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Thanks for any help or comments.

Suppose $G$ is a finite group and $t$ is an involution in $G$. Suppose also that $\frac{C_G(t)}{\langle t\rangle}\cong \mathbb{S}_{n-2}$ and $C_G(t)$ contains some subgroup isomorphic with $\mathbb{S}_{n-3}$. Is it true that $\langle t\rangle$ has a complement in $C_G(t)$? By complement I mean a subgroup $L$ such that $L\cap \langle t\rangle=1$ and $C_G(t)=L\langle t\rangle$.

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    $\begingroup$ I am not sure why you have written $S_{n-2}$ and $S_{n-3}$ rather than just $S_n$ and $S_{n-1}$. $\endgroup$ – Derek Holt Nov 15 '15 at 21:59
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There is another argument from Derek Holt's if $n$ is odd and at least $7$. The desired conclusion can be seen by a theorem of Gaschutz, which asserts that an Abelian normal subgroup $A$ of a finite group $G$ is complemented in $G$ if and only if $A \cap P$ is complemented in $P$ for some Sylow $p$-subgroup $P$ of $G$, for each prime $p$ dividing $|A|$. Apply with $A = \langle t \rangle$. The assumption about a subgroup isomorphic to $S_{n-3}$ (which does not contain $t$ as the symmetric group contains no central elements of order $2$) implies that the conditions of Gaschutz's theorem are satisfied. I see no such shortcut when $n$ is even though.

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If $M$ is the trivial module over ${\mathbb F}_2$ then, for $n \ge 4$, the dimension of $H^2(S_n,M)$ is $2$ and, for $n \ge 5$, the restriction $H^2(S_n,M) \to H^2(S_{n-1},M)$ is an isomorphism. So I think the answer to your question is yes for $n-2 \ge 5$, because non-split extensions of $M$ by $S_{n-2}$ restrict to nonsplit extensions of $M$ by $S_{n-3}$.

To justify those assertions, you could look at the presentations of the covering groups of $S_n$, which were computed by Schur originally. You can also find them, for example, in Huppert's book "Endliche Gruppen I", and you see from the presentations arising from the Coxeter presentation of $S_n$ that the covering groups of $S_n$ restrict to those of $S_{n-1}$ for $n \ge 5$.

The presentations of the extensions of $M$ by $S_n$ are as follows.

$\langle a_i\,(1 \le i \le n-1),\,t \mid t^2=[t,a_i]=1, a_i^2=x\, (1 \le i \le n-1),\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (a_ia_{i+1})^3=t\,(1\le i \le n-2),\,(a_ia_j)^2=y\, (4 \le i+2 \le j \le n) \rangle,$

where $x$ and $y$ are equal to $1$ or $t$, giving the four different isomorphism types of extensions (except that two of them are isomorphic when $n=6$).

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    $\begingroup$ In the case n = 6, the answer is "no". The group G = GL(2,3) has a central subgroup <t> of order 2, and the factor group is S_4. Also, G contains a copy of S_3, but G does not split over <t>. $\endgroup$ – Marty Isaacs Nov 18 '15 at 22:17

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