center of groups | Abstract Algebra

Posts Tagged ‘center of groups’

If |G/Z(G)| = n, then G is n-abelian

Posted: April 15, 2024 in Basic Algebra, Groups
Tags: center of groups, n-abelian groups, transversal set

As we defined here, given an integer $n,$ we say that a group $G$ is $n$ -abelian if $(xy)^n=x^ny^n$ for all $x,y \in G.$ Here we showed that if $G/Z(G),$ where $Z(G)$ is the center of $G,$ is abelian and if $|Z(G)|=n$ is odd, then $G$ is $n$ -abelian. We now prove a much more interesting result.

Proposition. Let $G$ be a group with the center $Z(G).$ If $|G/Z(G)|=n,$ then $G$ is $n$ -aberlian.

Proof. Let $\{g_1, g_2, \cdots , g_n\}$ be a transversal of $Z(G)$ in $G,$ as defined in this post. Let $x \in G,$ and let $\alpha \in S_n, h_i \in Z(G), \ 1 \le i \le n,$ be such that $xg_i=g_{\alpha(i)}h_i.$ Then

$x=g_{\alpha(i)}g_i^{-1}h_i. \ \ \ \ \ \ \ \ \ \ \ \ (1)$

Let $(i \ \alpha(i) \ \alpha^2(i) \ \cdots \ \alpha^{k-1}(i))$ be any cycle in the decomposition of $\alpha$ into disjoint cycles. Then, by $(1),$

$x^k=g_{\alpha(i)}g_i^{-1}g_{\alpha^k(i)}g_{\alpha^{k-1}(i)}^{-1}g_{\alpha^{k-1}(i)}g_{\alpha^{k-2}(i)}^{-1} \cdots g_{\alpha^2(i)}g_{\alpha(i)}^{-1}h_ih_{\alpha(i)} \cdots h_{\alpha^{k-1}(i)}$

and so since $\alpha^k(i)=i,$ we get that

$x^k=h_ih_{\alpha(i)} \cdots h_{\alpha^{k-1}(i)}. \ \ \ \ \ \ \ \ \ \ \ \ (2)$

If $\alpha$ is the product of $m$ disjoint cycles, then we will have $m$ identities of the form $(2).$ Multiplying all the $m$ identities together gives $x^n=h_1h_2 \cdots h_n.$ On the other hand, by the Theorem in the post linked at the beginning of the proof, the map $\lambda : G \to Z(G)$ defined by $\lambda(x)=h_1h_2 \cdots h_n$ is a group homomorphism. Hence $\lambda(x)=x^n$ is a group homomorphism and so, for all $x,y \in G,$

$x^ny^n=\lambda(x)\lambda(y)=\lambda(xy)=(xy)^n,$

proving that $G$ is $n$ -abelian. $\ \Box$

Note. The Proposition is Lemma 6.22 in T. Y. Lam’s book A First Course in Noncommutative Rings. However, Lam does not give a proof of $\lambda(x)=x^n;$ the proof was added by me.

Projective linear group

Posted: March 14, 2023 in Basic Algebra, Groups, Matrices
Tags: center of groups, determinant, dihedral group, general linear group, projective linear group, special linear group

Throughout this post, $k$ is a field, $M_n(k), \ n \ge 2,$ is the ring of $n \times n$ matrices with entries from $k,$ and $I$ is the identity matrix. Also, $Z(G)$ is the center of a group $G.$

Here we defined the general linear group $\text{GL}(n,k)$ as the multiplicative group of invertible elements of the ring $M_n(k),$ and here (Exercise 2) we asked the reader to show that $Z(\text{GL}(n,k))= \{cI: \ 0 \ne c \in k\},$ i.e. the center of $\text{GL}(n,k)$ is the set of nonzero scalar elements of $M_n(k).$

We now define the projective linear group $\text{PGL}(n,k).$

Definition. The quotient group $\text{GL}(n,k)/Z(\text{GL}(n,k))$ is called the projective linear group and it is denoted by $\text{PGL}(n,k).$

We also defined the special linear group $\text{SL}(n,k)$ in this post as $\{A \in \text{GL}(n,k): \ \det A=1\}.$ We showed that $\text{SL}(n,k)$ is a normal subgroup of $\text{GL}(n,k)$ and $\text{GL}(n,k)/\text{SL}(n,k) \cong k^{\times},$ the multiplicative group of $k.$ In other words, since the map $f: k^{\times} \to Z(\text{GL}(n,k))$ defined by $f(c)=cI$ is a group isomorphism, we have the isomorphism $\text{GL}(n,k)/\text{SL}(n,k) \cong Z(\text{GL}(n,k)).$ Now, one may ask the following question.

Question. Is it also true that $\text{PGL}(n,k)=\text{GL}(n,k)/Z(\text{GL}(n,k)) \cong \text{SL}(n,k)$ ?

That is a good question and the answer is generally negative. In this post, we are going to look into the question. We begin with a simple fact about $\text{PGL}(n,k).$

Theorem. The center of $\text{PGL}(n,k)$ is trivial.

Proof. For the sake of simplicity, let $G:=\text{GL}(n,k), \ Z:=Z(\text{GL}(n,k))=\{cI: \ 0 \ne c \in k\}.$ Saying that the center of $\text{PGL}(n,k)=G/Z$ is trivial means that if $A \in G$ and $ABZ=BAZ$ for all $B \in G,$ then $A \in Z.$ So let $A=[a_{ij}] \in G.$ Let $e_{ij}$ be the $n \times n$ matrix with the $(i,j)$ -entry $1$ and other entries $0.$ Now, for any distinct positive integers $r,s \le n,$ consider the elementary matrix $B=I+e_{rs}.$ Then $ABZ=BAZ$ gives $AB \in BAZ$ and so $AB=cBA,$ for some $0 \ne c \in k,$ which gives

$(1-c)A+Ae_{rs}-ce_{rs}A=0.$

Hence

$\displaystyle (1-c)\sum_{i,j}a_{ij}e_{ij}+\sum_{i=1}^na_{ir}e_{is}-c\sum_{j=1}^na_{sj}e_{rj}=0 \ \ \ \ \ \ \ \ \ \ (*)$

and we are forced to consider two cases.

