# Are normal subgroups transitive?

Suppose $G$ is a group and $K\lhd H\lhd G$ are normal subgroups of $G$. Is $K$ a normal subgroup of $G$, i.e. $K\lhd G$? If not, what extra conditions on $G$ or $H$ make this possible?

Applying the definitions, we know $\{ghg^{-1}\mid h\in H\}=H$ and $\{hkh^{-1}\mid k\in K\}=K$, and want $\{gkg^{-1}\mid k\in K\}=K$. Clearly, the best avenue for a counterexample is if $gkg^{-1}\not\in K$ for some $k\in K$ and $g\in G-H$.

If no such element exists, $\{gkg^{-1}\mid k\in K\}\subseteq K$ implies $\{gkg^{-1}\mid k\in K\}=K$ because if $k'\in K$, $gk'g^{-1}=k\in K\Rightarrow k'=g^{-1}kg\in\{gkg^{-1}\mid k\in K\}$.

The alternating group, $A_4$, has the set $H=\{I,(12)(34),(13)(24),(14)(23)\}\cong V_4$ as a subgroup. If $f\in S_4\supseteq A_4$ is a permutation, then $f^{-1}[(12)(34)]f$ has the effect of swapping $f(1)$ with $f(2)$ and $f(3)$ with $f(4)$. One of these is $1$, and depending on which it is paired with, the conjugated element may be any of $H-\{I\}$, since the other two are also swapped. Thus $H\lhd S_4$ is normal, so $H\lhd A_4$ as well. Similarly, $H$ has three nontrivial subgroups, and taking $K=\{I,(12)(34)\}\cong C_2$, this is normal because $V_4$ is abelian. But $K\not\lhd A_4$, since
Moreover, this is a minimal counterexample, since $|A_4|=12=2\cdot 2\cdot 3$ is the next smallest number which factors into three integers, which is required for $K\lhd H\lhd G$ but $\{I\}\subset K\subset H\subset G$ so that $[G\,:\,H]>1$, $[H\,:\,K]>1$, $|K|>1$ and
The smallest integer satisfying this requirement is 8, but the only non-abelian groups with $|G|=8$ are the dihedral group $D_4$ and the quaternion group $Q_8$, and neither of these have counterexamples. (Note that if $G$ is abelian, then all subgroups are normal.) Thus $A_4$ is a minimal counterexample. (Edit: Oops, $D_4$ has a counterexample, as mentioned in the comments: $\langle s\rangle\lhd\langle r^2,s\rangle\lhd \langle r,s\rangle=D_4$, but $\langle s\rangle\not\lhd D_4$.)
However, if $H\lhd G$ and $K$ is a characteristic subgroup of $H$, then $K$ is normal in $G$. This is because the group action $f$ defines an automorphism on $G$, $\varphi(g)=f^{-1}gf$, and because $H$ is normal, $\varphi(H)=H$ so that $\varphi|_H$ is an automorphism on $H$. Thus $\varphi(K)=K$ since $K$ is characteristic on $H$ and so $\{f^{-1}kf\mid k\in K\}=K\Rightarrow K\lhd G$.