Normal subgroup of a normal subgroup Let $F,G,H$ be groups such that $F\trianglelefteq G \trianglelefteq H$.
I am asked whether we necessarily have $F\trianglelefteq H$. I think the answer is no but I cannot find any counterexample with usual groups. Is there a simple case where this property is not true?
 A: Let $H=D_8$, the group of symmetries of a square under flips and rotations. Let $F$ be the subgroup of flips about the vertical axis of symmetry. Let $G$ the symmetries you can find by combinations of such flips and $180$ degree rotations. You can show that $F$ is normal in $G$, and $G$ is normal in $H$.
Now let $h$ be a 90 degree clockwise turn and let $f$ be a flip. You can show that $hfh^{-1}$ is not a flip or the identity, so $F$ is not a normal subgroup of $H$.
A: $$\{\,(1)\,,\,(12)(34)\,\}\lhd\left\{\;(1)\,,\,(12)(34)\,,\,(13)(24)\,,\,(14)(23)\,\right\}\lhd A_4\;,\;\;\{\,(1)\,,\,(12)(34)\,\}\not\!\triangleleft A_4$$
A: Take any finite solvable group $G$ which has a minimal normal subgroup $M$ which is not cyclic. Let $\langle x \rangle$ be any non-identity cyclic subgroup of $M.$ Then 
$\langle x \rangle \lhd M \lhd G,$ but $\langle x \rangle \not \lhd G$, since $M$ is minimal
normal, but not cyclic. Now it's a matter of finding a solvable group with a non-cyclic minimal normal subgroup, which is not difficult.
A: Let $p$ be a prime, and let $G$ be a $p$-group of order $p^3$. Let $H \leq G$ be a non-normal subgroup of order $p$ (equivalently, $H$ is of order $p$ and not central). Then $H$ is contained in a subgroup $K \leq G$ of order $p^2$. In this case $H \trianglelefteq K \trianglelefteq G$, but $H$ is not normal in $G$.
For example, $G$ could be the Heisenberg group, which is the set
$$\left\{ \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix} : a, b, c \in \mathbb{Z}_p \right\}$$
of matrices under multiplication. In the case $p = 2$ this group is isomorphic to $D_8$, which is the example given in another answer.
