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My attempt at a solution:
Because of the Sylow theorems, $G$ has either one or three Sylow $2$-subgroups. If it has one, then $G$ has just one Sylow $2$-subgroup (which has order $8$), so it is normal, and we are done.
Now suppose that $G$ has three Sylow $2$-subgroups. I tried to show that the center of the group, $Z(G)$, has order $4$ or $12$ if $G$ not Abelian. In this case the result follows again because of the existence of a Sylow $2$-subgroup in $Z(G)$.
Any hints?
Suppose there exists 3-Sylow subgroups. The Sylow subgroups are conjugated, so $G$ acts transitively on the set $S$ of Sylow subgroups and there exists a morphisms of groups $f:G\rightarrow S_3$ defined by this action. The cardinal of the image of $S$ divides $3!=6$, it cannot be $2$ since a transposition does not acts transitively on $\{1,2,3\}$ so it is 3 or 6, this implies that the cardinal of the kernel of $f$ is $8$ or $4$ since $G/ker f\simeq Imf$.
