Past exam of the algebra course of the University of Oxford / 2011 / 6 / a / ii / Solution

If Sylow subgroups $S,T$ share $t\ne 1$, they are abelian by the preceding deduction, so $S,T\le C_G(t)$. But $|C_G(t)|=8$ and both subgroups have order eight. Thus $S=C_G(t)=T$. Consequently distinct Sylow $2$-subgroups intersect only in the identity.