Is $S_2 \wr S_k$ contained in $(S_a \times S_b) \wr S_{k-1}$? I'm working on my thesis and I want to prove a theorem but I need the following to be true:
$S_2 \wr S_k$ is not isomorphic to a subgroup of $(S_a \times S_b) \wr S_{k-1}$ where $a,b < 2k$.
Does anybody have an idea if this is true?
EDIT:I am using this as a lemma to prove that every graph $G$ with automorphismgroup equal to $S_2 \wr S_k$ has at least $k$ disjunct edges.
It is not hard to prove that this holds for $k=2$ but for $k=3$ there is a counterexapmle since $S_2 \wr S_3 \cong S_2 \times S_4$, it seems to me that it probably would hold if $k$ is high enough, though I don't know how to proof that.
(P.S. The counterexample for $k=3$ also gives a counterexample to the graph theorem I'm trying to prove since you can just take an empty graph on 4 vertices with the disjunct union of a single edge.)
 A: I claim this is true for $k \ge 5$. I haven't thought about $k=4$.
Let's first look at the structure of $G = S_2 \wr S_k$. This has a  normal elementary abelian $2$-subgroup $N$ of order $2^k$ with $G/N \cong S_k$. Then $N$ can be identified with the permutation module of $S_k$ over the field of order $2$.
The structure of this module is well known. For $k$ odd it is the direct sum of irreducible modules of dimensions $1$ and $k-1$ whereas, for $k$ even, it is indecomposable and uniserial has submodules of degrees $1$ and $k-1$ whose quotient is irreducible of dimension $k-2$.
So, in either case, the derived group $D$ of $G$ is an extension of $M$ by $A_k$, where $|M| = 2^{k-1}$, and $M$ corresponds to the submodule of $N$ of dimension $k-1$.
Note that, since $A_k$ is simple for $k \ge 5$, the only normal subgroups of $D$ are $1$, $M$ and $D$, together with a normal subgroup of order $2$ when $n$ is even, and these form a single chain of subgroups; in particular, $D$ has a unique minimal normal subgroup.
I claim that, for $k \ge 5$, $D$ is not isomorphic to a subgroup of $(S_a \times S_b) \wr S_{k-1}$ with $a,b < 2k$.
If it were, then since $A_k$ is not isomorphic to a subgroup of $S_{k-1}$ (that's a known result), and no proper normal subgroup of $D$ projects onto the factor $A_k$, $D$ would necessarily be contained in the base group $S_a^{k-1} \times S_b^{k-1}$ of the wreath product.
Since $D$ has a unique minimal normal subgroup, and this subgroup must project nontrivially onto at least one of the direct factors $S_a$ or $S_b$,  it follows that that $D$ must project onto at least one of these direct factors.
So we have reduced to proving that $S_a$ has no subgroup isomorphic to $D$ for any $a < 2k$. We will prove that by induction on $a$, this being clear for $a=1$ or $2$.
Suppose it has such a subgroup, which we might as well call $D$. If $D$ is intransitive, then as above, since $D$ has a unique minimal normal subgroup, $D$ must project onto at least one of the orbits, so the result follows by inductive hypothesis.
If $G$ is transitive but imprimitive, then the number of blocks of imprimitivity is at most $k-1$, so by the same argument as above (that $D$ is in the base group of the wreath product), $D$ is in the kernel of the action on the blocks, so $D$ is intransitive, which we have already eliminated.
So we are just left with the case when $D$ is primitive, in which case its point stabilizer is a maximal subgroup $C$ of index $a$ in $D$.
If $C$ projects onto $A_k$, then $C \cap M$ is a normal subgroup of $D$. Since $C \ne D$, we cannot have $C \cap M = M$, so $|C \cap M| \le 2$, and then $|D:C| \ge 2^{k-1}$, which greater than $2k$ for $k \ge 5$.
Otherwise, $C$ projects onto a maximal subgroup of $A_k$. Now it is again a known result that the only maximal subgroups (in fact the only only subgroups) of $A_k$ of index less than $2k$ are isomorphic to $A_{k-1}$
and have index $k$ in $A_k$. So $D$ projects onto a subgroup of index $k$ in $A_k$, and since we cannot have $M \le C$ (or the imprimitive action would not be faithful), we have $|D:C| \ge 2k$, contrary to assumption.
