Let $G$ be a non-cyclic group of order $p^3$ for an odd prime $p$. Prove that $G \simeq H \rtimes_{\theta}K$, where $H$ is a normal subgroup of $G$ of order $p^2$, $K$ is a subgroup of order $p$, and $\theta : K \to Aut(H)$ is a homomorphism.
I managed to prove that there exists a normal subgroup $H$ of order $p^2$. Then I took some $g \in G-H$. If $g$ is of order $p$, I am done since $G \simeq H \rtimes \langle g \rangle $. But what if all $g \in G - H$ are of order $p^2 $?
The result follows immediately if all elements of $G$ have order $1$ or $p$, so we can assume that there exists an element of order $p^2$. This generates a cyclic subgroup of index $p$, which is normal in $G$, so we can assume that $H$ is cyclic of order $p^2$.
So let $g \in G \setminus H$, and assume that $g$ has order $p^2$ (if it has order $p$ then we are done). So $g^p = h^p$ for some $h \in H$. We claim that $(gh^{-1})^p=1$, and hence $gh^{-1}$ has order $p$ and we are done.
This is immediate if $G$ is abelian, so suppose not. Then $[G,G]=Z(G)$ has order $p$, so commutators are central, and $(xy^{-1})^p = x^py^{-p}[y^{-1},x]^{p(p-1)/2}$ (where $[a,b]$ denotes the commutator $a^{-1}b^{-1}ab$).
Then, since $p$ is odd and $[y^{-1},x]$ has order $p$, we get $[y^{-1},x]^{p(p-1)/2}=1$ and the claim follows. (Note that this result is false when $p=2$, and the quaternion group $Q_8$ is a counterexample.)
