Let $k$ be a positive integer. Is it true that any finite group $H$ of cardinal $4k+2$ whose center contains an element $h$ of order $2$ is isomorphic to the direct product $H=(\mathbb{Z}/2\mathbb{Z})\times G$, where $G=H/\{1,h\}$?
An equivalent statement would be: Let $G$ be a finite group of odd cardinal. Is it true that the second cohomology group $H^2(G,\mathbb{Z}/2\mathbb{Z})$ with respect to the trivial action of $G$ on $\mathbb{Z}/2\mathbb{Z}$, vanishes?
There is a reasonably simple argument that any group $H$ of twice odd order has a normal subgroup $N$ of index 2. Given that, if you know also that $H$ has a central subgroup $Z$ of order 2, then it is straightforward to show that $H \cong N \times Z$.
The argument goes like this. By Cayley's Theorem, there is an isomorphism $\phi$ that maps $G$ to a permutation group of degree $|G|$ in which all non-identity elements of $G$ act without fixed points. So the elements of order 2 in $G$ are mapped by $\phi$ onto odd permutations, and hence the inverse image $N$ in $G$ of the intersection of the image of $\phi$ with the alternating group has index 2 in $G$.
Yes, because $G$ and $Z/2Z$ have coprime order, every extension of one by the other splits by the Schur-Zassenhaus theorem (http://en.wikipedia.org/wiki/Schur-Zassenhaus_theorem). If you prefer, $H^2(G,M)$ is killed both by $|G|$ and by $|M|$, so it must be trivial in your case.
