# Does the specific condition on a normal subgroup of a finite group imply that it is a direct factor? v2.0

Suppose $$G$$ is a finite group, $$H \triangleleft G$$, such that $$\frac{G}{H}$$ is simple and $$Var(G) = Var(\frac{G}{H})$$ (Here $$Var(G)$$ stands for minimal group variety containing $$G$$). Does that imply that $$G \cong H \times \frac{G}{H}$$?

A similar statement for simple $$H$$ and $$Var(H) = Var(G)$$ (instead of simple $$\frac{G}{H}$$ and $$Var(G) = Var(\frac{G}{H})$$) was proved in the answer to the question: Does the specific condition on a normal subgroup of a finite group imply that it is a direct factor? However, this case seems to be quite different from that one, and thus it most likely can not be solved by exactly the same method.

The abelian case is still somewhat obvious:

If $$\frac{G}{H} \cong C_p$$ for some prime $$p$$, then $$G$$ is an abelian group of exponent $$p$$ for some prime $$p$$, which results $$G \cong C_p^n$$ for some natural $$n$$. So by classification of abelian finite groups $$H$$ is a direct factor of $$G$$. So $$G \cong H \times \frac{G}{H}$$.

Let $$S$$ be a finite simple group. I reformulate the question in a slightly stronger way as Question $$Q(S)$$:

If $$G$$ is a finite group such that

1. $$Var(G)=Var(S)$$, and
2. $$G$$ has a normal subgroup $$H$$ such that $$G/H\cong S$$,

then must $$H$$ have a normal complement?

Qeustion $$Q(S)$$ is slightly stronger than the question from the original post, since it asks if $$H$$ is equal to a direct factor of $$G$$ rather than whether it is merely isomorphic to a direct factor of $$G$$.

The answer to the stronger Question $$Q(S)$$ is: Yes.
I will only discuss the case where $$S$$ is not abelian, since the abelian case is discussed in the problem statement.

First, a lemma:

Lemma. Assume that $$S$$ is a nonabelian simple group, and that $$G\in Var(S)$$ is subdirectly irreducible. If $$G$$ has a chief factor isomorphic to $$S$$, then $$G\cong S$$.

Sketch of Proof. (Terminology and notation)

1. A homomorphic image of a subgroup of $$A$$ is called a section of $$A$$.
2. A finite set $${\mathcal T} = \{T_1,\ldots, T_k\}$$ of finite groups affords a representation of a finite group $$J$$ if $$J$$ is a homomorphic image of a finite subdirect product of the groups in $${\mathcal T}$$.
3. For a subdirectly irreducible group $$K$$, let $$K^*$$ be the monolith.

If the monolith $$G^*$$ of $$G$$ is nonabelian, then it follows from Theorem 10.1 of Commutator Theory for Congruence Modular Varieties that $$G$$ is a homomorphic image of a subgroup of $$S$$, hence $$|G|\leq |S|$$ with equality iff $$G\cong S$$. Since $$G$$ has $$S$$ as a chief factor, we have $$|G|\geq |S|$$, so indeed $$G\cong S$$.

We argue that we must be in the case of the preceding paragraph, by deriving a contradiction from the alternative case, which is the case where $$G^*$$ is abelian, $$G\in Var(S)$$ is subdirectly irreducible, and $$G$$ has $$S$$ as a chief factor. In this case, some finite set $${\mathcal T} = \{T_1,\ldots, T_k\}$$ of sections of $$S$$ affords a representation of $$G$$. If the sections in $${\mathcal T}$$ have been chosen so that they have minimal cardinality to afford a representation of $$G$$, then all $$T_i$$ will be subdirectly irreducible. Moreover, it follows from the modularity of normal subgroup lattices that the set $$\{T_1/T_1^*,\ldots, T_k/T_k^*\}$$ will afford a representation of $$G/G^*$$.