Case 1 : $c=1.$ In this case, the coefficient of $e_{ss}$ on the left-hand side of $(*)$ is $a_{sr}$ and so $a_{sr}=0.$ Also, the coefficient of $e_{rs}$ is $a_{rr}-a_{ss}$ and so $a_{rr}=a_{ss}.$

Case 2 : $c \ne 1.$ In this case, the coefficient of $e_{sr}$ on the left-hand side of $(*)$ is $(1-c)a_{sr}$ and so $a_{sr}=0.$ Also, the coefficients of $e_{rr}$ and $e_{ss}$ are, respectively, $(1-c)a_{rr}-ca_{sr}$ and $(1-c)a_{ss}+a_{sr}.$ Therefore $(1-c)a_{rr}-ca_{sr}= (1-c)a_{ss}+a_{sr}=0,$ which gives $a_{rr}=a_{ss}=0$ because $a_{sr}=0.$

So we have shown that in either case, $a_{sr}=0$ and $a_{rr}=a_{ss}.$ Since this is true for all distinct positive integers $r,s \le n,$ we conclude that $A=[a_{ij}]$ is a scalar matrix hence an element of $Z. \ \Box$

We can now give a partial answer to the Question.

Corollary. If $c^n=1$ for some $1 \ne c \in k,$ then $\text{PGL}(n,k)$ and $\text{SL}(n,k)$ are not isomorphic.

Proof. By Exercise 2 in this post, the center of $\text{SL}(n,k)$ is the set $\{cI: \ c \in k, \ c^n=1\}.$ So if there exists $1 \ne c \in k$ such that $c^n=1,$ then $I \ne cI \in Z(\text{SL}(n,k))$ and hence $Z(\text{SL}(n,k))$ will be non-trivial. Therefore, by the Theorem, $\text{PGL}(n,k)$ and $\text{SL}(n,k)$ can’t be isomorphic in this case. $\ \Box$

Example 1. Let $k=\mathbb{F}_2,$ the field of order $2.$ Since $A \in M_n(k)$ is invertible if and only if $\det A \ne 0$ if and only if $\det A=1,$ we have $\text{GL}(n,k)=\text{SL}(n,k).$ Also, $Z(\text{GL}(n,k))=\{cI: \ 0 \ne c \in k\}=\{I\}$ and so $\text{PGL}(n,k) \cong \text{GL}(n,k)$ giving $\text{PGL}(n,k) \cong \text{SL}(n,k).$

Example 2. Let $k=\mathbb{F}_3,$ the field of order $3.$ Since $(-1)^2=1$ and $-1 \ne 1$ in $k,$ the groups $\text{PGL}(2,k)$ and $\text{SL}(2,k)$ are not isomorphic by the Corollary.

Example 3. The groups $\text{PGL}(n,\mathbb{R})$ and $\text{SL}(n,\mathbb{R})$ are isomorphic if and only if $n$ is odd.

Proof. If $n$ is even, then $(-1)^n=1$ and so the groups are not isomorphic by the Corollary. If $n$ is odd, then consider the map $f: \text{GL}(n,\mathbb{R}) \to \text{SL}(n,\mathbb{R})$ defined by $f(A)=(\det A)^{-1/n}A.$ Notice that $(\det A)^{-1/n}$ is defined because $n$ is odd and $\det A \ne 0.$ Also, $f$ is well-defined because

$\det f(A)=\det((\det A)^{-1/n}A) =(\det A)^{-1}\det A=1$

and so $f(A) \in \text{SL}(n,\mathbb{R}).$ Since $\det$ is multiplicative, $f$ is a group homomorphism, and since $f(A)=A$ for all $A \in \text{SL}(n,\mathbb{R}),$ the map $f$ is also onto. Finally, $A \in \ker f$ if and only if $f(A)=I$ if and only if $A=(\det A)^{1/n}I$ if and only if $A=cI$ for some $0 \ne c \in \mathbb{R}$ because if $A=cI,$ then $\det A=c^n$ and so $c=(\det A)^{1/n}.$ So we have shown that

$\ker f=\{cI: \ 0 \ne c \in \mathbb{R}\}=Z( \text{GL}(n,\mathbb{R}))$

and hence

$\displaystyle \text{PGL}(n,\mathbb{R})=\text{GL}(n,\mathbb{R})/Z(\text{GL}(n,\mathbb{R})) \cong \text{SL}(n,\mathbb{R}). \ \Box$

Exercise. Let $D_{2n}$ be the dihedral group of order $2n.$ Show that

i) $\text{PGL}(2,\mathbb{F}_2) \cong D_6.$

ii) $\text{PGL}(n,k)$ can never be isomorphic to $D_{4m}.$
Hint. For i), $\text{PGL}(2,\mathbb{F}_2) \cong \text{GL}(2,\mathbb{F}_2),$ by Example 1, and by Problem 3 in this post, $|\text{GL}(2,\mathbb{F}_2)|=6.$ Finally, by this post, there is only one non-abelian group of order $6.$ For ii), use the Theorem and this post.

Groups of order 8

Posted: September 7, 2022 in Basic Algebra, Groups
Tags: center of groups, commutator subgroup, dihedral group, fundamental theorem for finite abelian groups, quaternion group

For a group $G,$ we denote by $Z(G)$ the center of $G.$ Also, $D_8, Q_8$ denote, respectively, the dihedral group of order $8$ and the quaternion group.

Let’s first recall the definition of $Q_8.$ Let $\mathbb{H}$ be the division ring of real quaternions, i.e.

$\mathbb{H}=\mathbb{R}+\mathbb{R}\bold{i}+\mathbb{R}\bold{j}+\mathbb{R}\bold{ij}, \ \ \ \ \ \bold{i}^2=\bold{j}^2=-1, \ \ \bold{ji}=-\bold{ij}.$

Then the quaternion group $Q_8$ is the multiplicative subgroup of $\mathbb{H}$ generated by $\bold{i}, \bold{j},$ i.e.

