Posts Tagged ‘nilpotency class’

Nilpotent groups (4)

Posted: January 16, 2022 in Basic Algebra, Groups
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Links to previous parts of this post: part (1), part (2), part (3). As usual, we denote by Z(G) the center of a group G.

In part (2), we defined the lower central series \gamma_n(G), \ n \ge 0, of a group G and showed that G is nilpotent if and only if \gamma_n(G)=(1). We now define the upper central series of G and prove a similar condition for a group to be nilpotent.

Definition. The upper central series of a group G is an ascending chain of subgroups \{Z_n(G)\}_{n \ge 0} of G defined inductively as follows

Z_0(G)=(1), \ \ \ \ Z_n(G)/Z_{n-1}(G)=Z(G/Z_{n-1}(G)), \ \ \ n \ge 1.

Remark 1. Note that since Z_0(G)=(1) is a normal subgroup of G, and Z(G/Z_{n-1}(G)) is a normal subgroup of G/Z_{n-1}(G), we get by induction that every Z_n(G) is normal in G.

Remark 2. If Z_k(G)=Z_{k+1}(G) for some k \ge 0, then Z_n(G)=Z_k(G) for all n \ge k. That’s because, by induction,

Z_n(G)/Z_{n-1}(G)=Z(G/Z_{n-1}(G))=Z(G/Z_k(G))=Z_{k+1}(G)/Z_k(G)=(1).

Lemma. Let G be a group, and let H be a subgroup of G. If N is a normal subgroup of G, then [H,G] \subseteq N if and only if HN/N \subseteq Z(G/N),

Proof. Clearly HN/N \subseteq Z(G/N) if and only if [HN/N,G/N]=1_{G/N} if and only if [HN,G] \subseteq N. Now, if [HN,G] \subseteq N, then [H,G] \subseteq [HN,G] \subseteq N. Conversely, suppose that [H,G] \subseteq N, and let h \in H, g \in G, a \in N. Then

\begin{aligned}[ha,g]=haga^{-1}h^{-1}g^{-1}=(hah^{-1})(hga^{-1}g^{-1}h^{-1})(hgh^{-1}g^{-1})=(hah^{-1})(hga^{-1}g^{-1}h^{-1})[h,g] \in N,\end{aligned}

because N is normal and [h,g] \in [H,G] \subseteq N. So [HN,G] \subseteq N. \ \Box

Theorem. Let G be a group.

i) [Z_n(G), G] \subseteq Z_{n-1}(G) for all integers n \ge 1.

ii) If G is nilpotent, and (1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G is a central series in G, then G_i \subseteq Z_i(G) for all 0 \le i \le n.

iii) If Z_n(G)=G, for some integer n \ge 0, then (1)=Z_0(G) \subseteq Z_1(G) \subseteq \cdots \subseteq Z_n(G)=G is a central series in G.

iv) G is nilpotent if and only if Z_n(G)=G for some integer n \ge 0.

v) \gamma_n(G)=(1) if and only if Z_n(G)=G.

Proof. i) Use the Lemma with H=Z_n(G), \ N=Z_{n-1}(G).

ii) The proof is by induction. For i=0, we have (1)=G_0=Z_0(G). Suppose now that the result holds for some i \ge 0 with i+1 \le n. Then, by Remark 2 in the first part of this post, [G_{i+1},G] \subseteq G_i and so, by our induction hypothesis, [G_{i+1},G] \subseteq Z_i(G). Hence, applying the Lemma with H=G_{i+1}, \ N=Z_i(G), we get that

G_{i+1}Z_i(G)/Z_i(G) \subseteq Z(G/Z_i(G))=Z_{i+1}(G)/Z_i(G)

and so G_{i+1} \subseteq G_{i+1}Z_i(G) \subseteq Z_{i+1}(G), which completes the induction.

iii) We have the normal series (1)=Z_0(G) \subseteq Z_1(G) \subseteq \cdots \subseteq Z_n(G)=G, which is clearly a central series because Z_{i+1}(G)/Z_i(G)=Z(G/Z_i(G)), by definition.

iv) If G is nilpotent and (1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G is a central series, then G=G_n \subseteq Z_n(G), by ii), which gives Z_n(G)=G. Conversely, suppose that Z_n(G)=G for some integer n \ge 0. Then, by iii), (1)=Z_0(G) \subseteq Z_1(G) \subseteq \cdots \subseteq Z_n(G)=G is a central series and so G is nilpotent.

