Constructing a group with given normal subgroups. Let $N_1,N_2,\dots,N_n$ be simple groups. Is there is a group $G$ with exactly $n$ nontrivial proper normal subgroups isomorphic to $N_1,N_2,\dots,N_n$?
 A: This does not hold for an arbitrary list of simple groups. To see this, note that if $H$ and $K$ are normal subgroups of a group $G$, then $HK = \{hk : h \in H, k \in K\}$ is a normal subgroup of $G$. Moreover, if $H$ and $K$ have coprime orders, then 
$$|HK|=|H||K|.$$ 
With this in mind, a list of $n \geq 3$ finite, nontrivial simple groups $N_1,\dots, N_n$ with pairwise coprime orders cannot be realized as the list of proper normal subgroups of a group. Indeed, if a group $G$ existed with precisely $N_1, \dots ,N_n$ as its normal subgroups, it must have at least $|N_1|\cdot|N_2| \cdots |N_n|$ elements. All the groups $N_i$ and $N_iN_j$, $i\neq j$, are proper nontrivial subgroups since $n \geq 3$. By the requirement that the orders are coprime, $N_iN_j \not\simeq N_k$ for any $k$.
To give a specific example, the above argument shows that no group $G$ can have only three normal subgroups that are isomorphic to $\mathbb Z/2\mathbb Z$ and $\mathbb Z/3 \mathbb Z$, and $\mathbb Z/5 \mathbb Z$.