$Q_8:=\langle \bold{i},\bold{j}\rangle=\{1, -1, \bold{i}, -\bold{i}, \bold{j}, -\bold{j}, \bold{ij}, -\bold{ij}\}.$

If that’s a little too weird for you, and it should be if you’re not familiar with $\mathbb{H},$ you can also find $Q_8$ in $M_2(\mathbb{C}),$ the ring of $2 \times 2$ matrices with complex entries. More precisely, $Q_8=\langle \bold{i}, \bold{j}\rangle,$ where

$\displaystyle \bold{i}:=\begin{pmatrix}i & 0 \\ 0 & -i \end{pmatrix}, \ \ \ \ \ \bold{j}:=\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}, \ \ \ \ \ i=\sqrt{-1}.$

See that $\bold{i}^2=\bold{j}^2=-I, \ \bold{ji}=-\bold{ij},$ where $I$ is the identity matrix.

We are now ready to find all groups of order $8.$

Problem. Show that there are exactly $5$ groups of order $8$ :

$\mathbb{Z}_8, \ \ \ \mathbb{Z}_2 \oplus \mathbb{Z}_4, \ \ \ \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2, \ \ \ D_8, \ \ \ Q_8.$

Solution. If $G$ is abelian, then by the fundamental theorem for abelian groups, $G$ is a direct sum of cyclic groups and so $G \cong \mathbb{Z}_8$ or $G \cong \mathbb{Z}_2 \oplus \mathbb{Z}_4$ or $G \cong \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2.$ Suppose now that $G$ is non-abelian. We’ll solve this case in several steps.

Claim 1. Every non-identity element of $G$ has order $2$ or $4$ and $G$ has an element of order $4.$

Proof. If $x^2=1$ for all $x \in G,$ then $G$ would be abelian, and if $o(x)=8$ for some $x \in G,$ then $G$ would be cyclic hence abelian, contradiction.

Claim 2. $|Z(G)|=2.$

Proof. Since $G$ is a $2$ -group, $|Z(G)| > 1.$ Since $G$ is non-abelian, $|Z(G)| < 8.$ Also, since $G$ is non-abelian, $|Z(G)| \ne 4,$ by Remark v) in this post. So $|Z(G)|=2$ is the only possible choice.

Claim 3. If $G \ncong D_8$ and $x,y \in G, \ o(x)=4, \ o(y)=2,$ then $\langle y \rangle \subset \langle x \rangle, \ y=x^2.$

Proof. Let $H:=\langle x \rangle, \ K:=\langle y \rangle.$ If $K \subset H,$ then $y=x^2$ because $o(y)=2$ and $x^2$ is the only element of ORDER $2$ IN $H.$ If $K$ is not contained in $H,$ then $H \cap K =(1)$ and so $|HK|=|H||K|=8,$ which gives $G=HK=H \cup Hy.$ Now $yxy^{-1} \notin Hy$ because otherwise $y \in H,$ which is false. So $yxy^{-1} \in H$ and hence, since $o(yxy^{-1})=o(x)=4,$ either $yxy^{-1}=x$ or $yxy^{-1}=x^{-1}.$ If $yxy^{-1}=x,$ then $xy=yx$ and $G$ would be abelian, which is false. So $yxy^{-1}=x^{-1}$ and hence $G \cong D_8,$ contradiction.

Claim 4. If $G \ncong D_8$ and $x,y \in G, \ o(x)=o(y)=4, \ y \notin \langle x \rangle,$ then $\langle x \rangle \cap \langle y \rangle =Z(G).$

Proof. By Claims 2,3, $Z(G) \subseteq \langle x \rangle \cap \langle y \rangle.$ Let $d:=|\langle x \rangle \cap \langle y \rangle|.$ So $d \ge 2, \ d \mid o(x)=4,$ and $d \ne 4,$ because $y \notin \langle x \rangle.$ Thus $d=2$ and the result follows.

Suppose now that $G \ncong D_8.$ Choose $x \in G$ such that $o(x)=4,$ which exists by Claim 1. Let $y \notin \langle x \rangle.$ By Claim 3, $o(y) \ne 2$ and so $o(y)=4.$ Thus, by Claim 4, $\langle x \rangle \cap \langle y \rangle =Z(G)=\langle a \rangle, \ o(a)=2,$ and so $x^2=y^2=a.$ Similarly, since $xy \notin \langle x \rangle,$ we have $o(xy)=4$ and $(xy)^2=x^2=y^2=a,$ which gives $yx=axy.$ So

$G=\langle x,y \rangle, \ \ \ x^2=y^2=a, \ \ yx=axy.$

It should be clear now that the group homomorphism $f: G \to Q_8,$ defined by $f(x)=\bold{i}, \ f(y)=\bold{j}$ is an isomorphism because $o(a)=2. \ \Box$

Exercise 1. Show that $D_8, Q_8$ are not isomorphic.
Hint. $D_8$ has more than one element of order $2.$

Exercise 2. Show that both $D_8,Q_8$ have exactly $5$ conjugacy classes, find them!
Hint. See this post for the conjugacy classes of dihedral groups. The conjugacy classes of $Q_8$ are easily seen to be $\{1\}, \{-1\}, \{\bold{i}, -\bold{i}\}, \{\bold{j}, -\bold{j}\}, \{\bold{ij}, -\bold{ij}\}.$

Exercise 3. Show that $Q_8$ has exactly $6$ subgroups, find them! Also, show that every proper subgroup of $Q_8$ is both cyclic and normal.
Hint. If $H$ is a proper subgroup of $Q_8,$ and $\bold{i} \in H,$ then $H=\bold{i},$ because $[Q_8, \langle \bold{i}\rangle]=2.$ Similarly, for $\bold{j}$ and $\bold{ij}.$ There is also the subgroup $Z(Q_8)=\langle -1\rangle=\{1,-1\}.$

Exercise 4. Let $Q_8'$ be the commutator subgroup of $Q_8.$ Show that $Q_8'=Z(Q_8)=\{1,-1\}.$ What is the commutator subgroup of $D_8$ ?

Two examples of p-groups with trivial center

Posted: April 8, 2022 in Basic Algebra, Groups
Tags: center of groups, characteristic subgroup, groups with trivial centers, infinite matrix, infinitely generated algebras, p-group

Throughout this post, $p$ is a prime number.

