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?
Best Answer
Yes this holds for Fermat primes. You need the following observation. This is sometimes called the $N/C$ Theorem.
Lemma Let $G$ be a group with a subgroup $H$, then $N_G(H)/C_G(H)$ embeds homomorphically into Aut$(H)$.
Here $N_G(H)=\{g \in G : gH=Hg\}$, the normalizer of $H$ in $G$ and $C_G(H)=\{g \in G : gh=hg \text { for all } h\in H\}$, the centralizer of $H$ in $G$.
Now let us apply this to the situation where $N=H$ is normal, $|N|=$ Fermat-prime and $|G|$ is odd. Then $N_G(N)=G$ because of the normality of $N$. And since $N$ is cyclic, |Aut$(N)|$ is a power of $2$ (in fact $|N|-1$). It follows from the $N/C$ theorem that $|G/C_G(N)|$ is a power of $2$. But obviously it also divides $|G|$, which is odd. This can only be when $G=C_G(N)$, that is $N \subseteq Z(G)$.
So how do you prove the lemma? Let me give a sketch and leave the details with you: $N_G(H)$ acts as automorphisms on $H$ by conjugation. The kernel of the action is $C_G(H)$.
Generalizations:
Proposition 1 Let $G$ be a finite group, $N$ a normal subgroup of prime order $p$, with gcd$(|G|,p-1)=1$, then $N \subseteq Z(G)$.
Note that this holds for $p=2$: a normal subgroup of order $2$ must be central.
Proposition 2 Let $G$ be a finite group, $N$ a normal subgroup of prime order $p$, where $p$ is the smallest prime dividing $|G|$. Then $N \subseteq Z(G)$.