Algebraic aspects of nilpotent groups I

March 08, 2025

Algebraic aspects of nilpotent groups I

In a previous post we discussed the classification of finite simple groups. One of the motivations for doing so was that to understand finite groups, it makes sense to break them up into the simplest pieces, akin to how prime number factorisations are key to number theory. It is in general not possible to tell the isomorphism type of $G$ from the isomorphism types of a normal subgroup $N$ and the quotient $G/N$ (consider for example cyclic and dihedral groups), but this process of breaking things up nonetheless yields lots of useful information about groups, so much so that it is fruitful to consider, for a group $G$ a finite series of subgroups

\[ \{e\} = G_r \unlhd G_{r-1}, \dots \unlhd G_0=G \]

where each $G_{i+1}$ is normal in $G_{i}$. This is known as a subnormal series, and is meaningful even for infinite groups. The quotients $G_{i}/G_{i+1}$ are known as the factors. One can also reverse the indexing and have $G_0$ be the trivial subgroup and $G_r=G$. By imposing conditions on the subnormal series, one obtains interesting classes of groups that admit subnormal series with these properties. In this post, following Keith Conrad's articles, we will consider the algebraic aspects of a class of groups known as nilpotent groups. Surprisingly, these groups have a lot of geometry, which will (hopefully) be covered in future posts.

Motivation

In many undergraduate mathematics courses the structure theorem for finitely generated abelian groups is proved. This is one of the many ways in which (finitely generated) abelian groups are well behaved. One can attempt to generalise this, but in order for these groups to have similar nice properties, we expect these groups to be similar to abelian groups in some sense, so we need to have some notion of how close a group is to being abelian.

Definition: Let $G$ be a group. Define $L_0=G$ and for $i \geq 0$

\[L_{i+1} =[G, L_i] \]. This is called the lower central series of $G$.

One can show by induction that $L_i$ are normal subgroups of $G$ by induction on $i$. This implies that $L_{i+1} \subset L_i$.

Since $L_1$ is the commutator subgroup $[G, G]$, which vanishes if and only if $G$ is abelian, we can ask more generally what happens if we instead require that $L_i=\{e\}$ for some $i$. Groups with this property are called nilpotent. The least such $i$ such that $L_i=\{e\}$is known as the nilpotency class of $G$. Abelian groups are thus exactly the groups of nilpotency class 1.

Theorem 1 Nilpotency is closed under subgroups, quotient groups, and direct products.

The proof is an exercise in applying the definitions. Note that it is not in general true that if $G$ has a nilpotent normal subgroup $N$ with nilpotent quotient $G/N$, then $G$ is nilpotent. Consider for example $S_3$. We say that nilpotency is not preserved under extensions.

Definition A subnormal series

\[ \{e\} = G_r \unlhd G_{r-1}, \dots \unlhd G_0=G \]

is said to be central if, for all $i$, $G_i$ is a normal subgroup of $G$ such that $G_{i}/G_{i+1} \subset Z(G/G_{i+1})$, with a corresponding definition for series with indices that go in the other direction.

The reason we call $\{L_i\}$ the lower central series is the following:

Theorem 2 Let $\{G_i\}$ be a central series for $G$. Then $L_i \subset G_i$.

The next theorem shows further similarities between nilpotent and abelian groups.

Definition Set $Z_0=\{e\}$ and $Z_1=Z(G)$, which are both normal. Having defined $Z_{i} \unlhd G$, write the centre of $G/Z_i$ as $Z_{i+1}/Z_i$, so $Z_{i+1}$. This is called the upper central series of $G$.

Similarly to theorem 2, it is true that for any central series $G_i \subset Z_i$.

Corollary 3 For every group $G, Z_i = G$ for some $i$ if and only if $L_i = \{e\}$ for some $i$, in which case the least $i$s for which these occur are the same.
Proof Suppose $Z_n = G$. By reversing the indices and setting $G_i = Z_{n−i}$ for $i = 0, 1, . . . , n$, theorem 2 implies $L_i \subset G_i = Z_{n−i}$, so $L_n$ is trivial. The exact same index reversing trick combined with the observation before the corollary yields the converse.

Hence nilpotent groups are also those for which the upper central series terminates in the whole group, and we use this viewpoint to prove

Theorem 4 If $ G $ is a nontrivial nilpotent group, then

For every nontrivial normal subgroup $ N \triangleleft G $, $ N \cap Z(G) \neq \{e\} $ and $[G, N] \neq N$
For every proper subgroup $ H $, $ H \neq N(H) $.