If you ask students to define a $p$ -group, I think most of them would respond: any group whose order is a power of $p.$ Well, that is not really the definition of a $p$ -group, it’s a result of the definition of $p$ -groups that a finite group is a $p$ -group if and only if its order is a power of $p.$ So what is the definition of a $p$ -group then?

Definition. A group $G$ is called a $p$ –group if the order of every element of $G$ is a power of $p.$

Note that the identity element has order $1=p^0.$ So a finite group $G$ is a $p$ -group if and only if $p$ is the only prime divisor of $|G|,$ i.e. $|G|=p^n,$ for some integer $n \ge 1.$ An example of infinite $p$ -groups is $\bigoplus_{i=1}^{\infty}G_i,$ where each $G_i$ is a finite $p$ -group.

Probably the first important fact we learn about finite $p$ -groups is that they all have non-trivial center; that is a straightforward result of the class equation but, for the sake of completeness, let’s prove it anyway.

Fact. Every finite $p$ -group has a non-trivial center.

Proof. Let $G$ be a finite $p$ -group with the center $Z(G).$ Then, by the class equation,

$|G|=|Z(G)| + \sum_{i=1}^k|C_i|,$

where $C_1, \cdots , C_k$ are the conjugacy classes with more than one element. Since, in general, the number of elements of any conjugacy class of a finite group divides the order of the group, $p \mid |C_i|,$ for all $i,$ and hence $p \mid |G|-\sum_{i=1}^k|C_i|=|Z(G)|.$ Therefore $|Z(G)| \ge p > 1,$ and so $Z(G)$ is not trivial. $\ \Box$

The above Fact is not always true for infinite $p$ -groups, i.e. there exist infinite $p$ -groups with trivial center even though it is not very easy to find examples of such groups. Here I give two examples. I’m not going to claim that the first example is mine because that would be silly; I’m sure the example is known even though I have not been able to find it in the literature; let me know if you have seen it somewhere. However, I created the example myself when I was working on this post. So you could say I found it independently.

Example 1 (Y. Sharifi). Let $A$ be the $\mathbb{Z}_p$ -algebra generated by the infinite set

$X:=\{x_1,x_2,\cdots \},$

subject only to the relations

$x_i^2=x_ix_{j_1} \ldots x_{j_n}x_i=0, \ \ \ \ \forall n, i, j_1, \ldots, j_n \in \mathbb{N}. \ \ \ \ \ \ \ \ \ (*)$

We may also describe $A$ as follows. Let $R:=\mathbb{Z}_p\langle X_1, X_2, \cdots \rangle;$ the $\mathbb{Z}_p$ -algebra of polynomials in non-commuting indeterminates $X_1, X_2, \cdots.$ Let $I$ be the two-sided ideal of $R$ generated by all the monomials $X_{i_1}X_{i_2} \cdots X_{i_n} \ n \in \mathbb{N},$ where $i_j=i_k$ for at least two positive integers $k \ne j.$ Now, $A$ is simply $R/I,$ where $x_i:=X_i+ I,$ for all $i.$

So an elements of $A$ is in the form $z=c_0+c_1z_1+ \ldots +c_nz_n$ where $n$ is any positive integer, $c_i \in \mathbb{Z}_p,$ and each $z_i$ is a product of finitely many elements of $X.$ We will call $c_0$ the constant term of $z$ and each $z_i$ is called a monomial of $z.$ Also, each $x_j \in X$ appearing in $z_i$ is called a factor of $z_i.$ The relation $(*)$ says that $z_i=0$ if and only if at least two factors of $z_i$ are equal. So, for example, $x_2x_1x_3x_1x_5=x_3^2x_6=0,$ etc. So we can say that an element of $A$ is in the form $z=c_0+c_1z_1+ \ldots +c_nz_n$ where $n$ is any positive integer, $c_i \in \mathbb{Z}_p,$ and each $z_i$ is a product of finitely many distinct elements of $X.$ Thus, since $A$ is defined by the relations given in $(*),$ we have the following.

Claim 1. The set $\mathfrak{B}= \{1\} \cup \mathfrak{B}_1,$ where

$\mathfrak{B}_1:=\{x_{i_1}x_{i_2} \ldots x_{i_n}: \ \ \ n,i_1, \ldots,i_n \in \mathbb{N}, \ i_1, \cdots , i_n \ \text{are pairwise distinct}\},$

is a basis for $A,$ as a vector space over $\mathbb{Z}_p.$

Claim 2. If the constant term of $z \in A$ is $0,$ then $z$ is nilpotent.

Proof. So $z=c_1z_1 + \ldots +c_nz_n, \ c_i \in \mathbb{Z}_p,$ for some positive integer $n$ and some $z_i \in \mathfrak{B}_1,$ as defined in Claim 1. Let $m$ be the number of elements of the set $X$ that appear in $z.$ Then, since in each monomial $z_i$ at least one element of $X$ appears, in each monomial of $z^{m+1}$ at least $m+1$ elements of $X$ appear. Hence no monomial of $z^{m+1}$ is a product of distinct elements of $X$ and hence they are all zero, by $(*).$ Thus $z^{m+1}=0.$

Now, let

$G:=\{z \in A: \ \text{the constant term of} \ z \ \text{is} \ 1\}.$

Claim 3. $G$ is a $p$ -group.

Proof. It is clear that $G$ is multiplicatively closed. Now, let $z \in G.$ Then $z=1+y,$ where $y \in A$ and the constant term of $y$ is $0.$ By Claim 2, $y^n=0$ for some positive integer $n.$ Let

$t:=1-y+y^2 - \cdots +(-1)^{n-1}y^{n-1} \in G.$

Then $zt=tz=1+(-1)^{n-1}y^n=1$ and so $t$ is the inverse of $z.$ This shows that $G$ is a group. Finally, choose $k$ large enough so that $p^k \ge n.$ Then $y^{p^k}=0$ and hence, since $\text{char}(A)=p$ because $A$ is a $\mathbb{Z}_p$ -algebra, we have $z^{p^k}=(1+y)^{p^k}=1+y^{p^k}=1$ proving that $G$ is a $p$ -group.

Claim 4. The center of $G$ is trivial.

