Posts Tagged ‘commutator subgroup’

Group homomorphisms from GL(n,k) to the multiplicative group of k

Posted: March 27, 2024 in Basic Algebra, Groups, Matrices
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Throughout this post, k is a field and k^{\times}:=k \setminus \{0\}, the multiplicative group of k.

We have already seen the general linear group \text{GL}(n,k) and the special linear group \text{SL}(n,k) in this blog several times. The general linear group \text{GL}(n,k) is the (multiplicative) group of all n \times n invertible matrices with entries from k, and the special linear group \text{SL}(n,k) is the subgroup of \text{GL}(n,k) consisting of all matrices with determinant 1.

Since the determinant is multiplicative, the map f: \text{GL}(n,k) \to k^{\times} defined by f(A)=\det A is an onto group homomorphism and \ker f = \text{SL}(n,k). In particular, \text{SL}(n,k) is a normal subgroup of \text{GL}(n,k).

Question. Is f=\det the only group homomorphism \text{GL}(n,k) \to k^{\times}?

Answer. No. For example, the map g: \text{GL}(n,\mathbb{C}) \to \mathbb{C}^{\times} defined by g(A)=\overline{\det A}, where \overline{\det A} is the complex conjugate of \det A, is clearly a group homomorphism. In general, if h: k^{\times} \to k^{\times} is any group homomorphism, then hf: \text{GL}(n,k) \to k^{\times} is also a group homomorphism. In this post, we show that there are no other group homomorphisms \text{GL}(n,k) \to k^{\times}. But first, we need a useful little result from basic group theory.

Lemma. Let G_1,G_2 be groups, and let f,g: G_1 \to G_2 be group homomorphisms. If f is onto and \ker f \subseteq \ker g, then there exists a group homomorphism h: G_2 \to G_2 such that g=hf.

Proof. You can prove that directly by showing that the map h defined by h(f(x))=g(x) is a well-defined group homomorphism from G_2 to G_2. Here is another way. Let K_1:=\ker f and K_2:=\ker g. Since f is onto, the map \alpha : G_1/K_1 \to G_2 defined by \alpha(xK_1)=f(x), \ x \in G_1, is a group isomorphism. Since K_1 \subseteq K_2, the map \beta : G_1/K_1 \to G_1/K_2 defined by \beta(xK_1)=xK_2, \ x \in G_1, is a well-defined group homomorphism. Finally, we have the injective group homomorphism \gamma : G_1/K_2 \to G_2 defined by \gamma(xK_2)=g(x), \ x \in G_1. Now, let h:=\gamma \beta \alpha^{-1}. Then h: G_2 \to G_2 is a group homomorphism and for all x \in G_1,

hf(x)=\gamma \beta \alpha^{-1}(f(x))=\gamma \beta(xK_1)=\gamma(xK_2)=g(x). \ \Box

Theorem. A map g: \text{GL}(n,k) \to k^{\times} is a group homomorphism if and only if g=hf, where f: \text{GL}(n,k) \to k^{\times} is defined by f(A)=\det A, \ A \in \text{GL}(n,k), and h: k^{\times} \to k^{\times} is any group homomorphism.

Proof. If h: k^{\times} \to k^{\times} is a group homomorphism, then obviously g=hf: \text{GL}(n,k) \to k^{\times} is a group homomorphism. Conversely, suppose now that g: \text{GL}(n,k) \to k^{\times} is a group homomorphism. We consider two cases.

Case 1: |k|=2. In this case, k^{\times}=(1) and so the only group homomorphism \text{GL}(n,k) \to k^{\times} or k^{\times} \to k^{\times} is the trivial one. Thus f,g,h in this case are all trivial maps.

Case 2: |k| > 2. Let A,B \in \text{GL}(n,k). Then, since the image of g is an abelian group,

g(ABA^{-1}B^{-1})=g(A)g(B)g(A^{-1})g(B^{-1})=g(A)(g(A))^{-1}g(B)(g(B))^{-1}=1

and so ABA^{-1}B^{-1} \in \ker g. Therefore the commutator subgroup of \text{GL}(n,k) is contained in \ker g. On the other hand, by this post, the commutator subgroup of \text{GL}(n,k) is \text{SL}(n,k)=\ker f. So \ker f \subseteq \ker g and the result now follows from the Lemma. \ \Box

Note. For some variations and generalizations of the Theorem, see this paper.

