# Groups satisfying the normalizer condition are nilpotent (without using Sylow theory)

Recall that a group $G$ satisfies the normalizer condition if for any proper subgroup $H$, its normalizer in $G$, $N_G(H)$ is a strictly larger group.

For finite groups, this property is equivalent to $G$ being nilpotent (that is, its lower central series terminates at the trivial group). The proof I know/found is by using yet another criteria for nilpotency: all Sylow subgroups are normal.

However, is there a proof that avoids mention of Sylow subgroups? I ask because both the condition of being nilpotent and having the normalizer condition are quite elementary and make no reference to Sylow subgroups.

Edit: I can at least show that the derived subgroup is a proper subgroup:

Let $M$ be a maximal subgroup of $G$, they exist by finiteness. The normalizer of $M$ properly contains $M$ and hence it is $G$, hence $M$ is normal in $G$. Moreover, $G/M$ has no subgroups and hence is prime cyclic. Now consider [G,G].

We will show that this is contained in $M$. This follows easily since [G/M,G/M] = [G,G]/M but $[G/M,G/M]$ is trivial since $G/M$ is abelian.

• "Without using Sylow theory" in group theory sounds to me like "without using divisibility" in elementary number theory. – Dietrich Burde Jul 6 '18 at 12:59
• I am quite comfortable with Sylow theory but not so much with central series. This question is an attempt to understand central series without the "crutch" of Sylow subgroups but your point is well taken anyways! – Asvin Jul 6 '18 at 13:00
• I understand. There are often questions about proving something without Sylow, or proving a limit without L'Hopital, etc. Then life would be much easier by just accepting Sylow or L'Hopital. But of course you are right, sometimes one wants to understand something differently. – Dietrich Burde Jul 6 '18 at 13:10

## 2 Answers

Personally, I don't believe that the theory of central series is more elementary than Sylow theory, and in my experience students find Sylow theory easier, probably because it does not involve technical calculations with quotient groups.

But if you insist, I think you can do it this way. By using induction on $|G|$, we can assume that $Z(G)=1$. Also all maximal subgroups $M$ of $G$ are normal, so they have prime index $p$. It is not hard to show that $M$ satisfies the normalizer condition, so $M$ is nilpotent by induction. So $Z(M) \ne 1$. Let $N$ be a minimal normal subgroup of $G$ that is contained in $Z(M)$, let $G = \langle M,g \rangle$ with $g^p \in M$, and let $H = \langle N,g \rangle$.

So $Z(H) \cap N \le Z(G)$ and hence $Z(H) \cap N = 1$. Since $\langle g \rangle \cap N \le Z(H) \cap N$, we have $\langle g \rangle \cap N =1$.

Now $H$ satifies the normalizer condition, by assumption if $H=G$, and by induction otherwise. So $\langle g \rangle < N_H(\langle g \rangle)$, and hence $N_N(\langle g \rangle) \ne 1$. But $[N_N(\langle g \rangle), \langle g \rangle] \le \langle g \rangle \cap N =1$, so $N_N(\langle g \rangle) \le Z(G)$, contradiction.

• For both $M$ and $N$ satisfying the normalizer condition, can't you just argue that the normalizer condition is preserved under taking subgroups? – Asvin Jul 6 '18 at 13:25
• I can see that the normalizer condition is preserved on passing to normal subgroups but not to arbitrary subgroups. – Derek Holt Jul 6 '18 at 15:52
• I was thinking of using the equivalence with the ascending chain condition but anyway if we believe that normalizer condition = nilpotent, then we should also believe that subgroup preserve the normalizer condition, no? – Asvin Jul 6 '18 at 15:54
• Anyway, I am not quite sure I follow the inequalities $Z(H) \cap N \leq Z(G)$ and $\langle g\rangle \cap N \leq Z(H) \cap N$, can you elaborate a little? – Asvin Jul 6 '18 at 15:55
• Both of those follow from the fact that any element in $Z(N)$ that centralizes $g$ lies in$Z(G)$. – Derek Holt Jul 6 '18 at 20:17

This is a reformulation of Derek Holt's answer in a way that makes sense to me:

First, let us make the preliminary observation that subgroups of $G$ also satisfy the normalizer property if $G$ does. This is because satisfying the normalizer property is equivalent to asserting that for all subgroups $H$, there exist a chain of ascending subgroups $H_0 = H \lhd H_1 \lhd \dots \lhd G$.

Now, to show that $H$ satisfies the normalizer property, let $K \subset H$ be a subgroup. As above, we can find a sequence of groups $K_i \lhd K_{i+1}$ with $K_0 = K$ and $K_{n} = G$ for some $N$. Consider now $K_i' = K_i \cap H$. Certainly $K_i' \lhd K_{i+1}'$ and moreover, $K_i' \subset H$, $K_0' = K$ and $K_n' = H$. The first $i$ such that $K_i' > K$ will then produce a non trivial normalizer of $K = K'_{i-1}$ within $H$.

Now, we come to question. It is clearly sufficient to prove that any finite group $G$ satisfying the normalizer condition contains a non trivial center (using the ascending central series definition of nilpotence). This is what we will do.

Consider a maximal subgroup $M$ of $G$. This is normal by the normalizer condition and moreover $G/M$ is prime cyclic. Let $g \in G$ be such that $g^p \in M$ and $\langle g,M \rangle = G$.

$M$ satisfies the normalizer condition and therefore, by induction, it has a non trivial center $Z(M)$.

Let us search for a center to $G$ within $Z(M)$. Consider $H = \langle Z(M), g\rangle$. Note that $\langle g\rangle \cap Z(M) \subset Z(G)$ and therefore, we can suppose that $\langle g\rangle \cap Z(M)$ is trivial.

Now, $N_H(\langle g \rangle) > \langle g\rangle$ by the normalizer property of $H$. Therefore, $A = N_{Z(M)}(\langle g\rangle)$ is non trivial.

However, $[A,\langle g\rangle] \subset \langle g\rangle \cap Z(M)$ since $Z(M)$ is normal in $G$ ($Z(M)$ is a characteristic subgroup of a normal subgroup). Since we assumed $\langle g\rangle \cap Z(M)$ is trivial, we have a contradiction!

• I am afraid that I still do not understand your preliminary observation that subgroups of $G$ satisfy the normalizer condition. Of course it is true because the property is equivalent to nilpotency , but that is what we are trying to prove, so we cannot use nilpotency at this point in the proof. At the moment you are just saying that it clearly holds. – Derek Holt Jul 7 '18 at 8:29
• @DerekHolt I edited my answer to expand on my argument, am I missing something? – Asvin Jul 7 '18 at 8:41
• Yes, your argument works OK, sorry! – Derek Holt Jul 7 '18 at 9:00
• No problem! Thanks a lot for answering my question. I learnt a lot. – Asvin Jul 7 '18 at 9:11