Answer with hints that must be checked and solved:
As already proved, a group $\;G\;$ of order $\;30\;$ must have either one unique subgroup of order five or one unique subgroup of order three.
Take now the unique such subgroup, say $\;N\;$ , and one of the (possibly several) groups of the other order, say $\;H\;$ . Since clearly $\;N\cap H=1\;$ , we get that $\;NH\;$ is a subgroup (why?) of order $\;15\;$ and thus $\;NH\lhd G\;$ (why?).
But a group of order $\;15\;$ is necessarily cyclic (why?), and since any subgroup of a normal cyclic subgroup of a big group is normal in the big group (why?), we get that both $\;N,H\lhd G\;$
Examine the action of $G$ on the set of Sylow 3-subgroups. This gives rise to a well-defined homomorphism $\phi: G\to S_4$. Since we know the Sylow 3-subgroups are all conjugate to one another, the image $\phi(G)$ must be a subgroup of $S_4$ that acts transitively on the set with four points. The only such subgroups of $S_4$ that could arise as a quotient a group of order 36 are isomorphic to either the cyclic group of order 4, the Klein 4 group, or the alternating group $A_4$.
Note that if $|\phi(G)| = 4$, then the kernel of $\phi(G)$ is a normal subgroup of order 9, contradicting our assumption that the Sylow 3-subgroups are not normal.
Now if $\phi(G)$ is isomorphic to $A_4$, then $\ker \phi$ is a normal subgroup of order 3. This implies that $\ker \phi$ is a subgroup of all Sylow 3-subgroups. Since all groups of order 9 are abelian, this means that the centralizer of $\ker \phi$ is at least order 27 (six unique elements from each of the four Sylow 3-subgroups as well as $\ker \phi$ itself) and thus $\ker \phi$ is actually central.
Let $x$ denote a nonidentity element of $\ker \phi$ and $y$ denote an element of order dividing 4. The subgroup generated by $y$ has trivial intersection with $\ker \phi$, so $y$ must actually have order $1$ or $2$, since the image $\phi(G)$ is a group with no elements of order 4.
Moreover, $(yx)^2 = y^2x^2 = x^2$, showing that $yx$ does not have order 2. Thus, the left cosets of $\ker \phi$ contain at most one element of order dividing 2. Since there are only four cosets that could contain such an element (as only four elements in the quotient group have order dividing 2), there are at most 4 such elements, meaning that the Sylow 2-subgroup is unique.
Best Answer
Hints: Show that $G/P_7$ is abelian. Use this to conclude that $N = P_3P_7$ is normal in $G$.