TITLE: Are normal subgroups transitive?
QUESTION [50 upvotes]: 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 $k'\in K$, $gk'g^{-1}=k\in K\Rightarrow k'=g^{-1}kg\in\{gkg^{-1}\mid k\in K\}$.

REPLY [5 votes]: 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 $N \lhd G$.