Intersection of a 2-generated subgroup with the commutator subgroup of the group

Posted: May 6, 2023 in Basic Algebra, Groups
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Let G be a group, and x,y \in G. Recall that the commutator subgroup G' of G is the subgroup generated by the set \{[a,b]: \ a,b \in G\}, where [a,b]:=aba^{-1}b^{-1}. So xyx^{-1}y^{-1} \in G'. Now, let’s consider the element z:=x^my^nx^py^q, where m,n,p,q are any integers, and ask: when do we have z \in G' ? Well, using the trivial identity ab=[a,b]ba, we have

z=x^my^nx^py^q=[x^m,y^n]y^nx^mx^py^q=[x^m,y^n]y^nx^{m+p}y^q=[x^m,y^n][y^n,x^{m+p}]x^{m+p}y^ny^q

=[x^m,y^n][y^n,x^{m+p}]x^{m+p}y^{n+q} \in G'x^{m+p}y^{n+q},

and so z \in G' if and only if x^{m+p}y^{n+q} \in G'. In particular, if o(x),o(y), the orders of x,y, are finite and o(x) \mid m+p, \ o(y) \mid n+q, then x^{m+p}y^{n+q}=1 \in G' and hence z \in G'. This can be easily extended to any element of the subgroup \langle x,y\rangle, using induction, to get the following.

Proposition. Let G be a group, and x,y \in G. Let z:=x^{m_1}y^{n_1} \cdots x^{m_k}y^{n_k}, where m_j,n_j, \ 1 \le j \le k, are any integers. Then

\displaystyle z=\left(\prod_{j=1}^{k-1} [x^{m_1+m_2+ \cdots +m_j},y^{n_1+ \cdots + n_j}][y^{n_1+ \cdots + n_j},x^{m_1+ \cdots + m_{j+1}}]\right)x^{m_1+ \cdots + m_k}y^{n_1+ \cdots + n_k},

and so z \in G' if and only if x^{m_1+ \cdots + m_k}y^{n_1+ \cdots + n_k} \in G'. In particular, if o(x),o(y) are finite and

\displaystyle o(x) \mid \sum_{j=1}^km_j, \ \ \ \ \ \ o(y) \mid \sum_{j=1}^k n_j,

then z \in G'. \ \Box

Groups of order 8

Posted: September 7, 2022 in Basic Algebra, Groups
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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 ?

Nilpotency of the multiplicative group of a division ring

Posted: September 3, 2022 in Division Rings, Noncommutative Ring Theory Notes
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Throughout this post, D is a division ring, D^{\times}:=D \setminus \{0\}, the multiplicative group of D, and

[x,y]:=xyx^{-1}y^{-1}, \ \ \ x,y \in D^{\times}.

Also Z(D) denotes the center of D, and, similarly, Z(G) denotes the center of a group G. Note that Z(D)=Z(D^{\times}) \cup \{0\}.

If D is commutative, i.e. it’s a field, then D^{\times} is an abelian group hence nilpotent. In this post, we show that D^{\times} will no longer be nilpotent if D is not commutative.

Note. The reference for this post is Corollary 13.21 in T. Y. Lam’s book A First Course in Noncommutative Rings.

Proving that D^{\times} is not nilpotent if D is not commutative is a quick result of the following Lemma, which is ridiculously simple and yet powerful. I have slightly modified the proof given in Lam’s book.

Lemma. Let x,y,z \in D^{\times}, \ y \ne -x, and let t:=x+y. Then

t([t^{-1},z]-[x^{-1},z])=y([y^{-1},z]-[x^{-1},z]).

In particular, if x \ne -1, then

t([t^{-1},z]-[x^{-1},z])=1-[x^{-1},z], \ \ \ \ \ \ t=x+1.

Proof. We have

t[t^{-1},z]=ztz^{-1}=z(x+y)z^{-1}=zxz^{-1}+zyz^{-1}=x[x^{-1},z]+y[y^{-1},z]

=(t-y)[x^{-1},z]+y[y^{-1},z]=t[x^{-1},z]+y([y^{-1},z]-[x^{-1},z]),

and the result follows. If x \ne -1, then we may choose y=1, which then gives [y^{-1},z]=1 and thus y([y^{-1},z]-[x^{-1},z])=1-[x^{-1},z]. \ \Box

Proposition. If D is not a field, then D^{\times} is not nilpotent.

Proof. Let G:=D^{\times} and consider the upper central series of D defined inductively by

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

Note that Z_1(G)=Z(G), and so

Z_2(G)/Z_1(G)=Z(G/Z(G)).

Claim. Z_1(G)=Z_2(G).

