Normal subgroups in groups of odd order I put the following question in my first-year algebra final this year: Suppose $G$ is a finite group of odd order and $N$ is a normal subgroup of order $5$. Show that $N\le Z(G)$. (By the way, this problem has been posed on this site before.)
The proof that I guided them through goes like this: all the conjugacy classes of $G$ have odd order; since $N$ is normal, it is a union of conjugacy classes. The only possibilities are $3$, $1$, $1$, and five $1$'s. In either case, $N$ contains a nonidentity element whose conjugacy class consists only of itself, so it is in $Z(G)$; but that element generates $N\cong \mathbb{Z}/5$ and the result follows. 
So let $S(p)$ be the following statement: If $G$ is a finite group of odd order and $N$ is a normal subgroup of order $p$, then $N\le Z(G)$. For which (odd) $p$ does this hold? The argument above shows that it holds for $p=5$, but pretty clearly that proof will not work for $p>5$. 
In fact, if $p$ is any prime that is not a Fermat prime, $S(p)$ is false; a counterexample follows. 
Let $q$ be an odd prime dividing $p-1$, and consider the nonabelian group $G$ of order $pq$. It must have only one $p$-Sylow subgroup, since the number of such subgroups divides $q$ is and $\equiv 1\mod{p}$, so it is normal. So $G$ satisfies the conditions of the theorem. But the center of $G$ is trivial since otherwise $G/Z(G)$ is cyclic and thus $G$ would be abelian. This counterexample does not work when $q=2$ since the group of order $pq$ has even order.
So my question is: does $S(p)$ hold when $p$ is a Fermat prime?

Let $N$ be a cyclic group of prime order, and let $A$ be any group of automorphisms of $N$. Then the orbits of $N\setminus\{e\}$ under $A$ all have the same size. (The proof is easy, based on the fact that every element of $N\setminus\{e\}$ is a generator of $N$, and hence every element of $N\setminus\{e\}$ is a power of every other element.) Note that this holds for your counterexample: all the orbits have size $q$.
In particular, if $N$ is a normal subgroup of $G$, then taking $A$ to be $G$ acting by conjugation, we see that all conjugacy classes in $N\setminus\{e\}$ have the same size. Furthermore, if $N$ has odd order, then all these conjugacy classes have odd size. Finally, if the order of $N$ is a Fermat prime, then the only odd divisor of $\#N-1$ is 1. Therefore every element of $N$ is its own conjugacy class, hence is in the center.
(PS: No way I could have answered this question if you hadn't written it so well!)
