Suppose that $G$ is a group with $H \lt G, \{H\} = Syl_2(G)$ (ie. H is the only 2-Sylow subgroup of $G$) and $H$ is also cyclic. Then the centre of $G$, $Z(G)$ is non trivial.
So, I know that $H$ being the only 2-Sylow subgroup gives us that it is normal in $G$. From here we can get that $\forall g \in G, H^g = H$. I'm not sure where to go from here to show that there must be an element in the centre.
Thanks!
Since $\;H\;$ is cyclic and of even order, it has a unique element of order two, say $\;h\;$, and since conjugated elements in a group have the same order, for any $\;x\in G\;$ we have that $\;x^{-1}hx=h\;$ and we're done.
Even more can be said: If $P$ is the Sylow $2$-subgroup then $P$ is contained in the center of the group.
To see this, note that since $P$ is normal, the statement that it is central is that same as the statement that $C_G(P) = N_G(P)$. But $P\leq C_G(P)$, so the order of $N_G(P)/C_G(P)$ is odd, and we also know by the normalizer/centralizer theorem that this quotient is isomorphic to a subgroup of $\operatorname{Aut}(P)$, which is a $2$-group since $P$ is a cyclic $2$-group. Combining these, we see that $N_G(P)/C_G(P)$ is trivial, which precisely means that $P$ is central as claimed.
In fact, $G$ will have a subgroup $Q$ such that $G = P\times Q$, as shown at Elements of odd order form a subgroup when the Sylow-$2$ subgroup is cyclic (see the comments for a link to an answer using less theory).
Further, note that the argument given by DonAntonio generalizes to the case where $P$ has a cyclic center, rather than just the case where $P$ itself is cyclic.
