Normal subgroup of a characteristic subgroup I came across the following question while reviewing for my qualifying exams:

Prove or provide a counterexample:
If $M$ is a normal subgroup of $N$, and $N$ is a characteristic subgroup of $G$, then $M$ is a normal subgroup of $G$.

Looking at our assumptions, it does not seem like we have enough information to deduce that $M$ is normal in $G$. However, coming up with a counterexample has proven difficult. I have tried letting $G = D_4$ and $N = \langle r\rangle$, but that did not prove fruitful. I also thought of using the quaternions, but all of its subgroups are normal, so that wouldn't be helpful here.
Any advice for this problem would be greatly appreciated.
 A: The permutations
$$\{(),(12)(34), (13)(24),(14)(23)\}\subset S_4$$
are a characteristic subgroup of $S_4$, isomorphic to $C_2\times C_2$.
As $C_2\times C_2$ is abelian, any subgroup is normal.  However The subgroup generated by $(12)(34)$ is not normal in $S_4$.
A: As some intuition, the socle is a subgroup worth considering. We restrict attention to finite groups. For a group $G$, the socle $\text{Soc}(G)$ is the subgroup generated by the minimal normal subgroups of $G$. As normal subgroups are closed under intersection, $\text{Soc}(G)$ is the direct product of the minimal normal subgroups. In particular, $\text{Soc}(G)$ is characteristic in $G$.
Now a minimal normal subgroup of $G$ is of the form $S^{k}$, where $S$ is a simple group. So if the minimal normal subgroups of $G$ are $N_{1}, \ldots, N_{m}$, where $N_{i} = S_{i}^{k_{i}}$, then we may write $\text{Soc}(G) = \prod_{i=1}^{m} \prod_{j=1}^{k_{i}} S_{i}.$ So each copy of $S_{i}$ is normal in $\text{Soc}(G)$.
Now if $k_{i} > 1$, $S_{i}$ is not normal in $G$. The way we see this is as follows. Consider the conjugation action of $G$ on $\text{Soc}(G)$. This induces a permutation on the direct factors of $\text{Soc}(G)$. In particular, the orbits of this action are precisely the minimal normal subgroups of $G$. Effectively, $N_{i}$ is the normal closure of $S_{i}$. That is, $N_{i} = \langle gS_{i}g^{-1} : g \in G \rangle$. So the conjugation action of $G$ on a fixed copy of $S_{i}$ (which we call $S$) effectively moves $S$ around to each copy of $S_{i}$ in $\text{Orb}(S)$.
For an infinite family of counter-examples, take $G = A_{5}^{n} \rtimes S_{n}$, where $S_{n}$ acts by permuting the factors of $A_{5}$.
If we assume that $G$ has no Abelian normal subgroups, then $G$ has a very rigid structure and is determined by (i) the isomorphism class of $\text{Soc}(G)$, and (ii) the conjugation action on $\text{Soc}(G)$. Effectively, building on the (spirit of) the counterexample yields an efficient isomorphism test. See Babai, Codenotti, and Qiao (https://people.cs.uchicago.edu/~laci/papers/icalp12.pdf) and its predecessor Babai, Codenotti, Grochow, and Qiao (https://people.cs.uchicago.edu/~laci/papers/soda11.pdf).
