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Introduction E9F5FC Questions FFFFC0 Software |
CT groups, Pin groups, Bott periodicity
Group Extensions Wikipedia: Metacyclic group A metacyclic group is an extension of a cyclic group by a cyclic group. A metacyclic group is metabelian and supersolvable. - Metabelian group A metabelian group is a group whose commutator subgroup is abelian. The commutator subgroup (also, the derived subgroup) is the group generated by the commutators, the elements of the form {$[g,h]=g^{-1}h^{-1}gh$}.
- Supersolvable group A supersolvable group is a group that has an invariant normal series where all factors are cyclic groups.
Math Stack Exchange. Extensions of Finite Cyclic Groups. - In the group extension {$1\rightarrow <i>\rightarrow Q_8 \rightarrow Q_8/<i>\rightarrow 1$}, the quaternion group {$Q_8$} is not a semi-direct product. Note here that {$Q_8/<i>$} has the cosets {$\{1,j,-1,-j\}\cong\mathbb{Z}_4$} and {$\{i,-k,-i,k\}\cong\mathbb{Z}_4i$}.
- https://math.stackexchange.com/questions/3460265/extension-of-a-group-g-by-a-commutative-group-f
- https://math.stackexchange.com/questions/1429542/loop-group-of-u1
- https://math.stackexchange.com/questions/3379017/general-formula-for-group-extensions-by-an-abelian-group
- https://math.stackexchange.com/questions/2601555/extensions-of-finite-groups-by-compact-lie-groups
Group Extension Example: Extending {$\mathbb{Z}_2$} by {$\mathbb{Z}_2$} Consider the group extension {$1\rightarrow\mathbb{Z}_2\rightarrow X\rightarrow\mathbb{Z}_2\rightarrow 1$}. This says that {$f:\mathbb{Z}_2\rightarrow X$} is injective, that {$g:X\rightarrow\mathbb{Z}_2$} is surjective, and that {$\textrm{Img}(f)=\textrm{Ker}(g)$}. Write {$\mathbb{Z}_2=\{1,-1\}$}, and similarly, write {$1, -1$} as the elements of the subgroup {$\mathbb{Z}_2$} embedded in {$X$}. We know that {$g(1)=1, g(-1)=1$}. Suppose {$x\in X,x\neq 1,-1$}. We know that either {$g(x)=1$} or {$g(x)=-1$} and so {$1=g(x)^2=g(x^2)$}. This means that {$x^2\in\textrm{Ker}(g)=\textrm{Im}(f)=\mathbb{Z}_2$}, which is embedded in {$X$}. We have two possibilities: {$x^2=1$} and {$x^2=-1$}. We know that {$(-1)x\neq 1$} (otherwise {$x=-1$}) and {$(-1)x\neq -1$} (otherwise {$x=1$}). Likewise {$x(-1)\neq 1,-1$}. Thus we know that {$((-1)x)^2=(x(-1))^2=1$}. Consequently, {$1=(-1)x(-1)x=x(-1)x(-1)$} and so {$[x(-1)](-1)x(-1)x=(-1)x(-1)x[x(-1)]$}, thus {$x(-1)=(-1)x$}. This means that we can write {$-x=(-1)x=x(-1)$} and the group {$\{1,-1,x,-x\}$} is commutative. We know that {$\mathbb{Z}_2=X/\mathbb{Z}_2$} and this means that {$|X|=4$}, thus {$X=\{1,-1,x,-x\}$}. There are two possibilities for {$X$}. - If {$x^2=–1$}, then {$x$} generates {$X=\{1,x,x^2=-1,x^3=-x\}=\mathbb{Z}_4$}.
- If {$x^2=1$}, then we can set {$X=\{1=(1,1),-1=(-1,1),x=(1,-1),-x=(-1,-1)\}$}, which is {$\mathbb{Z}_2\times\mathbb{Z}_2$}.