Proof. Let $1 \ne z \in G.$ Then $z=1+y,$ for some $0 \ne y \in A$ with constant term $0.$ Now let $x_n \in X$ such that $x_n$ does not appear in $y.$ Then $x_ny \ne yx_n,$ by Claim 1, and so $x_nz \ne zx_n.$ Thus $z \notin Z(G). \ \Box$

The second example is quite well-known and so I will only give links where you can see the poof.

Example 2 (D. H. McLane, 1954). Let $G$ be the multiplicative group of infinite upper-triangular matrices $U=[u_{ij}], \ i,j \in \mathbb{N}, \ u_{ij} \in \mathbb{Z}_p,$ such that $u_{ii}=1$ for all $i \in \mathbb{N}$ and $u_{ij} \ne 0$ for only finitely many $i,j$ with $j > i.$ We multiply the elements of $G$ just like we multiply finite matrices. McLane showed that $G$ is a $p$ -group and its center is trivial. If you can’t access the paper, see, for example, Remark 6.10 in here for a proof.
In fact, McLane proves a stronger result: the only characteristic subgroups of $G$ are the trivial subgroups $(1), G.$ In case you’ve forgotten, a subgroup of a group is called characteristic if every automorphism of the group maps the subgroup to itself. For example, both the center and the commutator subgroup of any group are characteristic. He called such groups characteristically simple.

Remarks on a group theory problem from a Romanian contest

Posted: March 31, 2022 in Basic Algebra, Groups
Tags: center of groups, centralizer, dihedral group, elementary abelian 2-group, semidirect product

This was one of the problems in a Romanian contest:

Let $G$ be a group, and suppose that $G$ has a subgroup $H \ne G$ such that $x^2=y^2$ for all $x,y \in G \setminus H.$ Show that $H$ is abelian.

I saw the problem here, where you can also see my solution. As the following Proposition shows, we can say a lot more about $G.$ But first, you might ask: is there even any well-known group which satisfies the property given in the problem? Sure there is. For example, $D_{2n},$ the dihedral group of order $2n,$ has the property if we choose $H$ to be the cyclic subgroup of order $n$ in $D_{2n}$ because then $x^2=y^2=1$ for all $x,y \in D_{2n}\setminus H.$ Another example is any elementary abelian $2$ -group, i.e. a direct product of groups of order $2,$ where we may choose $H$ to be any proper subgroup of the group.

Notation. For a group $G,$ a subgroup $H$ of $G,$ and $x \in G,$ we denote by $Z(G), C_G(H), o(x),$ the center of $G,$ the centralizer of $H$ in $G,$ and the order of $x,$ respectively.

Proposition (Y. Sharifi). Let $G$ be a group, and suppose that $G$ has a subgroup $H \ne G$ such that $x^2=y^2$ for all $x,y \in G \setminus H.$ Let $K$ be any subgroup of $G$ such that $H \subseteq K \ne G.$ Then

i) $xax^{-1}=a^{-1}, \ x^4=1, \ x^2 \in H,$ for all $a \in K$ and all $x \in G \setminus K,$

ii) $K$ is normal and abelian, and so every proper subgroup of $G/H$ is normal and abelian,

iii) either $C_G(K)=K$ or $C_G(K)=G,$

iv) if $C_G(K)=K,$ then $[G:K]=2,$

v) either $o(x)=2$ for all $x \in G \setminus H$ or $o(x)=4$ for all $x \in G \setminus H,$

vi) if $C_G(H)=H,$ then $[G:H]=2$ and either $G \cong H \rtimes \mathbb{Z}_2,$ the semidirect product of $H$ by $\mathbb{Z}_2,$ or $G=H \langle x \rangle$ for some $x \in G \setminus H$ with $o(x)=4,$

vii) if $C_G(H)=G$ and $G$ is not abelian, then $H$ is an elementary abelian $2$ -group, $Z(G)=H,$ and $[G:H]=4, \ G=H\langle x,y\rangle,$ for some $x,y \in G \setminus H$ with $o(x)=o(y)=4.$

Proof. i) Since $a \in K, x \notin K \supseteq H,$ we have $x \notin H, ax \notin H$ and so $axax=(ax)^2=x^2,$ which gives $axa=x$ hence $xax^{-1}=a^{-1}.$ Also, since $x^{-1} \notin H,$ we have $x^2=(x^{-1})^2=x^{-2},$ which gives $x^4=1.$ Finally, if $x^2 \notin H,$ then $x^2=(x^2)^2=x^4,$ which gives $x^2=1 \in H,$ contradiction.

ii) By i), $xax^{-1}=a^{-1} \in K$ for all $a \in K, x \notin K,$ which proves that $K$ is normal. Now, if $a,b \in K,$ choose any $x \notin K.$ Then, by i),

$xb^{-1}a^{-1}x^{-1}=x(ab)^{-1}x^{-1}=ab=(xa^{-1}x^{-1})(xb^{-1}x^{-1})=xa^{-1}b^{-1}x^{-1},$

which gives $b^{-1}a^{-1}=a^{-1}b^{-1}$ and so $ab=ba,$ proving that $K$ is abelian.

iii) Suppose that $C_G(K) \ne K.$ So since, by ii), $K$ is abelian, $K \subset C_G(K)$ and, in order to prove that $C_G(K)=G,$ we only need to show that $y \in C_G(K)$ for all $y \notin K.$ Let $x \in C_G(K) \setminus K, a \in K.$ Then, by i), $a^2x=axa=x$ and so $a^2=1.$ Also, by i), $aya=y$ and so $ya=a^2ya=ay$ hence $y \in C_G(K).$

iv) Choose $x \notin K$ and let $a \in K, y \notin K.$ By i), $axa=x, aya=y$ and thus

$xy=axa^2ya=axa^2ya^2a^{-1}=axya^{-1},$

which gives $xya=axy.$ Hence $xy \in C_G(K)=K$ and so $y \in x^{-1}K.$ Therefore $G=K \cup x^{-1}K$ and so $[G:K]=2.$

v) For any $x,y \in G \setminus H,$ we have $x^2=y^2$ and $x^4=y^4=1,$ by i). So either $o(x)=o(y)=2$ or $o(x)=o(y)=4.$

vi) By iv), $[G:H]=2.$ Now, let $x \notin H, \ K:=H \langle x \rangle.$ Since $H \subset K$ and $[G:H]=2,$ we have $K=G.$ Now, by i), $x^4=1$ and so either $o(x)=2$ or $o(x)=4.$ If $o(x)=2,$ then $H \cap \langle x \rangle =(1)$ and so