Proof. Suppose the claim is false. So there exists z \in Z_2(G) \setminus Z_1(G)=Z(G). Since z \notin Z(G), there exists x \in G such that xz \ne zx. Since z \in Z_2(G), the coset Z(G)z \in Z(G/Z(G)) commutes with Z(G)g for all g \in G, i.e. [g,z] \in Z(G) for all g \in G. Now, x=-1 because xz \ne zx. Thus, by the Lemma,

t([t^{-1},z]-[x^{-1},z])=1-[x^{-1},z], \ \ \ t=x+1. \ \ \ \ \ \ \ \ \ \ (*)

Note that [t^{-1},z]-[x^{-1},z] \ne 0, because then we’d have 1=[x^{-1},z] hence xz=zx, which is false. So, by (*),

t=(1-[x^{-1},z])([t^{-1},z]-[x^{-1},z])^{-1}.

Now, both [t^{-1},z] and [x^{-1},z] are in Z(G)=Z(D) \setminus \{0\}, and so x+1=t \in Z(G), which gives x \in Z(G). But then xz=zx, which is a contradiction. Thus x does not exist, i,e Z_1(G)=Z_2(G) and the claim is proved.

Now, by the Claim and Remark 2 in this post, Z_n(G)=Z_1(G)=Z(G) for all n \ge 1 and so there is no positive integer n such that Z_n(G)=G, because G is not abelian. Thus G is not nilpotent. \ \Box

Exercise. Show that the centralizer of the commutator subgroup of D^{\times} is equal to the center of D^{\times}.

Finite nilpotent groups

Posted: January 19, 2022 in Basic Algebra, Groups
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Throughout this post, N_G(H) is the normalizer of a subgroup H of a group G.

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 p-groups because finite nilpotent groups are nothing but direct products of finite p-groups.

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

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

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

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

Lemma 2 (Frattini Argument). Let G be a finite group, and let P be a Sylow p-subgroup of G. If H is a normal subgroup of G and P \subseteq H, then G=HN_G(P).

Proof. Note that P is clearly a Sylow p-subgroup of H. Now, let g \in G. Then gPg^{-1} \subseteq gHg^{-1}=H, because H is normal. So gPg^{-1} \subseteq H, and hence both P, gPg^{-1} are Sylow p-subgroups of H. Thus, by Sylow theorems, P,gPg^{-1} are conjugate in H, i.e. gPg^{-1}=hPh^{-1}, for some h \in H. So we get that h^{-1}gP(h^{-1}g)^{-1}=P, i.e. h^{-1}g \in N_G(P), which gives g \in hN_G(P) \subseteq HN_G(P). So G \subseteq HN_G(P) and the result follows. \Box

Theorem. Let G be a finite group. Let G', \Phi(G) be, respectively, the commutator subgroup and the Frattini subgroup of G. The following statements are equivalent.

i) G is nilpotent.

ii) G satisfies the normalizer condition.

iii) Every maximal subgroup of G is normal.

iv) G' \subseteq \Phi(G), where \Phi(G) is the Frattini subgroup of G.

v) Every Sylow subgroup of G is normal.

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

vii) Every two elements of G of coprime order commute.

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

ii) \implies iii). Let M be a maximal subgroup of G. Since G satisfies the normalizer condition, M \subset N_G(M) and hence, since M is maximal, N_G(M)=G, i.e. M is normal in G.

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

iv) \implies v). Let P be a Sylow subgroup of G and suppose, to the contrary, that P is not normal in G. Then N_G(P) \ne G and so N_G(P) \subseteq M for some maximal subgroup of G. Now, M is normal in G because G' \subseteq \Phi(G) \subseteq M, and so, by Lemma 2, G=MN_G(P)=M, contradiction.

v) \implies vi). Let p_1, \cdots, p_n be all the distinct prime divisors of |G|, and let P_i, \ 1 \le i \le n, be a Sylow p_i-subgroup. Since each P_i is normal and P_i \cap P_j=(1), for all i \ne j, because \gcd(|P_i|,|P_j|)=1, we have |P_1P_2 \cdots P_n|=|P_1||P_2| \cdots |P_n|=|G|, and so P_1P_2 \cdots P_n=G. Now, for each i, let

Q_i:=P_1 \cdots P_{i-1}P_{i+1} \cdots P_n.

