\documentclass[11pt]{article} \usepackage{amssymb,amsmath,amsthm} \usepackage[all]{xy} \newtheorem{lemma}{Lemma} \newtheorem{thm}{Theorem} \newcommand{\im}{\operatorname{im}} \newcommand{\K}{{\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}}} \newcommand{\R}{{\mathbb R}} \newcommand{\Q}{{\mathbb Q}} \newcommand{\Z}{{\mathbb Z}} \newcommand{\Aut}{\operatorname{Aut}} \parindent=0cm \title{Math 210A\\Homework 4} \author{Brett Hemenway} \begin{document} \maketitle \begin{itemize} \item [1.] Let $A$ and $C$ be groups, and $\theta : C \rightarrow \Aut(A)$ a homomorphism. Let $A \rtimes_{\theta} C$ be the set $A \times C$ with a binary operation defined by $(a,c)(a^\prime,c^\prime) = ((a \theta(c)(a^\prime),cc^\prime)$ for $a,a^\prime \in A$ and $c,c^\prime \in C$. This defines a group law because \begin{itemize} \item [closed:] $(a,c)(a^\prime,c^\prime) = (a \theta(c)(a^\prime),cc^\prime ) \in A \times C$ because $\theta(c)(a) \in A$, and $A$, and $C$ are closed. \item [identity:] \[ (a,c)(e_a,e_c) = (a \theta(c)(e_a), ce_c ) = ( a e_a, c ) = (a,c) \] We have $\theta(c)(e_a) = e_a$ because $\theta(c)$ is an automorphism, and \[ (e_a,e_c)(a,c) = (e_a \theta(e_c)(a), e_cc ) = (e_aa,c ) = (a,c) \] because $\theta$ is a homomorphism. \item [inverse:] \begin{align*} (a,c)(\theta(c^{-1})(a^{-1}),c^{-1}) &= (a\theta(c)(\theta(c^{-1})(a)),cc^{-1})\\ &= (a\theta(cc^{-1})a^{-1},cc^{-1})\\ &= (aa^{-1},e_c)\\ &= (e_a,e_c) \end{align*} and \begin{align*} (\theta(c^{-1})(a^{-1}),c^{-1})(a,c) &= (\theta(c^{-1})(a^{-1})\theta(c^{-1})(a),c^{-1}c)\\ &= (\theta(c^{-1}c)(a^{-1}a),e_c)\\ &= (\theta(e_c)(e_a),e_c)\\ &= (e_a,e_c) \end{align*} \item [associative:] \begin{align*} ((a_1,c_1)(a_2,c_2))(a_3,c_3) &= (a_1\theta(c_1)(a_2),c_1c_2)(a_3,c_3)\\ &= (a_1\theta(c_1)(a_2)\theta(c_1c_2)(a_3),c_1,c_2c_3)\\ &= (a_1\theta(c_1)(a_2)\theta(c_1)(\theta(c_2)a_3), c_1c_2c_3)\\ &= (a_1\theta(c_1)(a_2\theta(c_2)(a3)),c_1c_2c_3)\\ &= (a_1,c_1)((a_2\theta(c_2)(a_3),c_2c_3))\\ &= (a_1,c_1)((a_2,c_2)(a_3,c_3)) \end{align*} \end{itemize} So $A \rtimes_\theta C$ is a group.\\ Suppose $B \simeq A \rtimes_{\theta} C$, then let \begin{align*} f : A &\rightarrow B\\ a &\mapsto (a,e_c) \end{align*} and \begin{align*} g : B &\rightarrow C\\ (a,c) &\mapsto c \end{align*} $f$ is a homomorphism because \[ f(ab) = (ab,e_c) = (a\theta(e_c)b,e_ce_c) = (a,e_c)(b,e_c) = f(a)f(b) \] Where the second to last inequality holds because $\theta$ is a homomorphism. $f$ is clearly injective because \[ a = b \Leftrightarrow (a,e_c) = (b,e_c) \] $g$ is clearly a surjective homomorphism, and \[ \ker(g) = \{ (a,e_c) : a \in A \} = \im(f) = f(A) \] So we have the exact sequence \begin{equation} \xymatrix{ 1 \ar[r] & A \ar[r]^f &B \ar[r]^g &C \ar[r] &1 } \label{exact} \end{equation} If we define \begin{align*} g^\prime : C &\rightarrow B\\ c &\mapsto (e_a,c) \end{align*} then $gg^\prime(c) = g(g^\prime(c)) = g(e_a,c) = c$, so $gg^{-1}$ is the identity on $C$. For the converse, suppose the sequence (\ref{exact}) is exact, and $g^\prime : C \rightarrow B$ with $gg^\prime$ the identity on $C$. Let \[ \theta : C \rightarrow \Aut(A) \] by the rule \begin{align*} \theta(c) : A &\rightarrow A\\ a &\mapsto f^{-1}(g^\prime(c)f(a)g^\prime(c)^{-1}) \end{align*} $\theta(c)$ is an automorphism of $A$ because $f$ is an isomorphism from $A$ to $\im(f)$, and conjugation by $g^\prime(c)$ is an automorphism of $\im(f)$ because $\im(f) = \ker(g)$ is normal in $B$, and finally $f^{-1}$ is an isomorphism from $\im(f)$ to $A$, then since the composition of three isomorphisms is an isomorphism, $\theta(c)$ is an isomorphism from $A \rightarrow A$.\\ $\theta$ is a homomorphism because \begin{align*} \theta(c_1c_2)(a) &= f^{-1}(g^\prime(c_1c_2)f(a)g^\prime(c_1c_2)^{-1})\\ &= f^{-1}(g^\prime(c_1)g^\prime(c_2)f(a)g^\prime(c_2)^{-1}g^\prime(c_1)^{-1})\\ &= f^{-1}(g^\prime(c_1)f(f^{-1}(g^\prime(c_2)f(a)g^\prime(c_2)^{-1}))g^\prime(c_1)^{-1})\\ &= f^{-1}(g^\prime(c_1)f(\theta(c_2)(a))g^\prime(c_1)^{-1})\\ &= \theta(c_1)(\theta(c_2)(a)) \end{align*} Now, let \begin{align*} \phi : A \rtimes_\theta C &\rightarrow B\\ (a,c) &\mapsto f(a)g^\prime(c) \end{align*} We will show that $\phi$ is an isomorphism. $\phi$ is a homomorphism because \begin{align*} \phi((a,c)(a^\prime,c^\prime)) &= \phi( (a\theta(c)a^\prime,cc^\prime) )\\ &= \phi( (af^{-1}(g^\prime(c)f(a^\prime)g^\prime(c)^{-1}),cc^\prime ) )\\ &= f(af^{-1}(g^\prime(c)f(a\prime)g^\prime(c)^{-1}))g(cc^\prime)\\ &= f(a)g^\prime(c)f(a^\prime)g^\prime(c)^{-1}g^\prime(c)g^\prime(c^\prime)\\ &= (f(a)g^\prime(c))(f(a^\prime)g^\prime(c^\prime))\\ &= \phi(a,c)\phi(a^\prime,c^\prime) \end{align*} $\phi$ is injective because if $f(a)g^\prime(c) = f(a^\prime)g^\prime(c^\prime)$ then \begin{align*} g^\prime(c) &= f(a)^{-1}f(a^\prime)g^\prime(c^\prime)\\ &= f(a^{-1}a^\prime)g^\prime(c^\prime)\\ \end{align*} which gives \begin{equation} g^\prime(c(c^\prime)^{-1}) = f(a^{-1}a^\prime) \label{eqn:inj} \end{equation} but $\im(f) = \ker(g)$, so applying $g$ to both sides gives \[ c(c^\prime)^{-1} = e_c \] so $c = c^\prime$. Plugging this back in to equation (\ref{eqn:inj}) gives \[ g^\prime(e_c) = e_B = f(a^{-1}a^\prime) \] Since $f$ is injective, we have that $a = a^\prime$.\\ This shows that $\phi$ is injective. To show that $\phi$ is surjective, let $b \in B$, with $g(b) = c$. $g$ is an isomorphism from $B/\ker(g) = B/f(A)\rightarrow C$, so we have equality of right cosets \[ f(A)b = f(A)g^\prime(c) \] So $g^\prime(c)b^{-1} \in f(A)$. Suppose $g^\prime(c)b^{-1} = f(a)$, then $f(a^{-1})g^\prime(c) = b$. So $\phi$ is onto. \item [2.] \begin{itemize} \item [a.] Let $D_n$ be the dihedral group of order $2n$, i.e. $D_n = \langle \sigma, \tau \rangle$ where $\sigma^n = e = \tau^2$ and $\sigma\tau = \tau\sigma^{n-1}$. Let \begin{align*} f : \Z/n\Z &\rightarrow D_n\\ a &\mapsto \sigma^{a} \end{align*} Then $f$ is an injective homomorphism. Let \begin{align*} g : D_n &\rightarrow \Z/2\Z\\ \sigma^{a}\tau^c &\mapsto c \end{align*} Then $g$ is a surjective homomorphism. Finally, if we let \begin{align*} g^\prime : \Z/2\Z &\rightarrow D_n\\ c &\mapsto \tau^c \end{align*} We have that $gg^\prime$ is the identity on $\Z/2\Z$. By problem 1, we have that $D_n \simeq \Z/n\Z \rtimes_\theta \Z/2\Z$ where \begin{align*} \theta(c)(a) &= f^{-1}(g^\prime(c)f(a)g^\prime(c)^{-1})\\ &= f^{-1}(\tau^c\sigma^a\tau^c)\\ \end{align*} So \begin{align*} \theta(\bar{0})(a) &= f^{-1}(\tau^0\sigma^a\tau^0) = f^{-1}(\sigma^{a}) = a\\ \theta(\bar{1})(a) &= f^{-1}(\tau\sigma^a\tau) = f^{-1}(\sigma^{-a}) = -a \end{align*} \item [b.] Let $\mathbf{Q}$ the quaternion group of order 8, i.e. $\mathbf{Q} = \{ \pm 1, \pm i, \pm j, \pm k\}$. If $\mathbf{Q}$ were the semidirect product of groups, counting gives four possibilities \begin{align*} \mbox{(1) } \mathbf{Q} &\simeq \Z/4Z \rtimes_\theta \Z/2\Z\\ \mbox{(2) } \mathbf{Q} &\simeq \Z/2Z\times \Z/2\Z \rtimes_\theta \Z/2\Z\\ \mbox{(3) } \mathbf{Q} &\simeq \Z/2Z \rtimes_\theta \Z/4\Z\\ \mbox{(3) } \mathbf{Q} &\simeq \Z/2Z \rtimes_\theta \K \end{align*} Case (1) is ruled out because $\mathbf{Q} \not \simeq D_n \simeq \Z/4\Z \rtimes_{\theta} \Z/2\Z$\\ Case (2) is ruled out because $\mathbf{Q}$ has only $1$ element of order $2$ so there is no surjective map $g$ from $\mathbf{Q}$ to $\K$.\\ Case (3) is ruled out because if $f$ injects from $\Z/2\Z$ to $\mathbf{Q}$, then $f(\Z/2\Z) = \{\pm 1\}$ but then $\mathbf{Q}/\ker(g) = \mathbf{Q}/\{\pm 1\}$, but \[ \mathbf{Q}/\{\pm 1\} \simeq \K \not \simeq \Z/4\Z \] Case (4) is ruled out because there is no surjection from $\mathbf{Q}$ to $\K$ because $\mathbf{Q}$ has only one element of order 2.\\ So we conclude that $\mathbf{Q}$ is not the semidirect product of two non-trivial groups. \end{itemize} \item [3.] Let $H \subset G$ be a subgroup \begin{itemize} \item [a.] $Z_G(H) = \{g \in G : gh = hg \text{ for all } h \in H \}$. Clearly $Z_G(H) \subset N_G(H)$, $Z_G(H)$ is normal in $N_G(H)$ because it is the kernel of the map \begin{align*} \phi : N_G(H) &\rightarrow \Aut(H)\\ g &\mapsto \mbox{ conjugation by $g$ } \end{align*} $\phi$ is a homomorphism because \begin{align*} \phi(g_1g_2)(h) &= (g_1g_2)h(g_1g_2)^{-1}\\ &= g_1(g_2hg_2^{-1})g_1^{-1}\\ &= \phi(g_1) \circ \phi(g_2) (h) \end{align*} $Z(G)$ is the kernel of $\phi$ because $ghg^{-1} = h \Leftrightarrow gh = hg$. \item [b.] The monomorphism from $N_G(H)/Z_G(H)$ to $\Aut(H)$ was described in part (a). \item [c.] Let $G$ be infinite, and $H$ normal subgroup of $G$ with $|H| = n$. Then from part $b$ we have an injection from $N_G(H)/Z_G(H) = G/Z_G(H)$ into $\Aut(H) \subset S_n$. So $|G/Z_G(H)| \le n!$.