In particular, a nontrivial nilpotent group has a nontrivial center, and its commutator subgroup is a proper subgroup.

Proof (1): We have $ Z_i = G $ for some $ i \geq 1 $. Therefore, $ N \cap Z_i $ is nontrivial for some $ i $. Pick $ i $ minimal, so $ i \geq 1 $. Since $[G, Z_i] \subset Z_{i-1}$, we have $[G, N \cap Z_i] \subset N \cap Z_{i-1}$. By minimality of $ i $, $ N \cap Z_{i-1} $ is trivial, so $[G, N \cap Z_i]$ is trivial. Thus, $ N \cap Z_i \subset Z(G) = Z_1 $, so $ N \cap Z_i \subset N \cap Z_1 $. The reverse inclusion is clear, so $ N \cap Z_1 = N \cap Z_i \neq \{e\} $.

To prove $[G, N] \neq N$, assume $[G, N] = N$. Then $ N \subset [G, G] = L_1 $, so $ N = [G, N] \subset [G, L_1] = L_2 $. By induction, $ N \subset L_i $ for all $ i $, and taking $ i $ large enough implies $ N $ is trivial, a contradiction.

Taking $ N = G $ in (1) recovers the special case mentioned at the end of the theorem.

(2): Since $ Z_0 = \{e\} $ and $ Z_i = G $ for some $ i $, there is some $ i $ such that $ Z_i \subset H $. Then $[H, Z_{i+1}] \subset [G, Z_{i+1}] \subset Z_i \subset H $, so $ Z_{i+1} \subset N(H) $. Picking $ i $ maximal such that $ Z_i \subset H $, the fact that $ Z_{i+1} \not\subset H $ implies $ H \neq N(H) $. $\blacksquare$

The last result we present about nilpotent groups in general is the following:

Theorem 5 Let $G$ be a nilpotent group. Then the set of elements of $G$ of finite order is a subgroup of $G$ .

Note that this is false in general: $SL_2(\mathbb{Z})$ is generated by

\[ \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}.\] and

\[\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}.\]
,which are both finite order elements, but the group contains infinite order elements.

To prove the theorem, we need to lay some groundwork. By direct computation, one can show that for any group $G$ and any $x, y, z \in G$, $[xy,z] = [x,z]^y[y,z]$ and $[x,yz] = [x,z][x,y]^z$. Using these, we can deduce

Lemma 6 Let $ G $ be a group. If $[x, G] \subseteq Z(G)$, then $[xy, z] = [x, z][y, z]$ for all $ y, z \in G $.

Lemma 7 Let $ G $ be a group, and assume that $ G $ is generated by elements of finite order. Then for every $ c > 0 $, $ G^c / G^{c+1} $ is a torsion group.

Proof We proceed by induction on $ c $. If $ c = 1 $, then $ G^1 / G^2 = G^{\text{ab}} $, and an abelian group generated by elements of finite order is torsion.

Assume the result is true for $ c $, and consider $ G^{c+1} / G^{c+2} $. This is abelian, since $[G^{c+1}, G^{c+1}] \subseteq [G^{c+1}, G] = G^{c+2}$, so it suffices to show that the generators are of finite order.

$ G^{c+1} / G^{c+2} $ is generated by elements of the form $[x, g]$ with $ x \in G^c $ and $ g \in G $. By assumption, $ x $ is torsion modulo $ G^{c+1} $, so there exists $ n > 0 $ such that $ x^n \in G^{c+1} $. Moreover, since $ G^{c+1} / G^{c+2} $ is abelian, by Lemma 6 we have that $[x^n, g] \equiv [x, g]^n \pmod{G^{c+2}}$.

But since $ G^{c+1} / G^{c+2} $ is abelian, $[h, g] \equiv 1$ if $ h \in G^{c+1} $. Since $ x^n \in G^{c+1} $, then $[x, g]^n \equiv [x^n, g] \equiv 1 \pmod{G^{c+2}}$, so $[x, g]$ is torsion modulo $ G^{c+2} $, as desired. $\blacksquare$

Proof of Theorem 5: It suffices to show that the product of two torsion elements is again torsion. Consider two finite order elements $x,y$ which generate a subgroup $H$. Since nilpotency passes to subgroups, $H$ has a lower central series $H_i$, and by lemma 7 each of its factor groups is torsion, so all elements of $H$ are torsion as well. $\blacksquare$

Finite nilpotent groups

Finite nilpotent groups display much more structure which we explore in this section. We first exhibit many examples of these: finite $p-$ groups.

