Solution (source code)

= Solution

For a <group action> of a <finite group> $G$ on a set and a point $x$, the <orbit-stabilizer theorem> is
$$
\boxed{|Gx|=[G:G_x],\qquad |G|=|G_x||Gx|,}
$$
where $G_x=\{g:gx=x\}$ is the <stabilizer subgroup>. The map $gG_x\mapsto gx$ is well defined because elements of the same <coset> act identically on $x$. If $gx=hx$, then $h^{-1}g\in G_x$, so their <cosets> agree; every orbit point has the form $gx$, giving surjectivity. This is a <bijection>. <Lagrange's theorem>, $|G|=[G:H]|H|$ for any <subgroup> $H$, yields the counting formula.

Apply the <conjugation action> of the <finite p-group> $G$ to its nontrivial <normal subgroup> $N$. Normality ensures the action stays inside $N$. The singleton <orbits of a group action> are exactly the elements of $N\cap Z(G)$. Every larger <orbit of a group action> has size a positive power of $p$, by the <orbit-stabilizer theorem> and <Lagrange's theorem>. Therefore
$$
|N|\equiv |N\cap Z(G)|\pmod p.
$$
Since $N$ is nontrivial, its order is divisible by $p$. The intersection contains the identity and has cardinality a positive multiple of $p$, so it contains a nonidentity element. This proves the <central intersection property of normal subgroups of finite p-groups>.

For the proper <subgroup> $H$, let $H$ act by left multiplication on the left <cosets> $G/H$. A <coset> $gH$ is fixed by every $h\in H$ exactly when $g^{-1}Hg\subseteq H$. Both groups have the same finite size, so this inclusion is equality; the fixed <cosets> are precisely those with $g\in N_G(H)$. Hence their number is $[N_G(H):H]$. All non-singleton <orbits of a group action> have size divisible by $p$, so
$$
[G:H]\equiv[N_G(H):H]\pmod p.
$$
Since $H$ is proper, $[G:H]$ is a positive power of $p$ greater than one. The fixed-coset count is nonzero, because $H$ itself is fixed, and is divisible by $p$. Therefore \b[the <normalizer> strictly contains $H$]:
$$
\boxed{[N_G(H):H]\geq p,\qquad\exists g\in G\setminus H:\ g^{-1}Hg=H.}
$$
This is the <normalizer condition for finite p-groups>. The proof also covers $H=\{1\}$, whose action fixes all <cosets>.