normality of subgroups is not transitive

Let $G$ be a group.A subgroup $K$ of a subgroup $H$ of $G$ is obviously a subgroup of $G$.It seems plausible that a similar situation would also hold for normal subgroups, but in fact it does not:even when $K\mathrm{⊴}H$ and $H\mathrm{⊴}G$, it is possible that $K⋬G$. Here are two examples:

1. Let $G$ be the subgroup of orientation-preserving isometries (http://planetmath.org/Isometry) of the plane ${ℝ}^2$ ($G$ is just all rotations and translations), let $H$ be the subgroup of $G$ of translations, and let $K$ be the subgroup of $H$ of integer translations ${\tau }_{i,j}(x,y)=(x+i,y+j)$, where $i,j\in \mathbb{Z}$.

   Any element $g\in G$ may be represented as $g=r_1\circ t_1=t_2\circ r_2$, where $r_{1,2}$ are rotations and $t_{1,2}$ are translations. So for any translation $t\in H$ we may write

   $$g^{-1}\circ t\circ g=r^{-1}\circ t^{\prime }\circ r,$$

   where $t^{\prime }\in H$ is some other translation and $r$ is some rotation. But this is an orientation-preserving isometry of the plane that does not rotate, so it too must be a translation. Thus $G^{-1}HG=H$, and $H\mathrm{⊴}G$.

   $H$ is an abelian group, so all its subgroups, $K$ included, are normal.

   We claim that $K⋬G$. Indeed, if $\rho \in G$ is rotation by ${45}^{\circ }$ about the origin, then ${\rho }^{-1}\circ {\tau }_{1,0}\circ \rho$ is not an integer translation.

2. A related example uses finite subgroups. Let $G=D_4$ be the dihedral group with eight elements (the group of automorphisms of the graph of the square). Then

   $$D_4=⟨r,f\mid f^2=1,r^4=1,fr=r^{-1}f⟩$$

   is generated by $r$, rotation, and $f$, flipping.

   

   The subgroup

   $$H=⟨rf,fr⟩=\{1,rf,r^2,fr\}\cong C_2\times C_2$$

   is isomorphic to the Klein 4-group – an identity and 3 elements of order 2. $H\mathrm{⊴}G$ since $[G:H]=2$. Finally, take

   $$K=⟨rf⟩=\{1,rf\}\mathrm{⊴}H.$$

   We claim that $K⋬G$. And indeed,

   $$f\circ rf\circ f=fr\notin K.$$