v) Suppose first that \gamma_n(G)=(1), and let G_i:=\gamma_{n-i}(G), \ 0 \le i \le n. Then, by part iv) of the Theorem in the second part of this post, (1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n is a central series in G, and so, by ii),

G=\gamma_0(G)=G_n \subseteq Z_n(G),

which gives Z_n(G)=G. Conversely, if Z_n(G)=G, then, by iii),

(1)=Z_0(G) \subseteq Z_1(G) \subseteq \cdots \subseteq Z_n(G)=G

is a central series in G. Thus, by part iii) of the Theorem in part (2) of this post, \gamma_n(G) \subseteq Z_0(G)=(1) hence \gamma_n(G)=(1). \ \Box

Remark 3. In part (2) of this post, we defined the nilpotency class of a nilpotent group G as the smallest integer n \ge 0 such that \gamma_n(G)=(1). So, by part v) of the above theorem, G is a nilpotent group of class n if and only if Z_n(G)=G.

To learn about finite nilpotent groups, see this post!

Nilpotent groups (3)

Posted: January 13, 2022 in Basic Algebra, Groups
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See part (1) and part (2) of this post here and here, respectively.

In part (2) of this post, we defined the lower central series of a group G and proved that G is nilpotent if and only if \gamma_n(G)=(1) for some n \ge 0. We are now going to use this result to prove some facts about nilpotent groups.

Fact 1. Every subgroup of a nilpotent group is nilpotent.

Proof. Let G be a nilpotent group, and let H be a subgroup of G. We first prove that \gamma_i(H) \subseteq \gamma_i(G) for all i. The proof is by induction. That is clear for i=0 because \gamma_0(H)=H \subseteq G=\gamma_0(G). If the inclusion holds for i, then

\gamma_{i+1}(H)=[\gamma_i(H), H] \subseteq [\gamma_i(G),G]=\gamma_{i+1}(G),

which completes the induction. Now, since G is nilpotent, \gamma_n(G)=(1) for some integer n \ge 0. Thus \gamma_n(H) \subseteq \gamma_n(G)=(1) and so \gamma_n(H)=(1), implying that H is nilpotent. \Box

Fact 2. Let G be a group, and let H be a normal subgroup of G.

i) If K is a group, and f: G \to K is a group homomorphism, then \gamma_i(f(G))=f(\gamma_i(G)) for all i \ge 0.

ii) If G is nilpotent, then G/H is nilpotent.

Proof. i) The proof is by induction. We have \gamma_0(f(G))=f(G)=f(\gamma_0(G)) and so the equality holds for i=0. Now, if the equality holds for i, then

\gamma_{i+1}(f(G))=[\gamma_i(f(G)),f(G)]=[f(\gamma_i(G)), f(G)]=f([\gamma_i(G),G])=f(\gamma_{i+1}(G)),

which completes the induction.

ii) Let f: G \to G/H be the natural group homomorphism. Since G is nilpotent, \gamma_n(G)=(1) for some integer n \ge 0, and so, by i), \gamma_n(G/H)=f(\gamma_n(G))=f((1))=(1), implying that G/H is nilpotent. \Box

Remark 1. Let G be a group with a normal subgroup H. In Theorem 1, ii), in this post, we showed that G is solvable if and only if both H,G/H are solvable. In Remark 1 in the first part of this post, we showed that every nilpotent group is solvable. However, it is not true that if both H, G/H are nilpotent, then G is nilpotent too. For example, the symmetric group S_3 is not nilpotent, by Example 3 in the first part of this post, but both A_3,S_3/A_3 are nilpotent because they are abelian. However, if H is in the center of G, and G/H is nilpotent, then, as the next fact shows, G is nilpotent too.

Fact 3. Let G be a group with the center Z(G).

i) If H \subseteq Z(G) is a subgroup of G and G/H is nilpotent, then G is nilpotent.

ii) \gamma_n(G/Z(G))=(1) if and only if \gamma_{n+1}(G)=(1).

