6
$\begingroup$

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 $n\in N$

$\endgroup$
  • 4
    $\begingroup$ Hint: Conjugation gives a map from $ G \to Aut(N)$. What is the order of $Aut(N)$? $\endgroup$ – lulu May 5 '17 at 14:45
5
$\begingroup$

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 $N \subseteq Z(G)$.

$\endgroup$
4
$\begingroup$

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 $N\leq Z(G).$

$\endgroup$
0
$\begingroup$

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 $N \subseteq Z(G)$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Paramanand Singh May 6 '17 at 2:30
  • $\begingroup$ 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. $\endgroup$ – Nicky Hekster May 6 '17 at 19:35
  • $\begingroup$ 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. $\endgroup$ – Paramanand Singh May 7 '17 at 3:01
  • $\begingroup$ Paramanand, yes, if $N$ is cyclic, then it is easy. But this requirement can be dropped. $\endgroup$ – Nicky Hekster May 7 '17 at 14:15
0
$\begingroup$

$gng^{-1}=n^k$ . 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.