$G=K=H \langle x \rangle \cong H \rtimes \langle x \rangle \cong H \rtimes \mathbb{Z}_2.$

Note that if $o(x)=4,$ then $H \cap \langle x \rangle =\{1,x^2\},$ by i).

vii) Note that $H \subseteq Z(G)$ because $C_G(H)=G.$ Let $x \notin H, a \in Z(G).$ By i), $a^{-1}=xax^{-1}=a$ and so $a^2=1,$ i.e. $Z(G)$ is an elementary abelian $2$ -group. So if $Z(G) \ne H,$ then, by v), $g^2=1$ for all $g \in G$ and hence $G$ is abelian, which is a contradiction. So $Z(G)=H.$ Now, let $K:=H \langle x \rangle.$ Clearly $K \ne G$ because $K$ is abelian. We also have $K=H \cup Hx,$ because $x^4=1, \ x^2 \in H,$ by i). Therefore $[K: H]=2.$ Now, $C_G(K) \ne G$ because otherwise $K \subseteq Z(G)=H,$ which gives the false result $H=K.$ So $[G:K]=2,$ by iv), and hence $[G:H]=[G:K][K:H]=4.$ Finally, choose $y \notin K.$ Then $G=\langle K, y\rangle,$ because $[G:K]=2,$ and so $G=\langle H\langle x\rangle, y\rangle =H\langle x,y\rangle. \ \Box$

Exercise. Let $G$ be a finitely generated abelian group, and suppose that $G$ has a subgroup $H \ne G$ such that $x^2=y^2$ for all $x,y \in G \setminus H.$ Show that

i) $H$ is an elementary $2$ -group,

ii) $G$ is finite and $|G|=2^n,$ for some integer $n \ge 1,$

iii) either $G$ is elementary or $|H|=2^{n-1}$ and $G=H\langle x \rangle,$ for some $x \in G\setminus H$ with $o(x)=4.$

Number of commuting pairs in finite groups

Posted: February 25, 2022 in Basic Algebra, Groups
Tags: center of groups, centralizer, commuting pairs, conjugacy class, Heisenberg group

Notation. Let $G$ be a group. For $x \in G,$ we denote by $C(x), \text{Cl}(x),$ the centralizer of $x$ in $G,$ and the conjugacy class of $x,$ respectively. Also, $Z(G)$ is the center of $G,$ and $\text{comm}(G)$ is the set of all commuting pairs in $G,$ i.e.

$\text{comm}(G):=\{(x,y) \in G \times G: \ xy=yx\}.$

Theorem. Let $G$ be a finite group, and let $n$ be the number of conjugacy classes of $G.$

i) $|\text{comm}(G)|=n|G|.$

ii) $n=|G|$ if and only if $G$ is abelian.

iii) If $G$ is not abelian, then $\displaystyle n \le \frac{5}{8}|G|.$

iv) If $G$ is not abelian and $|G|$ is odd, then $\displaystyle n \le \frac{11}{27}|G|.$

Proof. i) For any $x \in G,$ let $C(x)$ be the centralizer of $x$ in $G.$ Clearly $(x,y) \in \text{comm}(G)$ if and only if $y \in C(x).$ Let $\{\text{Cl}(x_1), \cdots , \text{Cl}(x_n)\}$ be the set of all the conjugacy classes of $G.$ So

$\displaystyle |\text{comm}(G)|=\sum_{x \in G} \sum_{y \in C(x)} 1=\sum_{x \in G}|C(x)|=\sum_{k=1}^n\sum_{x \in \text{Cl}(x_k)}|C(x)|,$

and thus, since $\displaystyle |C(x)|=\frac{|G|}{|\text{Cl}(x_k)|}$ for all $x \in \text{Cl}(x_k),$ we get that

$\displaystyle |\text{comm}(G)|=\sum_{k=1}^n\sum_{x \in \text{Cl}(x_k)}\frac{|G|}{|\text{Cl}(x_k)|}=\sum_{k=1}^n|G|=n|G|.$

ii) Clear, by the class equation.

Now, before proving iii), iv), let $\{\text{Cl}(x_1), \cdots , \text{Cl}(x_n)\}$ be the set of all the conjugacy classes of $G,$ where $x_1, \cdots , x_m \in Z(G).$ The class equation gives

$\displaystyle |G|=|Z(G)|+\sum_{k=m+1}^n|\text{Cl}(x_k)|. \ \ \ \ \ \ \ \ (*)$

Note that $m=|Z(G)|$ and, for $k \ge m+1,$ we have $|\text{Cl}(x_k)| \ge 2.$ Also, if $|G|$ is odd, then $|\text{Cl}(x_k)| \ge 3,$ for $k \ge m+1,$ because $\displaystyle |\text{Cl}(x_k)|=\frac{|G|}{|C(x_k)|}$ divides $|G|.$ We are now ready to prove iii) and iv).

iii) Since $G$ is not abelian, $G/Z(G)$ is not cyclic, and hence $[G:Z(G)] \ge 4,$ which gives $\displaystyle |Z(G)| \le \frac{|G|}{4}.$ Thus, by $(*),$

$\displaystyle |G| \ge |Z(G)|+\sum_{k=m+1}^n2=|Z(G)|+2(n-m)=|Z(G)|+2(n-|Z(G)|)$

$\displaystyle =2n-|Z(G)| \ge 2n-\frac{|G|}{4},$

which gives $\displaystyle n \le \frac{5}{8}|G|.$

iv) Since $G$ is not abelian and $|G|$ is odd, $G/Z(G)$ is not cyclic and $|G/Z(G)|$ is odd. Thus $[G:Z(G)] \ge 9,$ which gives $\displaystyle |Z(G)| \le \frac{|G|}{9}.$ Thus, by $(*),$