\\ If $G$ has no non-trivial finite quotient, then it for any finite subgroup $H$, we have $G/Z_G(H)$ must be trivial, so $Z_G(H) = G$. This means that $H$ is abelian, and since $Z_G(H) \subset N_G(H)$, we have that $N_G(H) = G$, and hence $H$ is normal. \end{itemize} \item [4.] Let $G$ be a simple group and $H$ a subgroup of $G$ with index $n < \infty$. Then $G$ acts on the cosets $G/H$. This gives us a homomorphism from $G$ to $S_{G/H} = S_n$. The kernel of this homomorphism is a normal subgroup of $G$, but since $G$ is simple it must be $G$ or $\{e\}$. It cannot be $G$ since $G$ does not act trivially on the cosets $G/H$. So the kernel must be trivial. From this we conclude $G$ injects into $S_n$, so $|G| \le n!$. \item [5.] Let $H \subset G$. \begin{itemize} \item [a.] Suppose $G$ is solvable, then we have a sequence \[ 1 = H_0 \subset H_1 \subset \ldots \subset H_n = G \] with $H_{i+1}/H_i$ abelian. Consider the sequence \[ 1 = H_0 \cap H \subset H_1 \cap H \subset \ldots \subset H_n \cap H = H \] $(H_i\cap H) \triangleleft (H_{i+1} \cap H)$ because $g(H_i \cap H)g^{-1} \subset H_{i}$ for $g \in H_{i+1}$ and $g(H_i\cap H)g^{-1} \in H$ for $g \in H$.\\ Next, we apply the second isomorphism theorem to the group $(H\cap H_{i+1})/(H \cap H_i) = (H \cap H_{i+1})/((H \cap H_{i+1}) \cap H_i)$, this gives \[ (H \cap H_{i+1})/(H \cap H_i) = (H \cap H_{i+1})/((H \cap H_{i+1})\cap H_i) \simeq H_{i}(H \cap H_{i+1})/H_i \] But $H_i(H \cap H_{i+1})/H_i$ is a subgroup of $H_{i+1}/H_i$, so it must be abelian as well. Thus $H$ is solvable. \item [b.] Suppose $H$ and $G/H$ are solvable. So we have series \[ 1 = H^\prime_0 \subset \ldots \subset H^\prime_r = G/H \] with each $H^\prime_{i+1}/H^\prime_i$ abelian. Then by the correspondence theorem, we have a sequence \[ H = H_0 \subset \ldots \subset H_r = G \] By the third isomorphism theorem $H_{i+1}/H_i \simeq H^\prime_{i+1}/H^\prime_i$, so each $H_{i+1}/H_i$ is abelian. Since $H$ is solvable, we also have the sequence \[ 1 = K_0 \subset \ldots \subset K_t = H \] Splicing these two sequences together we have that $G$ is solvable because \[ 1 = K_0 \subset \ldots \subset K_t = H = H_0 \subset \ldots \subset H_r = G \] For the converse, if $G$ is solvable, by part (a), $H$ is solvable. Then suppose \[ 1 = H_0 \subset \ldots \subset H_n = G \] Consider the series \begin{equation} H = HH_0 \subset \ldots \subset HH_n = G \label{quotser} \end{equation} Each $HH_i$ is a subgroup of $G$ because $H$ is normal in $G$. $HH_i$ is normal in $HH_{i+1}$ because if $h \in H$, $h_{i+1} \in H_{i+1}$, we have \begin{align*} (hh_{i+1})HH_i(hh_{i+1})^{-1} &= (hh_{i+1})HH_i(h_{i+1}^{-1}h^{-1})\\ &= H(hh_{i+1})H_i(h_{i+1}^{-1}h^{-1})\\ &= (Hh)(h_{i+1}H_ih_{i+1}^{-1})h^{-1}\\ &= HH_ih^{-1}\\ &= H_iHh^{-1}\\ &= H_iH\\ &= HH_i \end{align*} We know each quotient $H_{i+1}/H_i$ is abelian, and we have surjective homomorphism from $H_{i+1}/H_i$ to $HH_{i+1}/HH_i$ given by $h_{i+1}H_i \mapsto h_{i+1}H_iH = h_{i+1}HH_i$ So each quotient $HH_{i+1}/HH_i$ is abelian because $H_{i+1}/H_i$ is. To see that $G/H$ is solvable, mod out each term in the series (\ref{quotser}) by $H$, and apply the third isomorphism theorem. \item [c.] \begin{itemize} \item [(i)] This is almost identical to the proof of part (a).