Csdrhrt

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

Applying the definitions, we know ${ghg^{-1}mid hin H}=H$ and ${hkh^{-1}mid kin K}=K$, and want ${gkg^{-1}mid kin K}=K$. Clearly, the best avenue for a counterexample is if $gkg^{-1}notin K$ for some $kin K$ and $gin G-H$.

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

Using some suggestions from the other commenters:

The alternating group, $A_4$, has the set $H={I,(12)(34),(13)(24),(14)(23)}cong V_4$ as a subgroup. If $fin S_4supseteq A_4$ 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 $Hlhd S_4$ is normal, so $Hlhd A_4$ 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 $Knotlhd A_4$, since $$[(123)][(12)(34)][(132)]=(13)(24)in H-K.$$

Moreover, this is a minimal counterexample, since $|A_4|=12=2cdot 2cdot 3$ is the next smallest number which factors into three integers, which is required for $Klhd Hlhd G$ but ${I}subset Ksubset Hsubset G$ so that $[G,:,H]>1$, $[H,:,K]>1$, $|K|>1$ and
$$|G|=[G,:,H]cdot[H,:,K]cdot|K|.$$
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: $langle sranglelhdlangle r^2,sranglelhd langle r,srangle=D_4$, but $langle sranglenotlhd D_4$.)

However, if $Hlhd G$ 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 ${f^{-1}kfmid kin K}=KRightarrow Klhd G$.

Look at $S_4$ and its following subgroups $A = langle (12)(34) rangle$ and $B={(12)(34),(13)(42),(23)(41),e }$. Try to show that $A$ is normal in $B$ and $B$ is normal in $S_4$ but $A$ is not normal in $S_4$.

We need a non-abelian group, since all subgroups of abelian groups are normal. One small candidate is $D_8$, the symmetries of a square, here in a little more detail about how we might go about finding examples:

Consider all the subgroups in $D_8$. It's useful to visualize the subgroups as a lattice:

(Picture of Dummit and Foote I found on the web)

Now we try to pick an $H$. For all the subgroups on the third row from the top, their only proper subgroup is the trivial subgroup, which is trivially normal to $G$, so it doesn't make sense to use any of the subgroups on the third row for $H$.

Our only options for $H$ now are the second row: $langle s, r^2 rangle$, $langle r rangle$, and $langle rs, r^2 rangle$. We observe that if $H = langle r rangle$, the proper subgroups $langle r^2 rangle$ and $1$ are both normal to $D_8$, so that case is excluded. Our candidates are $langle s, r^2 rangle$ and $langle rs, r^2 rangle$.

Take $H = langle s, r^2 rangle$ and $K = langle s rangle$. It's easy to verify that $langle s rangle$ is not normal to $D_8$. All that's left is to show $K lhd H$ and $H lhd G$.

This is not difficult if we remember that any element in $D_8$ can be written as $r^i s^j$ with $0 le i le 3$ and $j = 0, 1$. Also we have the identity $rs = sr^{-1}$ which can repeated as $r^k s = s r^{-k}$. So for $g in D_8$, we want to prove or disprove $r^i s^j n s^j r^{-i} in N$ to show $N lhd G$.