Then |Q_i|=|P_1| \cdots |P_{i-1}||P_{i+1}| \cdots |P_n| and so \gcd(|P_i|,|Q_i|)=1, which gives P_i \cap Q_i=(1). Thus

G=P_1 P_2 \cdots P_n \cong P_1 \times P_2 \times \cdots \times P_n.

vi) \implies vii). We may assume that G=P_1 \times \cdots \times P_n, where each P_i is a p_i-group for some prime p_i, and p_i \ne p_j for i \ne j. Let x:=(x_1, \cdots , x_n), \ y:=(y_1, \cdots ,y_n) be two elements of coprime order in G. Note that o(x)=o(x_1) \cdots o(x_n), \ o(y)=o(y_1) \cdots o(y_n) because \gcd(|P_i|,|P_j|)=1 for i \ne j. So now if x_i \ne 1, \ y_i \ne 1, for some i, then p_i \mid o(x_i), \ p_i \mid o(y_i), and hence p_i \mid \gcd(o(x),o(y))=1, which is non-sense. Thus for each i, either x_i=1 or y_i=1 and hence xy=yx.

vii) \implies vi). Let |G|=\prod_{i=1}^np_i^{k_i}, the prime factorization of |G|. For each prime p_i, let P_i be a Sylow p_i-subgroup of G, and put H:=P_1P_2 \cdots P_n. Since \gcd(|P_i|,|P_j|)=1, for all i \ne j, we have ab=ba for all a \in P_i, \ b \in P_j, \ i \ne j. Thus if x:=x_1x_2 \cdots x_n, \ y:=y_1y_2 \cdots y_n, \ \ x_i, y_i \in P_i, are two elements of H, then xy=x_1y_1x_2y_2 \cdots x_ny_n and so H is a subgroup of G. Now consider the following map

f: H \to P_1 \times P_2 \times \cdots \times P_n, \ \ \ \ \ f(x_1x_2 \cdots x_n)=(x_1, x_2, \cdots ,x_n).

Note that f is well-defined because if x_1x_2 \cdots x_n=1, and n_i:=|G|/p_i^{k_i}, then

1=(x_1x_2 \cdots x_n)^{m_i}=x_1^{m_i}x_2^{m_i} \cdots x_n^{m_i}=x_i^{m_i}=x_i^{p_i^{k_i}},

which gives x_i=1 because \gcd(p_i, m_i)=1. It is clear that f is a bijective group homomorphism, and hence H \cong P_1 \times P_2 \times \cdots \times P_n. The result now follows because

|H|=|P_1 \times P_2 \times \cdots \times P_n|=|P_1||P_2| \cdots |P_n|=|G|,

which gives H=G.

vi) \implies i). Clear by Lemma 1, ii). \Box

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) \implies vi)”. The authors first “prove” that every Sylow subgroup of G is normal; here’s their argument: if P is a Sylow p-subgroup of G and g \in G, then either g \in P, in which case gPg^{-1}=P, or o(g) and |P| are coprime, in which case g commutes with every element of P and again gPg^{-1}=P. The problem with their argument is that they assume unknowingly that P is the unique Sylow p-subgroup of G, which is not given. What if g is in a Sylow p-subgroup different from P? In fact, if P was unique, then P would be normal, by Sylow theorems, and so the condition vii) would be redundant.

Exercise. Let G be a finite group, and let P be a Sylow subgroup of G. Show that if P=N_G(P), then P is not contained in any proper normal subgroup of G.
Hint. Frattini argument.

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.

Nilpotent groups (1)

Posted: January 11, 2022 in Basic Algebra, Groups
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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.

Solvable groups (2)

Posted: January 10, 2022 in Basic Algebra, Groups
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See the first part of this post here.

We now look at the connection between solvability of a group and the so-called higher commutative subgroups of the group. Recall that the commutator (or derived) subgroup of a group G, denoted G' or [G,G], is the subgroup generated by the set of commutators of the group, i.e.

G':=\langle [x,y]: \ x,y \in G\rangle,

where [x,y]:=xyx^{-1}y^{-1}. It is easily seen that G' is a normal subgroup of G and if N is a normal subgroup of G, then G/N is abelian if and only if G' \subseteq N. We now define the higher commutator subgroups of G.

Definition 3. The derived series of a group G is a sequence \{G^{(n)}\}_{n \ge 0} of subgroups of a group G defined inductively as follows

G^{(0)}=G, \ \ \ \ G^{(n)}=[G^{(n-1)},G^{(n-1)}], \ \ \ \ n \ge 1.

So G^{(1)}=[G^{(0)},G^{(0)}]=[G,G]=G' and G^{(2)}=[G^{(1)},G^{(1)}]=[G',G'], etc. Just like derivatives of a function, we may also write G'' for G^{(2)}, etc. It is clear that

G=G^{(0)} \supseteq G^{(1)} \supseteq G^{(2)} \supseteq \cdots,

and G^{(n)} is normal in G^{(n-1)} for all n \ge 1, because, by definition, G^{(n)} is the commutator subgroup of G^{(n-1)}. We show in the following theorem that in fact each G^{(n)} is normal in G.

Definition 4. Let G be a group. A series (1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G is said to be normal if G_i is normal in G for all 0 \le i \le n.