\\ Suppose $G$ is polycyclic. Then we have a series \[ 1 = H_0 \subset \ldots \subset H_n = G \] with $H_{i+1}/H_i$ cyclic. Then consider the sequence \[ 1 = H_0\cap H \subset \ldots H_n \cap H = H \] $(H_i\cap H) \triangleleft (H_{i+1} \cap H)$ because $g(H_i \cap H)g^{-1} \in H_{i+1}$ for $g \in H_{i+1}$ and $g(H_i\cap H)g^{-1} \in H$ for $g \in H$.\\ Next, we apply the second isomorphism theorem to the group \[ (H\cap H_{i+1})/(H \cap H_i) = (H \cap H_{i+1})/((H \cap H_{i+1}) \cap H_i) \] this gives \[ (H \cap H_{i+1})/(H \cap H_i) \simeq H_{i}(H \cap H_{i+1})/H_i \] But $H_i(H \cap H_{i+1})/H_i$ is a subgroup of $H_{i+1}/H_i$, since any subgroup of a cyclic group is cyclic, we have that $(H\cap H_{i+1})/(H \cap H_i)$ is cyclic, thus $H$ is polycyclic. \item [(ii)] This is almost identical to the proof of part (b).\\ Suppose $H$ and $G/H$ are polycyclic. So we have series \[ 1 = H^\prime_0 \subset \ldots \subset H^\prime_r = G/H \] with each $H^\prime_{i+1}/H^\prime_i$ cyclic. Then by the correspondence theorem, we have a sequence \[ H = H_0 \subset \ldots \subset H_r = G \] By the third isomorphism theorem $H_{i+1}/H_i \simeq H^\prime_{i+1}/H^\prime_i$, so each $H_{i+1}/H_i$ is cyclic. Since $H$ is polycyclic, we also have the sequence \[ 1 = K_0 \subset \ldots \subset K_t = H \] Splicing these two sequences together we have that $G$ is polycyclic because \[ 1 = K_0 \subset \ldots \subset K_t = H = H_0 \subset \ldots \subset H_r = G \] For the converse, if $G$ is polycyclic, by part (c.i), $H$ is polycyclic. Then suppose \[ 1 = H_0 \subset \ldots \subset H_n = G \] Again, consider the series \begin{equation*} H = HH_0 \subset \ldots \subset HH_n = G \end{equation*} We have already seen that $HH_i$ is normal in $HH_{i+1}$ we know each quotient $H_{i+1}/H_i$ is cyclic, and we have surjective homomorphism from $H_{i+1}/H_i$ to $HH_{i+1}/HH_i$ given by $h_{i+1}H_i \mapsto h_{i+1}H_iH = h_{i+1}HH_i$ So each quotient $HH_{i+1}/HH_i$ is cyclic because $H_{i+1}/H_i$ is. To see that $G/H$ is polycyclic, mod out each term in the series by $H$, and apply the third isomorphism theorem. \end{itemize} \end{itemize} \item [6.] A group $G$ is said to have a composition series if there is a finite chain of subgroups \[ 1 \subset H_0 \subset H_1 \subset \ldots \subset H_n =G \] of $G$ with $H_i \triangleleft H_{i+1}$ and $H_{i+1}/H_i$ simple. \begin{itemize} \item [a.] The group of order $1$ obviously has a composition series\\ Suppose all groups of order less than $n$ have a composition series\\ Let $G$ be a group with $|G| = n$.\\ If $G$ is simple, then we have the composition series $1 \subset G$.