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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?

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    $\begingroup$ This is an example of a very good post, where the OP shows his thoughts, analysis and attempts. +1 from me! $\endgroup$ Dec 18, 2014 at 20:20

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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 $N \subseteq Z(G)$.

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  • $\begingroup$ Yes, the lemma is clear. Thanks. Nice proof. $\endgroup$
    – rogerl
    Dec 18, 2014 at 20:17
  • $\begingroup$ The proof of the proposition is the same isn't it? You get the order of $N_G(H)/C_G(H)$ divides the order of $\Aut(H)=p-1$ and the order of $N(G)$. Why does $H$ have to be normal though? Can't you get the same results without $H$ being normal? $\endgroup$
    – Asinomás
    Dec 18, 2014 at 20:33
  • $\begingroup$ proposition $2$ can be proven using the thing about normal subgroups being a union of conjugacy classes.the smallest non trivial ones have size $p$. Although prop 2 is not generalization. $\endgroup$
    – Asinomás
    Dec 18, 2014 at 20:37
  • $\begingroup$ @Jorge - yes or again using the lemma! $\endgroup$ Dec 18, 2014 at 20:39
  • $\begingroup$ I don't see why $H$ needs to be normal for the first proof. Couldn't we use the same argument without normality?That is: the order of $N_G(H)/C_G(H)$ divides the order of $N_G(H)$ and therefore the order of $G$. But it also divides the order of $Aut(H)$? since the order of $Aut(H)$ is relatively prime to the order of $G$ we get$C_G(N)=N_G(N)$. Oh, I see why you need it to be normal lol. $\endgroup$
    – Asinomás
    Dec 18, 2014 at 20:41
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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!)

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  • $\begingroup$ +1 for the uncalled-for flattery :) Both your proof and @NickyHekster's proof are specific to the Fermat prime case. I was expecting someone to post a proof that applied to all cases (except of course $p=5$). But perhaps no simple such proof exists. $\endgroup$
    – rogerl
    Dec 18, 2014 at 20:18
  • $\begingroup$ Well a generalization would be that $|N|=p$ is prime and gcd$(|G|,p-1)=1$. $\endgroup$ Dec 18, 2014 at 20:25
  • $\begingroup$ When you say a proof that applied to all cases, what do you mean? $S(p)$ is false for non-Fermat primes, as you pointed out. $\endgroup$ Dec 18, 2014 at 23:38
  • $\begingroup$ Yes, you're right: I guess there's a proof that it holds for Fermat primes and a counterexample for non-Fermat primes. $\endgroup$
    – rogerl
    Dec 19, 2014 at 14:19
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In addition to the things I answered here, I just want to add the dual of Proposition 1:

Proposition 1* Let $G$ be a finite group and $H$ a subgroup of prime index $p$, with gcd$(|G|,p-1)=1$. Then $G' \subseteq H$.

Note that this implies that $H \unlhd G$, and that it is in fact sufficient to prove that $H$ is normal, since then $G/H \cong C_p$ is abelian.

Proof This is more sophisticated than the proof below: firstly, we may assume by induction on $|G|$, that $H$ is core-free, that is core$_G(H)=\bigcap_{g \in G}H^g=1$. This means that $G$ can be homomorphically embedded in $S_p$. Let $P \in Syl_p(G)$ and note that because $|S_p|=p \cdot (p-1) \cdots \cdot 1$, $|P|=p$. By the $N/C$-Theorem, $N_G(P)/C_G(P)$ embeds in Aut$(P) \cong C_{p-1}$. By the assumption gcd$(|G|,p-1)=1$, we get that $N_G(P)=C_G(P)$. Since $P$ is abelian we have $P \subseteq C_G(P)$, whence $P \subseteq Z(N_G(P))$. We now can apply Burnside's Normal $p$-complement Theorem, which implies that $P$ has a normal complement $N$, that is $G=PN$ and $P \cap N=1$. Note that $|G/N|=p$.

Look at the image of $H$ in $G/N$. Then $G=HN$, or $HN=N$. In the latter case $H \subseteq N$, and $|G:H|=|G:N|=p$, whence $H=N$ and we are done if we can refute the first case. If $G=HN$, then $|G:H \cap N|=|G:N|\cdot|N:H \cap N|=|G:N|\cdot |G:H|=p \cdot p=p^2$, contradicting the fact that $|G| \mid |S_p|$. The proof is now complete.

Corollary 1 Let $G$ be a finite group and let $H$ be a subgroup with $|G:H|=p$, the smallest prime dividing the order of $G$. Then $G' \subseteq H$. In particular, $H$ is normal.

Corollary 2 Let $G$ be a finite group and let $H$ be a subgroup with $|G:H|=p$ and gcd$(|H|,p-1)=1$. Then $H$ is normal.

Observe that this last result renders a well-known result for $p=2$! Finally for fun:

Corollary 3 Let $G$ be a finite group of odd order and $H$ a subgroup with $|G:H|=65537$. Then $H$ is normal.

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