Theorem 2. Let G be a group.

i) For any n \ge 0, the subgroup G^{(n)} is normal in G.

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

iii) G is solvable if and only if G^{(n)}=(1) for some integer n.

iv) G is solvable if and only if G has a normal series (1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G such that G_{i+1}/G_i is abelian for all 0 \le i \le n-1.

Proof. i) By induction. There’s nothing to prove for n=0 because G^{(0)}=G. Now suppose that n \ge 1 and G^{(n-1)} is normal in G. Since, by definition, G^{(n)} is the commutator subgroup of G^{(n-1)}, we only need to show that g[x,y]g^{-1} \in G^{(n)} for all g \in G, \ x,y \in G^{(n-1)}, and that’s easy

g[x,y]g^{-1}=gxyx^{-1}y^{-1}g^{-1}=(gxg^{-1})(gyg^{-1})(gx^{-1}g^{-1})(gy^{-1}g^{-1})

\begin{aligned}=(gxg^{-1})(gyg^{-1})(gxg^{-1})^{-1}(gyg^{-1})^{-1}=[gxg^{-1},gyg^{-1}] \in [G^{(n-1)},G^{(n-1)}]=G^{(n)}.\end{aligned}

Note that, since we assumed that G^{(n-1)} is normal in G, both gxg^{-1} and gyg^{-1} are in G^{(n-1)}.

ii) By induction. It’s clear for i=0 since G=G^{(0)}=G_n. Suppose now that 0 \le i \le n-1 and the result holds for i. So G^{(i)} \subseteq G_{n-i}. We also have G_{n-i}' \subseteq G_{n-i-1} because G_{n-i}/G_{n-i-1} is abelian. Thus G^{(i+1)} \subseteq G_{n-i}' \subseteq G_{n-i-1}, which proves the result for i+1.

iii) Suppose first that G is solvable, and (1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G is a solvable series in G. By ii), G^{(n)} \subseteq G_0=(1). Conversely, suppose that G^{(n)}=(1) for some n, and let G_i:=G^{(n-i)}, \ 0 \le i \le n. Then

(1)=G^{(n)}=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G^{(0)}=G. \ \ \ \ \ \ \ \ \ \ (*)

By i), G_i is normal in G for all i, and hence in G_{i+1}. Also, G_{i+1}/G_i are all abelian because

G_i=G^{(n-i)}=(G^{(n-i-1)})'=G_{i+1}'.

Thus (*) is a solvable series in G and hence G is solvable.

iv) It is clear that if G has a normal series (1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G such that G_{i+1}/G_i is abelian for all i, then G is solvable because every normal series is obviously subnormal. Conversely, if G is solvable, then, by iii), G^{(n)}=(1) for some n and then, as explained in the proof of iii), the series (*) has the required properties. \Box

Frattini subgroup and Burnside’s basis theorem (1)

Posted: January 8, 2022 in Basic Algebra, Groups
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Let G be a finite group. We say that a subset S of G is a minimal generating set of G if S generates G but no proper subset of S generates G. This is like the definition of a basis for a finite dimensional vector space V but there’s an important difference, i.e. very two basis of V have the same number of elements but two minimal generating sets of G need not have the same number of elements. For example, both sets \{1\} and \{2,3\} are minimal generating sets of the group \mathbb{Z}_6. In 1912, the great British mathematician William Burnside proved that if G is a finite p-group, i.e. |G|=p^n for some prime number p and some integer n \ge 1, then every two minimal generating sets of G have the same number of elements. We will prove Burnside’s result in the second part of this post. But first, we need to define the Frattini subgroup of a group, named after the Italian mathematician Giovanni Frattini, and prove a couple of facts about the subgroup.

Recall that a proper subgroup H of a group G is said to be maximal if the only proper subgroup of G that contains H is H.

Definition 1. The Frattini subgroup \Phi(G) of a group G is the intersection of all the maximal subgroups of G. If G has no maximal subgroup, then we define \Phi(G)=G.

Remark 1. If G is finite, then clearly G has a maximal subgroup. But if G is infinite, then the existence of maximal subgroups is no longer guaranteed, as the following example shows.

Example. Let F be any field of characteristic zero, e.g. \mathbb{Q} or \mathbb{R} or \mathbb{C}, etc., and let G:=(F,+), the additive subgroup of F. We claim that G has no maximal subgroup. Suppose, to the contrary, that G has a maximal subgroup H. Then, since G is abelian, H is normal and G/H is a group with no non-trivial subgroup. Thus G/H is a finite group of (prime) order p. Hence px \in H for all x \in G. But then, since p \ne 0_F because the characteristic of F is zero, p is invertible in F and so x=p(p^{-1}x) \in H, for all x \in G, i.e. H=G, which is a contradiction.