Theorem 8 If $ G $ is a nontrivial finite $ p $-group and $ N $ is a nontrivial normal subgroup of $ G $, then $ N \cap Z(G) \neq \{e\} $.

In particular, they have non-trivial centre.

Proof Since $ N $ is a normal subgroup of $ G $, a conjugacy class in $ G $ that meets $ N $ lies entirely inside of $ N $ (that is, if $ g \in N $, then $ xgx^{-1} \in N $ for all $ x \in G $). Let $ K_{g_1}, \dots, K_{g_s} $ be the different conjugacy classes of $ G $ that lie inside $ N $, so
\[ |N| = |K_{g_1}| + \cdots + |K_{g_s}|. \]
(Note that elements of $ N $ can be conjugate in $ G $ without being conjugate in $ N $, so breaking up $ N $ into its $ G $-conjugacy classes in is a coarser partitioning of $ N $ than breaking it into $ N $-conjugacy classes.) The left side of is a power of $ p $ greater than 1. Each term on the right side is a conjugacy class in $ G $, so $ |K_{g_i}| = [G : Z(g_i)] $, where $ Z(g_i) $ is the centralizer of $ g_i $ in $ G $. This index is a power of $ p $ greater than 1 except when $ g_i \in Z(G) $, in which case $ |K_{g_i}| = 1 $. The $ g_i $’s in $ N $ with $ |K_{g_i}| = 1 $ are elements of $ N \cap Z(G) $. Therefore, if we reduce the equation modulo $ p $, we get
\[ 0 \equiv |N \cap Z(G)| \pmod{p}, \]
so $ |N \cap Z(G)| $ is divisible by $ p $. Since $ |N \cap Z(G)| \geq 1 $, the intersection $ N \cap Z(G) $ contains a non-identity element. $\blacksquare$

Corollary 9 If $ G $ is a nontrivial finite $ p $-group with size $ p^n $, then there is a normal subgroup of size $ p^j $ for every $ j = 0, 1, \dots, n $.

Proof: We first show this for $j=1$. By theorem 8, $Z(G)$ is a non-trivial abelian group, so contains a subgroup of order $p$. This is then normal in $G$.

We now do the general case by induction on $ n $. The result is clear if $ n = 1 $. Suppose $ n \geq 2 $ and the theorem is true for $ p $-groups of size $ p^{n-1} $. If $ |G| = p^n $, then it has a normal subgroup $ N $ of size $ p $ by the preceding corollary. Then $ |G/N| = p^{n-1} $, so for $ 0 \leq j \leq n-1 $, there is a normal subgroup of $ G/N $ with size $ p^j $. The pullback of this subgroup to $ G $ is normal and has size $ p^j \cdot |N| = p^{j+1} $. $\blacksquare$

Theorem 10 Every finite $ p $-group is nilpotent.

Proof We will construct a central series of a $ p $-group $ G $ from the bottom up. The numbering will be reversed: $ G_{n-k} = Z_k(G) $, which are defined as follows:

$ Z_0(G) = 1 $ (i.e., $ G_n $ where $ n $ is undetermined).
$ Z_1(G) = Z(G) $ (which is $ > 1 $ if $ G $ is a $ p $-group).
Given $ Z_k(G) $, let $ Z_{k+1}(G) $ be the subgroup of $ G $ that contains $ Z_k(G) $ and corresponds to the center of $ G/Z_k(G) $, i.e.,
    \[
    Z_{k+1}(G)/Z_k(G) = Z(G/Z_k(G)).
    \]

Since $ G/Z_k(G) $ is a $ p $-group, it has a nontrivial center, making $ Z_{k+1}(G) > Z_k(G) $ unless $ Z_k(G) = G $. Since $ G $ is finite, we must have $ Z_n(G) = G $ for some $ n $, making $ G $ nilpotent of class $ \leq n $. $\blacksquare$

Theorem 11 A finite group is nilpotent if and only if all of its Sylow subgroups are normal, or equivalently, the group is isomorphic to the direct product of its Sylow subgroups.

Proof: Suppose $ G $ is a finite nilpotent group. Let $ P $ be a Sylow subgroup. To show $ P \triangleleft G $, we will show the normalizer $ N(P) $ is $ G $. From Sylow theory, $ N(N(P)) = N(P) $. No proper subgroup of a nilpotent group equals its normalizer, so we must have $ N(P) = G $. Conversely, suppose all the Sylow subgroups of $ G $ are normal. Then $ G $ is isomorphic to a direct product of its Sylow subgroups, by the Sylow theorems. The Sylow subgroups are finite $ p $-groups, hence they are nilpotent, so their direct product is nilpotent too.

