1999 March

There is no simple group of order 328.

Every finite $p$group has a nontrivial center.

Suppose $G$ is a group and $H$ is a subgroup of $G$ and $N$ is a normal subgroup of $G$.
a. If $G$ is solvable, then $H$ is solvable.
b. If $G$ is solvable, then $G/N$ is solvable.
c. If $N$ and $G/N$ are solvable, then $G$ is solvable 
Suppose $H$ and $K$ are subgroups of the group $G$. Then $[H, K]$ is the subgroup generated by $\{hkh^{1}k^{1} \mid h\in H, \ k\in K \}$. We recursively define $G_n$ by $G_0 = G$ and $G_{n+1} = [G_n, G]$. A group is nilpotent iff for some $k$, $G_k = {1}$. Use this definition of nilpotent group in what follows. For the rest of the problem, $G$ is assumed to be a finite group.
a. If $P$ is a Sylow subgroup of $G$, then $N(N(P)) = N(P)$, where $N(P)$ denotes the normalizer of $P$.
b. If $G$ is a nilpotent group and $H$ is a subgroup of $G$ not equal to $G$, then $N(H) \neq H$.
c. If $G$ is nilpotent and $P$ is a Sylow subgroup of $G$, then $P$ is a normal subgroup of $G$.
d. If each of the Sylow subgroups of $G$ are normal in $G$, then $G$ is the direct product of its Sylow subgroups.
e. Every finite, nilpotent group is the direct product of its Sylow subgroups.