Let $ p $ be a fast number and $ (G, \cdot) $ a finite p-group. Show that if $H$ is a normal subgroup of $G$, then $ H \cap Z (G) \neq \left \{I_ {G}\right \}.$ (Here $Z (G)$ is the $G$ centralizer.)
I think there is a theorem that if a $p$-group is finite, then $| G |$ is it's a $p$-power. I think there is a theorem Sylow that if a $p$-group is finite, then $| G |$ is it's a $p$-power.
I assume you mean that $p$ is a prime number, and also assume that $H$ is nontrivial.
Let $G$ act on $H$ by conjugation. Then the orbit-stabilizer theorem gives us $$|H| = \text{# of one-point orbits} + \sum (\text{sizes of nontrivial orbits})$$ Now $h \in H$ is in a one-point orbit iff $ghg^{-1} = h$ for every $g \in G$, iff $gh = hg$ for every $g \in G$, if and only if $h \in H \cap Z(G)$. Therefore, $$|H| = |H \cap Z(G)| + \sum (\text{sizes of nontrivial orbits})$$ Now compute this equation modulo $p$. Since $G$ is a $p$-group, we have $|H|$ is a power of $p$, so $|H| \equiv 0\ (\operatorname{mod} p)$. Also, the size of any orbit is a divisor of $|G|$, so the size of each nontrivial orbit is also $0$ mod $p$. Thus the equation reduces to $$0 \equiv |H \cap Z(G)| + 0\ (\operatorname{mod} p)$$ Since $H \cap Z(G)$ is nonempty (it contains the identity), this means that $|H \cap Z(G)|$ is at least $p$, hence $H \cap Z(G)$ is nontrivial.