Each $$T_i/T_i^*$$ is a proper section of $$S$$, hence cannot have $$S$$ as a chief factor. It follows that $$\{T_1/T_1^*,\ldots, T_k/T_k^*\}$$ cannot afford a representation of any group that has $$S$$ as a chief factor. But $$G/G^*$$ does have $$S$$ as a chief factor, since (i) $$G$$ had $$S$$ as a chief factor, (ii) $$S$$ is nonabelian, and (iii) $$G/G^*$$ has the same nonabelian chief factors as $$G$$. This contradiction completes the proof of the lemma. \\\

Assume that the answer to Question $$Q(S)$$ is No for some simple $$S$$. Let $$G$$ be a minimal counterexample. That is, $$G$$ is a finite group with $$|G|$$ minimal which satisfies $$Var(G)=Var(S)$$ and $$\exists H(G/H\cong S)$$), but $$H$$ does not have a normal complement in $$G$$. I will argue that such a minimal $$G$$ must be subdirectly irreducible, and then apply the lemma to derive a contradiction.

Claim 1. There is a smallest normal subgroup $$L\lhd G$$ such that $$HL=G$$.

Proof. To show that there is a smallest, it suffices to to show that $$X, Y\lhd G$$ and $$HX=HY=G$$ together imply $$H(X\cap Y)=G$$, since then $$L$$ can be taken to be the intersection of $$\{X\lhd G\;|\;HX=G\}$$.

If $$X, Y\lhd G$$ and $$HX=HY=G$$, then(Some details are being skipped here.) $$G':=[G,G]=[HX,HY]\leq H[X,Y]$$. $$G'\not\leq H$$, since $$G/H\cong S$$ is nonabelian, so from $$G'\leq H[X,Y]$$ we derive that $$[X,Y]\not\leq H$$. Since $$X\cap Y\supseteq [X,Y]$$, it follows that $$X\cap Y\not\leq H$$, and therefore that $$H(X\cap Y)=G$$. \\\

Claim 2. Any minimal normal subgroup of $$G$$ is contained in $$H$$. Hence, there is only one minimal normal subgroup of $$G$$.

Proof. Suppose that $$A$$ is a minimal normal subgroup of $$G$$. If $$A\not\leq H$$, then $$A$$ is a normal complement to $$H$$, contrary to the assumption that $$G$$ is a counterexample to Question $$Q(S)$$. This proves the first sentence of the claim.

To prove the second sentence of the claim, suppose that $$A, B\lhd G$$ are distinct minimal normal subgroups of $$G$$. By the previous paragraph, $$A, B\leq H$$, and hence $$AB\leq H$$. Since $$G$$ is a minimal counterexample to Question $$Q(S)$$, the group $$G/A$$ is not a counterexample. Since $$H/A\lhd G/A$$ and $$(G/A)/(H/A)\cong G/H\cong S$$, we derive that $$H/A$$ has a normal complement in $$G/A$$. That is, there is some normal subgroup $$A'\supseteq A$$ such that $$A'/A$$ is complementary to $$H/A$$ in the normal subgroup lattice of $$G/A$$. Hence $$HA'=G$$ in the normal subgroup lattice of $$G$$. By Claim 1, $$L\leq A'$$. Since $$L$$ is not an atom in the normal subgroup lattice of $$G$$, and since $$A'$$ covers $$A$$ in the normal subgroup lattice of $$G$$ (and therefore has height $$2$$ in this lattice), we derive that $$A'=L$$.

Similarly $$B'=L$$, so $$A\neq B$$ both have height $$1$$ and $$L=A'=B'$$ covers both of them and has height $$2$$, from which it follows that $$L=AB$$ is the join of $$A$$ and $$B$$. But this is impossible, since $$A, B\leq H$$ and $$L\not\leq H$$. \\\

To complete the answer to Question $$Q(S)$$, it follows from the claims that if $$G$$ is a minimal counterexample, then $$G\in Var(S)$$ is subdirectly irreducible. Moreover, $$G/H\cong S$$, so $$Var(G)=Var(S)$$ and $$G$$ has $$S$$ as a chief factor. By the lemma, $$G\cong S$$, so it is not a counterexample at all. \\\