Maximal abelian subgroups in a $p$-group are always normal? Does all the maximal abelian subgroups in a given $p$-group have to be normal? I doubt the correctness of this claim however couldn't manage to find a counterexample. Can anyone give me some advice? Or would someone be able to disprove it?
 A: Take $p > 3$, say, and consider the semidirect product $G$ of an elementary abelian group of order $p^{3}$, so a vector space of basis $v_{3}, v_{2}, v_{1}$, extended by the automorphism $a$ of order $p$ that acts as $v_{i}^{a} = v_{i} v_{i-1}$, where we mean $v_{1}^{a} = v_{1}$.
Then $\langle a, v_{1} \rangle = C_{G}(a)$ is an abelian subgroup, which is maximal with respect to inclusion (if this is your meaning of maximal here), because it's the centralizer of $a$, but which is not normal, as $v_{3}$ does not normalize it.
I am pretty sure a similar example can be constructed with an abelian subgroup of maximal order among abelian subgroups.
A: Consider the dihedral group $G = D_{2^n} = \langle x, y: x^2 = y^{2^{n-1}} = 1, x^{-1}yx = y^{-1} \rangle$, where $n \geq 4$.
Let $H = \langle x, y^{2^{n-2}} \rangle$. Now $H$ is not a normal subgroup, but $C_G(H) = C_G(x) = H$, so $H$ is a maximal abelian subgroup.
A: If you mean maximal with respect to order (ie, $A$ is a maximal abelian subgroup of a $p$-group $P$ if it is abelian and there is no abelian subgroup $B$ of $P$ with $|B|>|A|$) then there are indeed $p$-groups $P$ that contain maximal abelian subgroups not normal in $P$ (indeed I believe there are groups $P$ where no maximal abelian subgroup is normal in $P$) though I'm afraid I don't have examples to hand.
For $p=2,3$ I think it is an open conjecture whether, for any given $p$-group $P$, there exists a maximal abelian subgroup $A$ that is normal in its normal closure in $P$ (ie, $A\unlhd\langle A^P\rangle$). Clearly if $A$ were necessarily normal in $P$ this would make the conjecture somewhat trivial.
