D+F 4.5.22

$|G|=132 \Longrightarrow G$ is not simple.

$\textit{Proof}$ : The prime factorization of 132 is $132=2^2 \cdot 3 \cdot 11$. We will again use the 3rd part of Sylow's theorem to count the possible numbers of Sylow p-subgroups.
Writing $132=11^1 \cdot 12$ we note that $11 \nmid 12$ so Sylow (iii) gives $$n_{11} \equiv 1 (mod11)$$ and, $$n_{11} \mid 12$$ So, $n_{11} = 1,12$. Similarly, $$n_3 \equiv 1(mod3)$$ $$n_3 \mid 44$$ So, the only positive integers satisfying these constraints is 1 and 4, so $n_3=1,4$.
$$n_2 \equiv 1(mod2)$$ $$n_2 \mid 33$$ So, $n_3 = 1,3,11,33$
Now, suppose to the contrary that $G$ is a simple group. So, this means that $G$ has no non-trivial normal subgroups. By Corollary 20 p.142, this means that $n_p \neq 1 \quad \forall p$ in the prime factorization of $|G|=132$. So, this forces $n_{11}=12$, $n_3=4$ and $n_2=3,11,33$. The Sylow 11-subgroups are subgroups of order 11, thus isomorphic to $\mathbb{Z}_{11}$, which has 10 elements of order 11. So, with $n_{11}=12$, this means that in $G$, there are $12 \cdot 10 = 120$ elements of order 11. The Sylow 3-subgroups have order 3, so they are isomorphic to $\mathbb{Z}_3$, which has 2 elements of order 3. With $n_3 = 4$, this means that there are $4 \cdot 2 = 8$ elements of order 3 in $G$.
So far, we have counted $120+8=128$ non-identity elements of $G$. This leaves only 4 remaining. However, the Sylow 2-subgroups have order $2^2=4$. Thus, the remaining 4 elements must be exactly the Sylow 2-subgroup (which is either $\mathbb{Z}_4$ or $\mathbb{Z}_2 \bigoplus \mathbb{Z}_2$, but it doesn't actually matter at this point). So, $n_2=1$ giving us a non-trivial, normal subgroup by Corollary 20 p.142. This contradicts the assumption that $G$ was simple. In any case, we must have that $n_p=1$ for at least one $p$ in the factorization of 132.
So, any group of order 132 is not simple. $\blacksquare$