Remark 2. For any group G, the Frattini subgroup \Phi(G) is a normal subgroup of G. That’s because if H is a subgroup, and x \in G, then clearly H is maximal if and only if xHx^{-1} is maximal.

Definition 2. Let G be a group. We say that x \in G is a non-generator if it can be omitted from any generating set of G, i.e. if S is a subset of G and G=\langle x, S \rangle, then G=\langle S\rangle.

The following is the most fundamental fact about the Frattini subgroup.

Theorem 1. Let G be a group. The Frattini subgroup \Phi(G) is the set of non-generators of G.

Proof. Suppose first that x \in G is a non-generator. If G has no maximal subgroup, then x \in G=\Phi(G) and we’re done. Suppose now that H is a maximal subgroup of G. If x \notin H, then G=\langle x,H\rangle because \langle x,H\rangle properly contains H. So, since x is a non-generator, we get that H=G, which is false. So x \in H for any maximal subgroup H of G and hence x \in \Phi(G).
Conversely, suppose that x \in \Phi(G) and G=\langle x,S\rangle for some S \subseteq G. Let H:=\langle S\rangle. Then clearly G=\langle x,H\rangle because S \subseteq H and G=\langle x,S\rangle. We need to show that H=G. So we consider two cases.

Case 1: x \in H. In this case, G=\langle x,H \rangle =H, which gives H=G, and we’re done.

Case 2: x \notin H. We show that this case is impossible. Let T be the set of all subgroups K \supseteq H of G such that x \notin K. We have T \ne \emptyset because H \in T. Also, it’s clear that any totally ordered subset of (T, \subseteq) has an upper bound in T (the upper bound is the union of all the elements of the chain). Thus, by Zorn’s lemma, (T, \subseteq) has a maximal element M. So x \notin M and therefore M is not a maximal subgroup of G, because x \in \Phi(G). Hence there exists a subgroup N of G such that M \subset N \subset G. Thus, by the maximality of M in (T,\subseteq), we must have N \notin T and so x \in N. But then N=\langle x,N \rangle \supseteq \langle x,M\rangle \supseteq \langle x,H\rangle =G, which gives N=G, contradiction. \Box

Theorem 2. Let G be a group with the commutator subgroup G'.

i) If every maximal subgroup of G is normal, then G' \subseteq \Phi(G).

ii) If G is a finite p-group, then every maximal subgroup of G is normal and hence, by i), G' \subseteq \Phi(G).

iii) If G is a finite p-group of order p^n, then a subgroup H of G is maximal if and only if |H|=p^{n-1}.

Proof. i) There is nothing to prove if G has no maximal subgroup because in this case \Phi(G)=G. Now, suppose, to the contrary, that G' \nsubseteq \Phi(G). So there exists a maximal subgroup M of G such that G' \nsubseteq M. Hence, since G' is generated by all the elements xyx^{-1}y^{-1}, \ x,y \in G, there exist x,y \in G such that xyx^{-1}y^{-1} \notin M, and so x \notin M. Now, H:=\langle x,M\rangle properly contains M and so H=G. Also, since M is normal, H=M\langle x \rangle and so G=M\langle x \rangle. Thus y=gx^n for some g \in M and some integer n. But then xyx^{-1}y^{-1}=xgx^nx^{-1}x^{-n}g^{-1}=xgx^{-1}g^{-1} \in M, contradiction.

ii) We prove this part by induction over |G|. If |G|=p, there is nothing to prove since G is abelian. Now suppose that G is a p-group of order \ge p^2 and the claim is true for any p-group of order < |G|. Let M be a maximal subgroup of G, and let Z be the center of G. It’s an immediate result of the class equation that |Z| > 1. If G is abelian, every subgroup is normal and we’re done. So we may assume that Z \ne G. We now consider two cases.

Case 1: Z \nsubseteq M. In this case, ZM is a subgroup that properly contains M, and so ZM=G. Hence if g \in G, then g=zx for some z \in Z, x \in M, and so gMg^{-1}=zxMx^{-1}z^{-1}=xMx^{-1} =M, proving that M is normal.

