Nonabelian group of order $p^4$

Question

Let $P$ be a nonabelian group of order $p^4$, where $p$ is a prime, and let $A$ be a subgroup of $P$ maximal with the property of being normal and abelian. Prove that $A$ is of order $p^3$.

Thanks a lot.

Answer 1

Since $A$ is abelian and normal, there is an morphism $\rho : P/A \rightarrow {\rm Aut}(A)$ given by $$\rho(\overline{\sigma})a:= \sigma a \sigma^{-1},$$ for all $\sigma \in P$ and $a \in A$.

If $A$ has order $\leq p^2$, then $\rho$ has a kernel. If $\overline{\sigma}_0 \in P$ is in the kernel, then the subgroup generated by $\sigma_0$ and $A$ is normal abelian (I let you fill in the details).