If $H \trianglelefteq K$ and $K \trianglelefteq D_4$, does it follow that $H \trianglelefteq D_4$? PROBLEM

If $H \trianglelefteq K$ and $K \trianglelefteq D_4$, does it follow that $H \trianglelefteq D_4$?  ($D_4$ is a dihedral group with $|D_4|=8$.)

MY ATTEMPT
My hunch is that the answer to the problem is NO.  To construct a specific counterexample, I shall use the results in this Wikipedia article.
In particular, I have $H \trianglelefteq K$ and $K \trianglelefteq D_4$.
So the indices
$$[D_4 : K] = \frac{|D_4|}{|K|}$$
and
$$[K : H] = \frac{|K|}{|H|}$$
are (positive) integers.
In order to have
$$H \not\trianglelefteq D_4,$$
we need to have
$$H \subset K \subset D_4$$
where the inclusion of sets is strict.  In particular, since $|D_4| = 8$, we need to choose $|K| = 4$ and $|H| = 2$.
Since $H \trianglelefteq K$ and $K \trianglelefteq D_4$, if I try to set
$$K = \langle a^2, b \rangle = \{e, a^2, b, {a^2}b\} \trianglelefteq D_4,$$
$$H = \langle b \rangle = \{e, b\} \trianglelefteq K,$$
then $H$ is not normal in $D_4$.
Lastly, I checked this Wikipedia article section and verified that


A normal subgroup of a normal subgroup of a group need not be normal in the group. That is, normality is not a transitive relation. The smallest group exhibiting this phenomenon is the dihedral group of order $8$.


QUESTION

Does my counterexample indeed contradict the conditions of my problem?

 A: Assuming your notation means what I think it does, then "yes". I'm assuming that...
$$D_4 = \langle a,b \;|\; a^4=b^2=(ab)^2=1 \rangle = \{ e,a,a^2,a^3,b,ab,a^2b,a^3b \}$$
...or in other words $a$ plays the role of a 90-degree rotation and $b$ plays the role of a reflection. For example: $ba=a^{-1}b=a^3b$.
$K=\{e,a^2,b,a^2b \}$ is a subgroup ($a^2$ is in the center of the group so it commutes with everything this makes closure easy to check).
$H = \langle b \rangle = \{e,b \}$ is a (cyclic) subgroup.
Both $H$ is normal in $K$ and $K$ is normal in $D_4$ because they are subgroups of index 2.
All that you lack is showing that $H$ is not normal in $D_4$. You could either show that left and right cosets fail to coincide or your could show that it is not closed under conjugation. To make conjugation closure fail, we'll have to pick something in $D_4$ which isn't in $K$, $a$ should work.
$aea^{-1}=e$ is ok, but $aba^{-1} = aab = a^2b \not\in \langle b \rangle =H$. Thus $aHa^{-1} \not= H$ so that $H$ is not normal in $D_4$.
Good job. :)