\\ Otherwise let $H \triangleleft G$. By our induction assumption $G/H$ has a composition series \[ 1 = H^\prime_0 \subset \ldots \subset H^\prime_r \subset G/H \] with $H^\prime_{i+1}/H^\prime_i$ simple. Then by the correspondence theorem, we have a sequence of subgroups \[ H = H_0 \subset \ldots \subset H_r = G \] with $H_{i+1}/H_i$ simple. By our induction assumption, $H$ has a composition series, \[ 1 = K_0 \subset \ldots \subset K_t \subset H \] Then, splicing these two series together we obtain a composition series for $G$ \[ 1 = K_0 \subset \ldots \subset K_t = H = H_0 \subset \ldots \subset H_r = G \] So any finite group $G$ has a composition series. \item [b.] First, note that in any composition series, the first subgroup, $H_0$ must be simple.\\ Every subgroup of $\Z$ has the form $n\Z$, but $n\Z$ is not simple because it has a normal subgroup $2n\Z$. So $\Z$ does not have a composition series. \item [c.] Let $G$ be a finite solvable group with \[ 1 = H_0 \subset \ldots \subset H_n = G \] Each quotient $H_{i+1}/H_i$ is finite, so by part (a) has a composition series \[ 1 = K_0 \subset \ldots K_{r_i} = H_{i+1}/H_i \] with $K_{j+1}/K_j$ simple. But each $K_j$ is abelian because $H_{i+1}/H_i$ is abelian. Each quotient $K_{j+1}/K_j$ is cyclic, because the only finite abelian simple groups are cyclic, and the correspondence theorem gives us a composition series \[ H_i = \widehat{K}_0 \subset \ldots \widehat{K}_{r_i} = H_{i+1} \] and $\widehat{K}_{j+1}/\widehat{K}_j \simeq K_{j+1}/K_j$.\\ Continuing this process for each neighboring $H_{i+1},H_i$, we can ``fill out" our solvable series to make a chain all of whose quotients are abelian and simple, and hence cyclic. So we conclude that any finite solvable group is polycyclic. \item [d.] We will show that every polycyclic group is finitely generated. Let $G$ be a polycyclic group with composition series \[ 1 = H_0 \subset H_1 \subset \ldots \subset H_n = G \] We proceed by induction on $n$.\\ if $n = 1$, then $G$ is cyclic, so $G$ is finitely generated.\\ Assume all groups with a composition series of less than $n$ terms are finitely generated.\\ $G/H_{n-1}$ is cyclic, so let $\langle aH_{n-1} \rangle = G/H_{n-1}$. Then for any $g \in G$ there exists an $k$ such that $gH_{n-1} = a^kH_{n-1}$. So $a^{-k}g = h \in H_{n-1}$, or $g = a^kh$ but by our induction assumption $H_{n-1}$ is finitely generated, so any element in $G$ can be written as a combination of the generators of $H_{n-1}$ and $a$. So $G$ is finitely generated.\\ To show that $\Q$ is not polycyclic, it remains only to show that $\Q$ is not finitely generated. This is clear though because the set $\langle \frac{p_1}{q_1},\ldots,\frac{p_n}{q_n} \rangle$ contains no elements with denominator greater than lcm$( q_1,q_2,\ldots,q_n )$, so it cannot be all of $\Q$. \end{itemize} \end{itemize} \end{document}