Case 2: Z \subseteq M. In this case, M/Z is a maximal subgroup of G/Z, because M is a maximal subgroup of G. So, since G/Z is a finite p-group and |G/Z| < |G, the induction hypothesis gives that M/Z is a normal subgroup of G/Z, and hence M is a normal subgroup of G.

iii) If |H|=p^{n-1} and H \subset K, for some subgroup K of G, then |K| > |H|=p^{n-1}, and |K| \mid |G|=p^n, which gives |K|=p^n=|G|. Thus K=G and so H is maximal. Conversely, if H is maximal, then, by ii), H is normal and G/H is a p-group with no non-trivial subgroup. Thus |G/H|=p and so |H|=p^{n-1}. \ \Box

Exercise. Find \Phi(G) for any cyclic group G.
Hint. A maximal subgroup of \mathbb{Z} is in the form p\mathbb{Z}, where p is any prime number. So \Phi(\mathbb{Z})=\bigcap_{p}p\mathbb{Z}=(0). For finite cyclic groups, let G:=\mathbb{Z}/n\mathbb{Z}, and let p_1, \cdots , p_k be all the distinct prime divisors of n. Show that \Phi(G)=m\mathbb{Z}/n\mathbb{Z}, where m=p_1p_2 \cdots p_k.

See the second part of this post here.

Subgroups of SL(2,3)

Posted: December 8, 2021 in Basic Algebra, Groups
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Recall that the special linear group \text{SL}(2,3)=\text{SL}(2,\mathbb{Z}_3) is the multiplicative group of all 2 \times 2 matrices with entries from the field \mathbb{Z}_3 and with determinant 1 (see this post!). Here we’re going to find all subgroups of this group.

Notation. Throughout this post, G:=\text{SL}(2,3) and Z(G), G' are, respectively, the center and the commutator subgroup of G. Also, as usual, I is the identity matrix, and e_{ij} is the matrix whose (i,j)-entry is 1 and all other entries zero.

Problem 1. Show that

i) |G|=24=2^3 \times 3,

ii) Z(G)=\{I,-I\}.

Solution. i) By Exercise 1 in this post, \displaystyle |G|=\frac{1}{3-1}\prod_{i=1}^2(3^2-3^{i-1})=24.

ii) It is clear that \{I,-I\} \subseteq Z(G). Now, let A:=(a_{ij}) \in Z(G). Hence, since both I+e_{12} and I+e_{21} are in G, we must have A(I+e_{12})=(I+e_{12})A and A(I+e_{21})=(I+e_{21})A. So Ae_{12}=e_{12}A and Ae_{21}=e_{21}A, which give a_{11}=a_{22}, \ a_{12}=a_{21}=0. Thus a_{11}^2=\det A=1 and so a_{11}=\pm 1, which give A=\pm I. \ \Box

Problem 2. Show that

i) G has exactly 4 Sylow 3-subgroup, and they’re all isomorphic to \mathbb{Z}_3,

ii) if P is a Sylow 3-subgroup of G, then N(P)=C(P), where N(P), C(P) are, respectively, the normalizer and the centralizer of P in G.

Solution. By Problem 1, i), |G|=24=2^4 \times 3 and so every Sylow 3-subgroup of G has order 3 hence isomorphic to \mathbb{Z}_3. Let n be the number of Slow 3-subgroups of G. By Sylow theorems, n \equiv 1 \mod 3 and n \mid 24. So either n=1 or n=4. But we can easily find at least two Sylow 3-subgroup of G, here they are: \{I+ae_{12}: \ a \in \mathbb{Z}_3\} and \{I+ae_{21}: \ a \in \mathbb{Z}_3\}. So n \ne 1 and hence n=4.

ii) By i), G has exactly 4 Sylow 3-subgroups and by Sylow theorems, the number of Sylow 3-subgroups is [G:N(P)]. So 4=[G:N(P)]=[24:N(P)], and hence |N(P)|=6. Now, since, in every group, the centralizer of a subgroup is a subgroup of the normalizer of the subgroup, we have C(P) \subseteq N(P) and hence |C(P)| \le 6. Also, clearly Z(G)P \subseteq C(P) and so |C(P)| \ge |Z(G)P|=|Z(G)||P|=6, by Problem 1, ii). Hence |C(P)|=6=|N(P)| and so C(P)=N(P). \ \Box

Problem 3. Show that

i) G has only 1 Sylow 2-subgroup, say Q, and so Q is normal in G,

ii) Q =G' \cong Q_8, the quaternion group.

Solution. i) Let P be a Sylow 3-subgroup of G, and let N(P), C(P) be, respectively, the normalizer and the centralizer of P in G. By Problem 2, ii), N(P)=C(P) and therefore, by Burnside’s theorem, there exists a normal subgroup Q of G such that P \cap Q=(1) and PQ=G. Thus 24=|G|=|P||Q|=3|Q|, which gives |Q|=8. So Q is a normal, hence the unique, Sylow 2-subgroup of G.

ii) First note that since G/Q \cong \mathbb{Z}_3 is abelian, G' \subseteq Q. Now, let

A:=I+ae_{12}, \ \ \ B:=I+be_{21}, \ \ \ \ a,b \in \mathbb{Z}_3.