Hence the non-commutativity of finite nilpotent groups is contained within every prime, i.e. within its Sylow subgroups.

Theorem 12 For a nontrivial finite group $ G $, the following are equivalent to nilpotency:

For every proper subgroup $ H $, $ H \neq N(H) $,
If $ d \mid |G| $, then there is a normal subgroup of size $ d $,
Every nontrivial quotient group of $ G $ has a nontrivial center,
Elements of relatively prime order in $ G $ commute.

Proof: The strategy in most cases is to show that the condition implies all the Sylow subgroups are normal.

We saw nilpotency implies (1) in theorem 4. Now assume (1) is true. Let $ P $ be a Sylow subgroup of $ G $. From the Sylow theorems, $ N(N(P)) = N(P) $, so $ N(P) $ is its own normalizer. Therefore, $ N(P) = G $ by (1), so $ P \triangleleft G $. Thus, all the Sylow subgroups of $ G $ are normal, so $ G $ is nilpotent.

Note that we have shown that (2) holds for finite $p$ groups in corollary 9. Next, in a direct product of $ p $-groups for different $ p $, say $ P_1 \times \cdots \times P_m $ where $ |P_i| = p_i^{e_i} $, to construct a normal subgroup with size $ d $, write $ d = p_1^{f_1} \cdots p_m^{f_m} $ and use $ H_1 \times \cdots \times H_m $ where $ H_i $ is normal of size $ p_i^{f_i} $ in $ P_i $. Conversely, if (2) is true, then letting $ d $ be the maximal power of a prime dividing $ |G| $ shows the Sylow subgroups are normal.

Since nilpotency of $ G $ implies nilpotency of every quotient, (3) is immediate. Conversely, if (3) is true, then $ G/Z_i $ has a nontrivial center when $ Z_i \neq G $. Therefore, $ Z_{i+1} \neq Z_i $ when $ Z_i \neq G $. This can’t continue indefinitely, so some $ Z_i $ equals $ G $.

If (4) is true, let $ P_1, \dots, P_m $ be a set of nontrivial Sylow subgroups of $ G $ for the different primes dividing $ |G| $. Elements in $ P_i $ and $ P_j $ commute when $ i \neq j $, by (4), so the map $ P_1 \times \cdots \times P_m \to G $ given by $ (x_1, \dots, x_m) \mapsto x_1 \cdots x_m $ is a homomorphism between groups of equal size. Since the $ x_i $’s commute and have relatively prime order, they multiply to the identity only when each $ x_i $ is trivial, so this map is injective and thus is an isomorphism. This implies $ G $ is a direct product of finite $ p $-groups, so it is nilpotent. Conversely, to see that in a finite nilpotent group (4) is true, write $ G $ as a direct product of its Sylow subgroups. In this product decomposition, two elements in $ G $ with relatively prime order live inside products of disjoint Sylow subgroups, so they commute.

Definition The Frattini subgroup of a finite group is the intersection of its maximal subgroups:
\[\Phi(G) = \bigcap_{\text{max. } M} M. \]
We set the trivial group to have trivial Frattini subgroup.

Theorem 13 For every finite group $ G $, $ \Phi(G) $ is nilpotent.

Proof: We will show all the Sylow subgroups of $ \Phi(G) $ are normal subgroups of $ \Phi(G) $, so $ \Phi(G) $ is nilpotent. Let $ P $ be a Sylow subgroup of $ \Phi(G) $. Then $ G = \Phi(G) N_G(P) $ by the Frattini argument. If $ P $ is not normal in $ G $, then $ N_G(P) \neq G $. Let $ M $ be a maximal subgroup of $ G $ containing $ N_G(P) $, so $ \Phi(G) \subset M $. Therefore, $ G = \Phi(G) N_G(P) \subset M $, a contradiction, so $ P \triangleleft G $, which implies $ P \triangleleft \Phi(G) $.

A natural question to ask at this point is which finite nilpotent groups arise as Frattini subgroups of finite groups. This was completely characterised by Eick: Let $H$ be a finite group and $G$ be a normal subgroup contained in the Frattini subgroup of $H$. Then the group of inner automorphisms of $G$ is isomorphic to a subgroup of Frattini subgroup of the automorphism group of $G$.

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