# Let $N$ be a normal subgroup of a finite group $G$

Let $N$ be a normal subgroup of a finite group $G$. Suppose $|N|=5$ and that $|G|$ is odd. Prove $N$ is contained in $Z(G)$ , the center of $G$

because the order of $N$ is prime, then N is cyclic, but I'm more concerned with how I can derive for $g\in G$ and $g^{-1}ng=n$ for all $n\in N$

• Hint: Conjugation gives a map from $G \to Aut(N)$. What is the order of $Aut(N)$? – lulu May 5 '17 at 14:45

Consider $G$ acting on $N$ via conjugation. Every element $g \in G$ defines an homomorphism of groups $N \to N$. Now, it is known that $\operatorname{Aut}(C_5)=C_4$, so the above action defines a map $G \to C_4$.
But, since $|G|$ is odd, every element of $G$ has odd order.
It follows that the homomorphism defined by conjugation has to be the trivial one, so every element of $G$ acts trivially on $N$, that is, for any $x \in N, g \in G$ $$gxg^{-1}=x$$ which implies $$gx = xg$$ so $N \subseteq Z(G)$.

By using N/C Lemma, $G/C_G(N)$ is isomorphic to a subgroup of $Aut(N)$.
Note that $|Aut(N)|=4$.
Since $|G|/|C_G(N)|$ divides $|Aut(N)|$ and $|G|$ is odd, we have $G=C_G(N)$, that is $N\leq Z(G).$

In general: if $N \unlhd G$ with $|G|$ is odd and $|N|$ is a product of different Fermat prime numbers (there are $32$ possibilities here!), then $N \subseteq Z(G)$.

• This appears to be based on a difficult theorem which forces $N$ to be cyclic. If $|N|=n$ then $\phi(n) =2^{k}$ for some $k$ and then $(n, \phi(n)) =1$ which forces $N$ to be cyclic and then $|Aut(N) |= \phi(n) =2^{k}$ and we can argue like answer from Alan Wang. – Paramanand Singh May 6 '17 at 2:30
• Not necessarily: if $G=C_{p_1} \times C_{p_2} \cdots \times C_{p_k}$ with the $p_i$'s different prime numbers, then $Aut(G) \cong C_{p_{1}-1} \times \cdots \times C_{p_{k}-1}$. This follows from the Chinese Remainder Theorem. – Nicky Hekster May 6 '17 at 19:35
• But in your last comment $G$ is cyclic because each $p_{i}$ is distinct. It is not necessary that a group $G$ is cyclic if it's order is a product of distinct odd primes. Take a group of order $55=5\cdot 11$ and then $G$ may be non cyclic in which case $|Aut(G) |\neq 40$. If the distinct primes are Fermat primes then $G$ will necessarily be cyclic. – Paramanand Singh May 7 '17 at 3:01
• Paramanand, yes, if $N$ is cyclic, then it is easy. But this requirement can be dropped. – Nicky Hekster May 7 '17 at 14:15

$gng^{-1}=n^k$ . This implies $g^lng^{-l}=n^{k^l} \forall l.$ Putting $l=G$ we have $n=n^{k^{G}}$ Thus we have 5 divides $k^{G}-1$. Since $G$ is odd we have therefore 5 divides $k-1$ and hence $n^k=n$ .This completes the proof.