center of a group>. A nontrivial intersection $S\cap S'$ is a nontrivial finite $2$-<group>, so by <Cauchy's theorem for finite groups> it contains an <involution>. This must be both the unique <involution> $t$ of $S$ and the unique <involution> of $S'$. Thus $t\in Z(S')$.
It follows that $S'\leq C_G(t)=N_G(S)$. But $S$ is a normal <Sylow subgroup> of its <normalizer>, so it is that <normalizer>'s unique <Sylow subgroup>. Hence $\boxed{S'=S}$. Distinct <Sylow subgroups> therefore intersect trivially. From the preceding result,
$$
n_2\equiv1\pmod8,\qquad n_2\mid168/8=21.
$$
Of $1,3,7,21$, only $1$ is congruent to $1$ modulo $8$. Therefore \b[$n_2=1$, so $S$ is normal in $G$].
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