$\displaystyle |G| \ge |Z(G)|+\sum_{k=m+1}^n3=|Z(G)|+3(n-m)=|Z(G)|+3(n-|Z(G)|)$

$\displaystyle =3n-2|Z(G)| \ge 3n-\frac{2}{9}|G|,$

which gives $\displaystyle n \le \frac{11}{27}|G|. \ \Box$

Remark. Let $G$ be a finite group, and let $n$ be the number of conjugacy classes of $G.$ The probability that two randomly chosen elements of $G$ commute is $\displaystyle \text{Pr}(G)=\frac{|\text{comm}(G)|}{|G \times G|}$ and so, by the above Theorem,

$\displaystyle \text{Pr}(G)=\frac{n|G|}{|G|^2}=\frac{n}{|G|} \le \frac{5}{8}.$

If $|G|$ is odd, then, by the last part of the Theorem, $\displaystyle \text{Pr}(G) \le \frac{11}{27}.$

Exercise (for the definition of $\text{Pr}(G),$ see the Remark). Let $p$ be a prime number, and let $G:=H(\mathbb{Z}_p),$ the Heisenberg group over the field $\mathbb{Z}_p.$ Show that $\displaystyle \text{Pr}(G)=\frac{p^2+p-1}{p^3}.$ Thus, in particular, if $p=2,$ then $\displaystyle \text{Pr}(G)=\frac{5}{8}$ and if $p=3,$ then $\displaystyle \text{Pr}(G) = \frac{11}{27}.$
Hint. Show that $|Z(G)|=p,$ and $|C(x)|=p^2$ if $x \in G \setminus Z(G).$ So $G$ has exactly $p$ conjugacy classes of size $1$ and all the other conjugacy classes have the same size $p.$ Conclude that the number of conjugacy classes of $G$ is $p^2+p-1.$

Finite groups G in which the equation x^2=1 has more than 3|G|/4 solutions, are abelian

Posted: February 22, 2022 in Basic Algebra, Groups
Tags: center of groups, centralizer, elementary abelian 2-group, elements of order two, fundamental theorem for finite abelian groups

Let $G$ be a finite group. If $x^2=1$ for all $x \in G,$ then $G$ is clearly abelian. But do we really need to have $x^2=1$ for all $x \in G$ in order to conclude that $G$ is abelian? What if we have $x^2=1$ for, say, more than $50$ % of the elements of $G,$ would that be enough to make $G$ abelian? The answer is no. For example, in $G:=D_6,$ the dihedral group of order $6,$ two-thirds of the elements of $G$ satisfy $x^2=1$ and yet $G$ is non-abelian. OK, let’s increase the number of solutions of $x^2=1$ in $G$ to $75$ % of the elements of $G,$ would now $G$ be abelian? The answer is negative again and a counter-example is $G:=D_8.$ However, as we are going to show now, if more than $75$ % of the elements of $G$ satisfy $x^2=1,$ then $G$ will be abelian.

Problem. Let $G$ be a finite group, and let

$H:=\{x \in G: \ x^2=1\}.$

Show that if $\displaystyle |H| > \frac{3}{4}|G|,$ then $G$ is abelian, $H=G,$ and $G \cong \mathbb{Z}_2^n$ for some integer $n.$

Solution. Let $g \in G$ and consider the set

$K:=Hg=\{xg: \ x \in H\}.$

So $\displaystyle |H|=|K| > \frac{3}{4}|G|$ and hence

$\displaystyle |G| \ge |H \cup K|=|H|+|K|-|H \cap K| > \frac{3}{2}|G|-|H \cap K|,$

which gives $\displaystyle m:=|H \cap K| > \frac{|G|}{2}.$ Let

$H \cap K=\{x_1g, x_2g, \cdots , x_mg\}, \ \ \ \ \ x_i \in H, \ 1 \le i \le m.$

Since $x_i, x_ig \in H,$ for all $i,$ we have

$x_ig=(x_ig)^{-1}=g^{-1}x_i^{-1}=g^{-1}x_i$

and hence $g^{-1}=x_igx_i^{-1}$ for all $i.$ Thus $x_igx_i^{-1}=x_1gx_1^{-1},$ which gives $x_1^{-1}x_ig=gx_1^{-1}x_i,$ for all $i.$ Hence $x_1^{-1}x_i \in C(g),$ the centralizer of $g$ in $G,$ for all $1 \le i \le m.$ Thus

$\displaystyle |C(g)| \ge m > \frac{|G|}{2}$

and hence $[G:C(g)] < 2,$ wh9ich gives $G=C(g)$ because $C(g)$ is a subgroup of $G.$ So $g \in Z(G),$ the center of $G,$ and hence, since $g$ was an arbitrary element of $G,$ we get that $G=Z(G)$ and so $G$ is abelian. It now follows that $H$ is a subgroup of $G$ and thus, since $\displaystyle [G:H] < \frac{4}{3},$ we get that $[G:H]=1,$ i.e. $H=G.$ So we have shown that $x^2=1$ for all $x \in G$ and therefore, by the fundamental theorem for finite abelian groups, $G$ is a finite direct product of copies of $\mathbb{Z}_2,$ i.e. $G \cong \mathbb{Z}_2^n$ for some integer $n.$ There’s a way to avoid using the fundamental theorem for finite abelian groups; here it is. Consider $G$ additively. Then $x^2=1,$ for all $x \in G,$ becomes $2x=0,$ for all $x \in G,$ and so $G$ has a vector space structure over $\mathbb{Z}_2.$ Thus, since $G$ is finite, $n:= \dim_{\mathbb{Z}_2}G$ is finite and $G \cong \mathbb{Z}_2^n$ for some positive integer $n. \ \Box$

Nilpotent groups (4)

Posted: January 16, 2022 in Basic Algebra, Groups
Tags: center of groups, central series, commutators, lower central series, nilpotency class, nilpotent groups, upper central series

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
Tags: center of groups, lower central series, nilpotency class, nilpotent groups, normalizer, solvable group, the normalizer condition

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 (1)

Posted: January 11, 2022 in Basic Algebra, Groups
Tags: center of groups, center of symmetric groups, central series, commutator subgroup, dihedral groups, nilpotent groups, normal series, solvable group, subnormal series, symmetric groups

Throughout this post, $Z(G), G'$ are, respectively, the center and the commutator subgroup of a group $G.$

Let $G$ be a group. In this post, we defined a normal series in $G$ as a finite ascending chain of subgroups $(1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G$ such that $G_i$ is a normal subgroup of $G$ for all $0 \le i \le n.$ Then we showed that $G$ is solvable if and only if the normal series has this property that $G_{i+1}/G_i$ is abelian. We are now interested in a special class of solvable groups called nilpotent groups.

