# 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.

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 $$S_4$$.

• Awesome, thank you! As a follow up question, is there a quick way to prove that the permutations you have listed comprise a characteristic subgroup? Sep 16, 2021 at 22:47
• The only elements of order $2$ are the ones of the form $(**)(**)$ or $(**)$. An automorphism cannot take an element of the first type to one of the second type, as $A_4$ is characteristic in $S_4$.
– tkf
Sep 16, 2021 at 22:52
• Or alternatively note that any automorphism of $S_4$ will map conjugacy classes to conjugacy classes, and the only conjugacy class of size $3$ is the one consisting of elements of the form $(**)(**)$.
– tkf
Sep 16, 2021 at 22:59
• Also note that the statement is true the other way round: Every characteristic subgroup of a normal subgroup is a normal subgroup.
– tkf
Sep 16, 2021 at 23:02
• Awesome, thank you! Sep 16, 2021 at 23:03

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).

• This is a brilliant answer! (+1) Jun 30, 2022 at 18:11