Proof. i) Note that H is normal in G because H \subseteq Z(G). Now, let f: G \to G/H be the natural group homomorphism. Since G/H is nilpotent, \gamma_n(G/H)=(1) for some integer n \ge 0. But, as we showed in the proof of Fact 2, (1)=\gamma_n(G/H)=f(\gamma_n(G)) and so \gamma_n(G) \subseteq \ker f=H. Thus

\gamma_{n+1}(G)=[\gamma_n(G),G] \subseteq [H,G]=(1),

because H \subseteq Z(G), and so \gamma_{n+1}(G)=(1).

ii) We showed in i) that if \gamma_n(G/Z(G))=(1), then \gamma_{n+1}(G)=(1). Suppose now that \gamma_{n+1}(G)=(1). Then [\gamma_n(G),G]=\gamma_{n+1}(G)=(1) and so \gamma_n(G) \subseteq Z(G). Hence, if f: G \to G/Z(G) is the natural group homomorphism, then, by Fact 2, i),

\gamma_n(G/Z)=\gamma_n(f(G))=f(\gamma_n(G)) \subseteq f(Z(G))=(1),

and so \gamma_n(G/Z)=(1). \ \Box

Remark 2. In part (2) of this post, we defined the nilpotency class of a nilpotent group G as the smallest integer n \ge 0 such that \gamma_n(G)=(1). So, by Fact 3, ii), G \ne (1) is a nilpotent group of class n if and only if G/Z(G) is a nilpotent group of class n-1. As the next Fact shows, this result may be used to prove nilpotency certain groups are nilpotent.

Fact 4. Let G_1, \cdots, G_n, \ n \ge 2, be groups and let G:=G_1 \times G_2 \times \cdots \times G_n. Then G is nilpotent if and only if each G_i is nilpotent.

Proof. By induction over n, we only to prove the result for n=2. Suppose first that G is nilpotent, and let H_1:=G_1 \times (1), \ H_2:=(1) \times G_2. Then H_1,H_2 are normal subgroups of G and we have the group isomorphisms G/H_1 \cong G_2, \ G/H_2 \cong G_1. Thus, by Fact 2, ii), both G_1,G_2 are nilpotent.
Conversely, suppose that G_1,G_2 are nilpotent of class n_1,n_2, respectively, and let n:=\max\{n_1,n_2\}. We prove, by induction over n, that G is nilpotent of class n. It is clear for n=0 because then G_1=G_2=(1), and so G=(1). Also, if n=1, then both G_1,G_2 are abelian and at least one of them is non-trivial. Thus G is a non-trivial abelian group hence nilpotent of class 1. Suppose now that n \ge 2, and consider the natural group homomorphism

f: G=G_1 \times G_2 \to G_1/Z(G_1) \times G/Z(G_2).

Clearly f is onto and \ker f=Z(G_1 \times G_2). Thus G/Z(G) \cong G_1/Z(G_1) \times G_2/Z(G_2). By Remark 2, G_1/Z(G_1), G_2/Z(G_2) are nilpotent of class n_1-1,n_2-1, respectively. So, by induction, G/Z(G) is nilpotent of class \max\{n_1-1,n_2-1\}=\max\{n_1,n_2\}-1, and hence, by Remark 2, G is nilpotent of class \max\{n_1,n_2\}. \ \Box

Fact 5 (The Normalizer Condition). Let G \ne (1) be a nilpotent group, and let H be a subgroup of G. If H \subset G, then H \subset N(H), where N(H) is the normalizer of H in G.

Proof. Suppose, to the contrary, that H=N(H). Since G is nilpotent, \gamma_n(G)=(1) for some integer n \ge 1 (note that we can’t have n=0 because G is non-trivial). Thus \gamma_n(G) \subseteq H. We also have that \gamma_0(G)=G \ne H. So there exists an integer k \ge 1 which is minimal with respect to this property that \gamma_k(G) \subseteq H. Hence

[\gamma_{k-1}(G),H] \subseteq [\gamma_{k-1}(G),G]=\gamma_k(G) \subseteq H,

and so \gamma_{k-1}(G) \subseteq N(H)=H, contradicting minimality of k. \ \Box

See part (4) of this post here.

Nilpotent groups (2)

Posted: January 13, 2022 in Basic Algebra, Groups
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See part (1) of this post here.

