There is no core free subgroup of order $p^2$ in a group of order $p^4$ By the classification of group of order $p^4$ ($p$ odd prime) from Burnside's book it seems to me that there is no core free subgroup of order $p^2$ in a group of order $p^4$. If I am not wrong there must be some argument to prove this. Can someone help me in proving this? 
 A: This is a partial answer.
Let $K<H<G, |K|=p^2, |H|=p^3$. Suppose that $H$ is non-abelian. Since $K$ is normal in $H$, then $Z(H)<K$ (otherwise $H=Z(H)\times K$, what is impossible). Further, $Z(H)$ is characteristic in $H$, hence it is normal in $G$. So the core of $K$ is not trivial.
A: Here is the other half of Boris Novikov's answer:
Let $p$ be an odd prime, and let $H=\langle h_1, h_2, h_3 \rangle$ be an elementary abelian subgroup of order $p^3$. Let $x$ act on $H$ as the matrix $$\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$$ so that $x$ has order $p$ (here we use $p$ odd) and the semidirect product $G=\langle x \rangle \ltimes H$ has order $p^4$.
Then $\langle h_1, h_2 \rangle$ is a core-free subgroup of order $p^2$ (numbering chosen so that $\langle h_3\rangle$, $\langle h_2, h_3 \rangle$, and $\langle h_1, h_2, h_3\rangle$ are normal subgroups of $G$).
In other words, just choose an indecomposable module of length 3 and a vector space complement to its socle.
