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Let $G$ be a group of order $36$ and given that $G$ has $4$ Sylow $3$-subgroups. Then I have to show that Sylow $2$-subgroup is unique.

So far we proved that $G$ has a normal subgroup of order $3$. Couldn't solve that it has a unique Sylow $2$-subgroup. Help me. Thanks.

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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.

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Alternatively, since a group of order $36$ is finite solvable, it has a subgroup $F$, called the Fitting subgroup, that is the unique maximal nilpotent normal subgroup. If $G$ has neither a normal $2$-Sylow nor a normal $3$-Sylow, then $|F|$ cannot be divisible by $4$ or $9$ lest it contain a normal Sylow subgroup of $G$, so $|F|=2$, $3$, or $6$, and since it is square-free nilpotent, it is cyclic, so $|Aut(F)|$ is $1$ (when $|F|=2$) or $2$ (when $|F|=3$ or $6$). But for a finite solvable group, the Fitting subgroup is self-centralizing, i.e., $C_G(F)\le F$. Since $N_G(F)=G$, this gives $6= 36/6 \le |G|/|F| \le |N_G(F)|/||C_G(F)| \le |Aut(F)| \le 2$, which is a massive contradiction.

I should add that one advantage of this argument is that it generalizes to show that any group of order $p^2q^2$, with $p$ and $q$ prime, has a normal Sylow subgroup for at least one of $p$ and $q$--because assuming it had no normal Sylows would imply $pq \le (p-1)(q-1)$.

In fact, one can use the same idea to show that any solvable finite group of cube-free order has at least one normal Sylow.

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