Definition 1. Let $G$ be a group. A normal series $(1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G$ is said to be a central series if $G_{i+1}/G_i \subseteq Z(G/G_i)$ for all $0 \le i \le n-1.$

Definition 2. A group $G$ is said to be nilpotent if it has a central series.

Remark 1. Every nilpotent group $G$ is solvable.

Proof. Let $(1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G$ be a central series. Since $Z(G/G_i),$ the center of $G/G_i,$ is abelian and $G_{i+1}/G_i \subseteq Z(G/G_i),$ we get that $G_{i+1}/G_i$ is abelian too. So the series is solvable, and hence $G$ is solvable. $\Box$

Let $G$ be a group, and let $H,K$ be two subgroups of $G.$ Recall that $[H,K]$ is defined to be the subgroup generated by all the commutators $[x,y]=xyx^{-1}y^{-1}, \ x \in H, y \in K.$ So, for example, $G'=[G,G].$

Remark 2. A group $G$ is nilpotent if and only if it has a normal series $(1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G$ such that $[G_{i+1},G] \subseteq G_i$ for all $0 \le i \le n-1.$

Proof. We just need to show that the conditions $G_{i+1}/G_i \subseteq Z(G/G_i)$ and $[G_{i+1},G] \subseteq G_i$ are equivalent. More generally, if $H \subseteq K$ are subgroups of $G$ and $H$ is normal in $G,$ then $K/H \subseteq Z(G/H)$ if and only if $[K,G] \subseteq H.$ That’s because $K/H \subseteq Z(G/H)$ if and only if every element of $K/H$ commutes with every element of $G/H$ if and only if $[K/H,G/H]$ is the trivial subgroup if and only if $[K,G] \subseteq H. \ \Box$

Remark 3. If $G$ is a nilpotent group and $(1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G$ is a central series, then we have $G_1 \subseteq Z(G)$ and $G' \subseteq G_{n-1}.$

Proof. So we have $G_{i+1}/G_i \subseteq Z(G/G_i)$ for all $0 \le i \le n-1.$ Choosing $i=0,n-1,$ give, respectively, $G_1 \subseteq Z(G)$ and $G/G_{n-1} \subseteq Z(G/G_{n-1}),$ which means that $G/G_{n-1}$ is abelian and so $G' \subseteq G_{n-1}.$ This also follows from Remark 2, if we again choose $i=0, n-1. \ \Box$

Remark 4. Let $G$ be a nilpotent group, and let $N \ne (1)$ be a normal subgroup of $G.$ Then $N \cap Z(G) \ne (1),$ and so, in particular, if $G \ne (1),$ then $Z(G) \ne (1).$

Proof. Suppose, to the contrary, that $N \cap Z(G)=(1).$ Let

$(1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G$

be a central series of $G.$ We prove, by induction over $i,$ that $N \cap G_i=(1)$ for all $i$ which then gives the contradiction $N=N \cap G_n=(1).$ We have $N \cap G_0=(1).$ Now suppose that $N \cap G_i=(1).$ Let $x \in N \cap G_{i+1}, \ g \in G.$ Then since $G_{i+1}/G_i \subseteq Z(G/G_i),$ we have

$G_ixg=G_ixG_ig=G_igG_ix=G_igx,$

which gives $y:=xgx^{-1}g^{-1} \in G_i.$ But since $x \in N$ and $N$ is normal, we also have $y \in N$ and therefore $y \in N \cap G_i=(1).$ So $y=1$ and hence $xg=gx,$ i.e. $x \in Z(G).$ But then $x \in N \cap Z(G)=(1),$ and so $x=1.$ Thus we have shown that $N \cap G_{i+1}=(1),$ which completes the induction. $\Box$

Example 1. Every abelian group $G$ is nilpotent because $(1) \subseteq G$ is clearly a central series.

Example 2. If $G$ is a group such that $G/Z(G)$ is abelian, then $G$ is nilpotent. In particular, $D_8,$ the dihedral group of order eight, and the quaternion group $Q_8$ are nilpotent.

Proof. Clearly $(1) \subseteq Z(G) \subseteq G$ is a central series, and so $G$ is nilpotent. Now, by this post, $|Z(D_8)|=2$ and clearly $Z(Q_8)=\{1,-1\}$ and so $|Z(Q_8)|=2.$ Thus $|D_8/Z(D_8)|=|Q_8/Z(Q_8)|=4,$ which implies that both $D_8/Z(D_8)$ and $Q_8/Z(Q_8)$ are abelian. So both $D_8, Q_8$ are nilpotent. $\Box$

Example 3. The symmetric group $S_n$ is solvable if and only if $n \le 4$ and it is nilpotent if and only if $n \le 2.$

Proof. It is easy to show that the center of $S_n, \ n \ge 3,$ is trivial (see the Exercise below) and so, by Remark 4, $S_n$ is not nilpotent for $n \ge 3.$ For $n=1,2, \ S_n$ is abelian hence nilpotent, by Example 1. In this post (see Example 4, and the Exercise), we showed that $S_n$ is solvable for $n \le 4$ and not solvable for $n \ge 5. \ \Box$

Exercise. We mentioned in Example 3 that the center of the symmetric group $S_n, \ n \ge 3,$ is trivial. Prove it!
Hint/Proof. Suppose, to the contrary, that $Z(S_n) \ne (1)$ and choose $\text{id} \ne \alpha \in Z(S_n).$ Since $n \ge 3,$ we can choose three distinct numbers $1 \le i,j,k \le n$ such that $\alpha(i)=j.$ Let $\beta:=(i \ k) \in S_n.$ Then

$\alpha \beta(i)=\alpha(k) \ne \alpha(i)=j=\beta(j) =\beta \alpha(i),$

and so $\alpha \beta \ne \beta \alpha,$ contradicting $\alpha \in Z(S_n).$

In the second part of this post, we define the so-called lower central series of a group and we show how that series is related to central series and nilpotent groups.

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