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
- $Var(G)=Var(S)$, and
- $G$ 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)
- A homomorphic image of a subgroup of $A$ is called a
section of $A$.
- 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 ${\mathcal T}$.
-
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.
\\\
Let's return to the problem.
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. \\\
