Show that $K$ is a normal subgroup of $H$ and $H$ is a normal subgroup of $D_4$, but $K$ is not a normal subgroup of $D_4$ Say I have the dihedral group $D_4 = \{1, \sigma, \sigma^2
, \sigma^3
, \tau, \tau\sigma, \tau\sigma^2
, \tau\sigma^3
\},$ where I know $o(\sigma) = 4$, $o(\tau ) = 2$, and $\sigma\tau = \tau\sigma^{−1} = \tau\sigma^3$
Let $K = \{1, \tau \}$ and $H = \{1, \sigma^2, \tau, \tau\sigma^2\}$. Show that $K$ is a normal subgroup of $H$ and $H$ is a normal subgroup of $D_4$, but $K$ is not a normal subgroup of $D_4$.
I know that a subgroup of index 2 is always normal, so proving the first part should be easy enough if I can show $K$ is a subgroup of $H$. I am struggling with the subgroup test on this part, but it's theoretically possible.
My next thought on the strategy would be to attempt to identify all of the subgroups of $D_4$.
For example, $\tau\sigma^2 = \sigma^2\tau$, so $\langle \sigma^2,\tau \rangle = H = \{1, \sigma^2, \tau, \tau\sigma^2\}$ forms a subgroup of $D_4$ easily enough. I know I need to prove that $H$ is normal in this case. I think it could be done using the given equalities at the top.
If I have all of the subgroups, I can say that $K$ is not a subgroup of $D_4$, hence it can't be a normal subgroup of $D_4$.
Does this look right?
 A: Since $|K|$ is finite, to show $K\le H$, it suffices to show that $K$ is a nonempty subset of $H$ that is closed under the operation of $H$; but $1^2=1$, $1\tau=\tau 1=\tau$, and $\tau^2=1$.
The index of $H$ in $D_4$ is two.
Note that $K\le D_4$ (by the same argument as above; simply replace $H$ by $D_4$).
Also note that $\langle \sigma^2\tau \rangle \neq \{1, \sigma^2, \tau, \tau\sigma^2\}$, since the latter is not cyclic (because it has order four while each of its nontrivial elements has order two).
To show $K\not\unlhd D_4$, consider
$$\begin{align}
\sigma\tau\sigma^{-1}&=(\tau\sigma^{-1})\sigma^{-1}\\
&=(\tau\sigma^3)\sigma^{-1}\\
&=\tau\sigma^2\\
&\notin K.
\end{align}$$
A: Using $o|D_4| = 8$, combined with the fact that subgroups of index 2 are always normal. When our group is finite (in this case $D_4$ is certainly finite), for any group $G$ and subgroup $H$, $$|G:H| = \frac{|G|}{|H|}$$
Thus, we need $|H:K| = \frac{|H|}{|K|} = 2 $, $|D_4:H| = \frac{8}{|H|} = 2$, and you will find that $|D_4:K| \neq 2$ for $o|H|=4$.
