Throughout this post, is the normalizer of a subgroup of a group

I already have four posts on nilpotent groups in this blog; you can see them here: (1), (2), (3), (4). In this post, the focus is on finite nilpotent groups. As you are going to see, asking how much we know about finite nilpotent groups is the same as asking how much we know about finite -groups because finite nilpotent groups are nothing but direct products of finite -groups.

**Lemma 1**. i) Every finite -group is nilpotent.

ii) Every finite direct product of finite -groups is a nilpotent group.

*Proof*. i) Let be a finite -group. The proof is by induction over If then is abelian hence nilpotent. Now, for the general case, since is a finite -group, is non-trivial, and hence is a finite -group with Thus, by induction, is nilpotent. Therefore, by Fact 3, i), in this post, is nilpotent.

ii) It follows from i) and Fact 4 in this post.

**Lemma 2** (Frattini Argument). Let be a finite group, and let be a Sylow -subgroup of If is a normal subgroup of and then

*Proof*. Note that is clearly a Sylow -subgroup of Now, let Then because is normal. So and hence both are Sylow -subgroups of Thus, by Sylow theorems, are conjugate in i.e. for some So we get that i.e. which gives So and the result follows.

**Theorem**. Let be a finite group. Let be, respectively, the commutator subgroup and the Frattini subgroup of The following statements are equivalent.

i) is nilpotent.

ii) satisfies the normalizer condition.

iii) Every maximal subgroup of is normal.

iv) where is the Frattini subgroup of

v) Every Sylow subgroup of is normal.

vi) is isomorphic to a direct product of its Sylow subgroups.

vii) Every two elements of of coprime order commute.

*Proof*. i) ii). This is true for any nilpotent group, not just finite ones. See Fact 5 in this post.

ii) iii). Let be a maximal subgroup of Since satisfies the normalizer condition, and hence, since is maximal, i.e. is normal in

iii) iv). Let be a maximal subgroup of Then is a group with no non-trivial proper subgroup and hence it is a cyclic group (of prime order). In particular, is abelian and so The result now follows since, by definition, is the intersection of all maximal subgroups of

iv) v). Let be a Sylow subgroup of and suppose, to the contrary, that is not normal in Then and so for some maximal subgroup of Now, is normal in because and so, by Lemma 2, contradiction.

v) vi). Let be all the distinct prime divisors of and let be a Sylow -subgroup. Since each is normal and for all because we have and so Now, for each let

Then and so which gives Thus

vi) vii). We may assume that where each is a -group for some prime and for Let be two elements of coprime order in Note that because for So now if for some then and hence which is non-sense. Thus for each either or and hence

vii) vi). Let the prime factorization of For each prime let be a Sylow -subgroup of and put Since for all we have for all Thus if are two elements of then and so is a subgroup of Now consider the following map

Note that is well-defined because if and then

which gives because It is clear that is a bijective group homomorphism, and hence The result now follows because

which gives

vi) i). Clear by Lemma 1, ii).

**Note**. The main reference for this post is the book *The Theory of Nilpotent Groups* by Clement, Majewicz, and Zyman. Regarding the above theorem, I should mention here that unfortunately the book (see page 49) gives a false proof of “vii) vi)”. The authors first “prove” that every Sylow subgroup of is normal; here’s their argument: if is a Sylow -subgroup of and then either in which case or and are coprime, in which case commutes with every element of and again The problem with their argument is that they assume unknowingly that is the unique Sylow -subgroup of which is not given. What if is in a Sylow -subgroup different from ? In fact, if was unique, then would be normal, by Sylow theorems, and so the condition vii) would be redundant.