a group of order $p^3$ contains a normal subgroup of order $p^2$ Let $p\neq2$ be a prime number. 
Let G be a group s.t. $|G|=p^3$.
Prove that there exists $H\triangleleft G$, a normal subgroup of order $p^2$.
 A: We know there exists a subgroup by Sylow's theorem. It is normal because its index is the smallest prime dividing order of the group.
A: In fact we can go further: every finite $p$-group $P$ has a  normal subgroup of every possible order (allowed by Lagrange's theorem: so any of $1,p,\cdots,p^n$). There are two key facts that go into proving this: the lattice correspondence theorem, and the fact that any finite $p$-group $P$ has a nontrivial center $Z(P)$.
Indeed, the proof can proceed inductively. Since $Z(P)\le P$ is nontrivial, it has a nontrivial element say $z\in Z(P)$ of order $p$ (which is fairly easy to show for $p$-groups), in which case $P/\langle z\rangle$ has order $p^{n-1}$ and by induction hypothesis has a normal subgroup $\overline{N}\trianglelefteq P/\langle z\rangle$ of prescribed order $p^{k-1}$, which corresponds to a normal subgroup $N\trianglelefteq P$ of order $p^k$.
The lattice correspondence theorem is sometimes called the "Fourth" Isomorphism Theorem, and the fact a finite $p$-group $P$ follows from the class equation (which itself follows from the orbit-stabilizer theorem) and basic modular arithmetic.