Clearly A,B \in G, and see that

\displaystyle [A,B]=ABA^{-1}B^{-1}=\begin{pmatrix}1+ab+a^2b^2 & -a^2b \\ ab^2 & 1-ab \end{pmatrix} \in G'.

So choosing a=b=1, \ a=1, b=-1, we get the following three elements \bold{i}, \bold{j}, \bold{k} \in G'

\displaystyle \begin{aligned}\bold{i}:=[I+e_{12}, I+e_{21}]=\begin{pmatrix}0 & -1 \\ 1 & 0 \end{pmatrix}, \ \ \ \ \bold{j}:=[I+e_{12},I-e_{21}]=\begin{pmatrix}1 & 1 \\ 1 & -1 \end{pmatrix}, \ \ \ \ \bold{k}:=\bold{i}\bold{j}.\end{aligned}

See that \bold{i}^2=\bold{j}^2=\bold{k}^2=-I, \ \bold{j}\bold{i}=-\bold{i}\bold{j}=-\bold{k}. So Q_8 \cong \langle \bold{i},\bold{j}\rangle \subseteq G' \subseteq Q. Thus, since |Q|=|Q_8|=8, we get Q=G' \cong Q_8. \ \Box

Problem 4. Show that

i) G has only 1 subgroup of order two, which is Z(G),

ii) G has exactly 3 subgroups of order 4, and they’re all isomorphic to \mathbb{Z}_4.

Solution. By Sylow theorems, every 2-subgroup is contained in a Sylow 2-subgroup. So since, by Problem 3, G has a unique Sylow 2-subgroup Q, every subgroup of order 2 or 4 of G is contained in Q. So since, by Problem 3, Q \cong Q_8, the quaternion group, we only need to find subgroups of order 2 or 4 in Q_8. Let

Q_8=\langle \bold{i}, \bold{j}\rangle=\{\pm I, \pm \bold{i}, \pm \bold{j}, \pm \bold{k}\}, \ \ \ \ \bold{i}^2=\bold{j}^2=\bold{k}^2=-I, \ \bold{k}=\bold{i}\bold{j}=-\bold{j}\bold{i}.

i) Since clearly -I is the only element of Q_8 which has order two, Q_8 has only one subgroup of order 2, and that’s \{I,-I\}=Z(G), by Problem 1, ii).

ii) It is clear that \pm \bold{i}, \pm \bold{j}, \pm \bold{k} all have order four, and so we have at least three subgroups of order four:

\langle \bold{i} \rangle =\langle -\bold{i}\rangle \cong \mathbb{Z}_4, \ \ \langle \bold{j} \rangle =\langle -\bold{j}\rangle \cong \mathbb{Z}_4, \ \ \langle \bold{k} \rangle =\langle -\bold{k}\rangle \cong \mathbb{Z}_4.

Finally, if H is any subgroup of order four in Q_8, then it must contain at least one of the elements \pm \bold{i}, \pm \bold{j}, \ \pm \bold{k}, which all have order four, and so H equals one of the above three groups. \Box

Problem 5. Show that

i) G has exactly 4 subgroups of order 6, and they’re all isomorphic to \mathbb{Z}_6,

ii) G has no subgroup of order 12.

Solution. i) Let H be a subgroup of order 6 in G. Then H has an element of order two, and an element of order three. By Problem 4, i), G has only one element of order two, which is -I, and the subgroup generated by that element is Z(G). Also, by Problem 2, i), G has exactly four elements A_i, \ 1 \le i \le 4, of order three. Since Z(G) is a normal abelian subgroup of G, we get four distinct subgroups \langle A_i\rangle Z(G) \cong \mathbb{Z}_6 of order six in G. Since |H|=6 and H must contain at least one of those four subgroups of order six, H must be equal to one of them.

ii) Suppose, to the contrary, that G has a subgroup K of order 12. Then [G:K]=2 and so K is normal in G. Now, since G/N \cong \mathbb{Z}_2 is abelian, we get G' \subseteq K, which is impossible because, by Problem 3, ii), |G'|=8, which does not divide |K|=12. \ \Box

Remark. So we have shown that G is a group of order 24 whose center has order two, G has no subgroup of order 12 but does have subgroups of other orders, all proper subgroups of G are cyclic except for its unique Sylow 2-subgroup Q, which is equal to G', and isomorphic to the quaternion group Q_8. More precisely, we showed in Problem 3, ii), that Q=\langle \bold{i},\bold{j}\rangle, where

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

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