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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 $K\trianglelefteq H$ and $H\trianglelefteq G$, it is possible that $K\ntrianglelefteq G$. Here are two examples:

1. 1.

   Let $G$ be the subgroup of orientation-preserving isometries of the plane $\mathbb{R}^{2}$ ($G$ 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 $H\trianglelefteq G$.

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

   We claim that $K\ntrianglelefteq G$. 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. 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

   $$D_{4}=\left\langle r,f\mid f^{2}=1,r^{4}=1,fr=r^{{-1}}f\right\rangle$$

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

   The subgroup

   $$H=\langle rf,fr\rangle=\left\{1,rf,r^{2},fr\right\}\cong C_{2}\times C_{2}$$

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

   $$K=\langle rf\rangle=\left\{1,rf\right\}\trianglelefteq H.$$

   We claim that $K\ntrianglelefteq G$. And indeed,

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