# Counterexamples in elementary Group Theory

Here are some silly things I keep forgetting.

Suppose we have a group $G$ and two subgroups $K \subseteq H \subseteq G$, we consider three possible hypotheses $(i)\ H \unlhd G$, $(ii)\ K \unlhd G$, $(iii)\ K \unlhd H$ and $(iv)\ K \unlhd H \unlhd G$. How do these properties relate to each other?
In the following table are summarized all the possible implications (except the ones that are trivially true which are indicated by dots) Let’s start with the second row, it’s pretty easy to see that $(ii) \implies (iii)$ and that by letting $K = \{1\}$ and $H$ any non normal subgroup of $G$ one finds that $(ii) \kern.6em\not\kern-.6em\implies (i)$.

Now the first row, we want to prove $(i) \kern.6em\not\kern-.6em\implies (ii), (iii)$. It suffices to find a group $H$ which has a non normal subgroup $K$, then using the property of direct products we find that letting $G = C_2 \times H$ (where $C_2$ is a cyclic group of order $2$) is a group with the desired property.
The smallest example satisfying the first step appears when $H \cong D_3 = \langle \sigma, \tau \ |\ \sigma^3 = \tau^2 = 1, \tau\sigma\tau = \sigma^2 \rangle$ the dihedral group of degree $3$ and $K = \langle \tau \rangle$; indeed $\sigma^{-1}\tau\sigma = \sigma^2\tau\sigma = (\tau\sigma\tau)\tau\sigma = \tau\sigma^2 \notin K$.
(technically we could have used a semidirect product $D_3 \rtimes C_2$ but it’s definitely easier this way).

Now let’s focus on $(iii)$, is it true that $K \unlhd H \implies K \unlhd G$? This one is easy to disprove, just take $K \unlhd H$ (with $K \neq \{1\}$) and a group $G$ in which $H$ is not normal. For example $K \cong C_2 \unlhd C_4 \cong H$, we then use the Cayley embedding to find a copy of $H$ in $S_4$ the symmetric group on $4$ letters which will fill the role of $G$. Since the only nontrivial normal subgroup of $S_4$ are the alternating group $A_4$ and one of its subgroups isomorphic to $C_2 \times C_2$ (thanks GroupNames!) this proves $(iii) \kern.6em\not\kern-.6em\implies (ii), (i)$ (this is probably not a minimal example but I like the reasoning).

We are almost done, we just need to find an example of a chain of subgroups $K \unlhd H \unlhd G$ such that $K\not\kern-.3em\unlhd \ G$,  proving that sadly (?) being a normal subgroup of a subgroup is not a transitive relation.
This was a bit harder to do, for starters it’s reasonable to pick an Abelian group for $H$ as all its subgroups will be normal.
Here’s a solution: take $D_4 \unrhd \langle \tau, \sigma^2 \rangle \unrhd \langle \tau \rangle$ for $G \unrhd H \unrhd K$, it’s easy (although rather boring) to prove the desired relations: for starters one should prove $H \cong \langle \tau \rangle \times \langle \sigma^2\rangle \cong C_2\times C_2$, then $K \unlhd H$ follows from the properties of the direct product. Then since $\sigma^{-1}\tau\sigma = \tau\sigma^2 \not\in K$ one has that $K\not\kern-.3em\unlhd \ G$. Finally to prove $H \unlhd G$ it a matter of checking wether $H^\sigma = H$ and $H^\tau = H$.

Here’s a much more fun way to find the last example that came to my mind after writing the boring one above: the idea is that a normal subgroup is fixed by conjugation, but those are not the only automorphisms a group can have! Indeed if we can find an automorphism $\rho : H \to H$ such that $\rho(K) \neq K$ we can form the semi-direct product $H \rtimes \langle \rho \rangle$ thus finding the desired chain $K \unlhd H \unlhd H \rtimes \langle \rho \rangle$.

Here’s an easy example following the above scheme. Take any group $A$, it’s clear that, identifying $A$ with the subgroup $A \times \{1\}$ we have $A\unlhd A \times A$, moreover it’s easy to show that the map $\rho : (a_1, a_2) \mapsto (a_2, a_1)$ is an automorphism of order $2$. By the reasoning above we conclude that $A \unlhd A \times A \unlhd (A\times A)\rtimes \langle \rho \rangle$ is a chain of subgroups with the desired property.

Taking $A \cong C_2$ the group $(C_2 \times C_2) \rtimes C_2$ has order $8$ and is nonabelian meaning it must be isomorphic to one of $D_4$ or $Q_8$ the Quaternion group. However all subgroups of the Quaternion group are normal so it must be that $(C_2 \times C_2) \rtimes C_2 \cong D_4$.