Recall that given subgroups H,K of a group G, we denote by [H,K] the subgroup of G generated by all the commutators [x,y], \ x \in H, \ y \in K, where, as always, [x,y]:=xyx^{-1}y^{-1}.

Definition 1. The lower central series of a group G is a descending chain of subgroups \{\gamma_n(G)\}_{n\ge 0} of G defined inductively as follows

\gamma_0(G)=G, \ \ \ \ \gamma_{n}(G)=[\gamma_{n-1}(G), G], \ \ \ \ \forall n \ge 1.

So \gamma_1(G)=[\gamma_0(G),G]=[G,G]=G', \ \gamma_2(G)=[\gamma_1(G),G]=[G',G], etc.

Theorem. Let G be a group.

i) \gamma_n(G) is a normal subgroup of G for all integers n \ge 0.

ii) \gamma_n(G) \subseteq \gamma_{n-1}(G) for all integers n \ge 1.

iii) If G is nilpotent, and (1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G is a central series in G, then \gamma_i(G) \subseteq G_{n-i} for all 0 \le i \le n.

iv) If \gamma_n(G)=(1), for some integer n \ge 0, and G_i:=\gamma_{n-i}(G), \ 0 \le i \le n, then the following is a central series in G: \ (1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G.

v) G is nilpotent if and only if \gamma_n(G)=(1) for some integer n \ge 0.

Proof. i) We prove it by induction over n. It obviously holds for n=0 because \gamma_0(G)=G. Now suppose that n \ge 1 and \gamma_{n-1}(G) is normal in G. To show that \gamma_n(G) is normal in G, note that since, by definition, \gamma_n(G)=[\gamma_{n-1}(G), G], we only need to show that a[x,b]a^{-1} \in \gamma_n(G) for all a,b \in G, \ x \in \gamma_{n-1}(G), which is easy because

\begin{aligned}a[x,b]a^{-1}=axbx^{-1}b^{-1}a^{-1}=(axa^{-1})(aba^{-1})(ax^{-1}a^{-1})(ab^{-1}a^{-1})=[axa^{-1},aba^{-1}] \in [\gamma_{n-1}(G),G],\end{aligned}

because \gamma_{n-1}(G) is normal in G which gives axa^{-1} \in \gamma_{n-1}(G).

ii) Since \gamma_n(G)=[\gamma_{n-1}(G),G], we only need to show that [x,a] \in \gamma_{n-1}(G) for all x \in \gamma_{n-1}(G) and a \in G. We have

[x,a]=xax^{-1}a^{-1}=x(ax^{-1}a^{-1}) \in \gamma_{n-1}(G),

because, by i), \gamma_{n-1}(G) is normal in G and so ax^{-1}a^{-1} \in \gamma_{n-1}(G).

iii) We prove it by induction over n. It obviously holds for i=0 because G_n=G=\gamma_0(G). Now suppose that the inclusion holds for some i \ge 0, and suppose that i+1 \le n. Then

\gamma_{i+1}(G)=[\gamma_i(G),G] \subseteq [G_{n-i},G] \subseteq G_{n-(i+1)},

by Remark 2 in the first part of this post.

iv) By i), ii), (1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G is a normal series. We also have that

[G_{i+1},G]=[\gamma_{n-i-1}(G),G]=\gamma_{n-i}(G)=G_i

for all 0 \le i \le n-1. Thus, by Remark 2 in the first part of this post, the series is central.

v) If G is nilpotent, and (1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G is a central series in G, then, by iii), we have \gamma_n(G) \subseteq G_0=(1), and so \gamma_n(G)=(1). Conversely, suppose that \gamma_n(G)=(1) for some integer n \ge 0. Then, by iv), (1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G, where G_i:=\gamma_{n-i}(G), \ 0 \le i \le n, is a central series in G and hence G is nilpotent. \Box

Definition 2. Let G be a nilpotent group. By the last part of the above Theorem, \gamma_n(G)=(1) for some integer n \ge 0. The smallest integer n \ge 0 such that \gamma_n(G)=(1) is called the nilpotency class of G.

Remark. Since for a group G, we have \gamma_0(G)=G, \ \gamma_1(G)=[G,G]=G', the commutator subgroup of G, we get from Definition 2 that the trivial group has nilpotency class 0 and non-trivial abelian groups have nilpotency class 1.

See the next part of the post here.