Group Extension Example: Extending {$\mathbb{Z}_2$} by {$\mathbb{Z}_4$} Group extension example: Extending {$\mathbb{Z}_n$} by {$\mathbb{Z}$} Math Stack Exchange. Group extensions of cyclic groups. Given a short exact sequence {$1\rightarrow N\overset{f}{\rightarrow}G\overset{g}{\rightarrow}\mathbb{Z}_2\rightarrow 1$} what are the possibilities for group $G$? In particular, I want to show that when $N=\mathbb{Z}_{2n}$, the only possibilities are the dihedral group and the dicyclic group, and when $N=U(1)$, the only possibilities are the pin groups $\textrm{Pin}_+(2)$ and $\textrm{Pin}_-(2)$. Group Extension Example: Extending {$\mathbb{Z}_2$} by {$\mathbb{Z}_{2n}$} Consider the group extension {$1\rightarrow\mathbb{Z}_{2n}\overset{f}{\rightarrow} G\overset{g}{\rightarrow}\mathbb{Z}_2\rightarrow 1$} These are cyclic groups {$\mathbb{Z}_{2n}=<\bar{z}\mid\bar{z}^{2n}=1>$} and {$\mathbb{Z}_2=<\bar{a}\mid\bar{a}^2=1>$}. The surjectivity of {$g$} implies there exists an {$a\in G$} such that {$g(a)=\bar{a}, g(a^2)=\bar{a}^2=1, a^2\in f(N)$}. Thus there exists {$0\leq i<2n$} such that {$a^2=z^i$}. Note that if {$x\in G$}, then {$g(x)=\bar{a}^k$} for some integer {$k$}, thus {$g(x)=g(a^k)$} and {$g(xa^{-k})=1$}, {$xa^{-k}\in N$}, {$x\in Na^k$}. Since {$z$} generates {$N$}, consequently {$z,a$} generate {$G$}. {$N$} is a normal subgroup of {$G$}, thus there exists a {$z'\in f(N)$} such that {$za=az'$}, where {$z'\neq 1$}, {$z'=z^j$} for some {$0<j<2n$}. We have {$za=az^j$} and we can calculate {$za^2=az^ja=a^2z^{j^2}$} and, in general, {$za^k=a^kz^{j^k}$}. When {$k=2$}, {$a^2=z^i$} commutes with {$z$} so {$z^{1+2i}=z^{2i+j^2}$}. Thus {$j^2\equiv 1 \mod 2n$}. We can solve this writing {$j^2-1\equiv 0 \mod 2n, (j-1)(j+1)\equiv 0\mod 2n$}, thus {$j=\pm 1 \mod 2n$}. Combinatorially, the number {$j=-1=2n-1 \mod 2n$}, if added {$2n-1$} times, shifts around the clock one tick smaller, starting from {$0$}, yielding {$0 - (2n-1) = 1 \mod 2n$}. Algebraically, {$(2n-1)(2n-1)=(4n^2-4n+1)=1\mod 2n$}. We have that the generators {$a$}, {$z$} are such that {$z^{2n}=1$} and {$za=az$} or {$za=az^{-1}$}. If {$za=az$}, then the two generators commute, so we have an abelian group, indeed, a finite abelian group of order {$4n$} with a cyclic subgroup of order {$2n$}. This group must be either {$\mathbb{Z}_{4n}$} or {$\mathbb{Z}_{2n}\times\mathbb{Z}_2$}. If {$a^2=z^{2m+1}$}, then consider the element {$a^{-1}z^{m+1}$} and note that, since {$a$} and {$z$} commute, we have {$(a^{-1}z^{m+1})^2=a^{-2}z^{2m+2}=z^{-2m-1+2m+2}=z$}. Recall that {$z$} has order {$2n$}, so {$a^{-1}z^{m+1}$} has order {$4n$} and generates {$\mathbb{Z}_4$}. If {$a^2=z^{2m}$}, then consider the element {$a^{-1}z^m$} and note that {$(a^{-1}z^m)^2=a^{-2}z^{2m}=z^{-2m+2m}=1$}. Note that {$a = z^m(a^{-1}z^m)^{-1}$} is generated by the elements {$z$} and {$a^{-1}z^m$}, which indeedy generate the group {$\mathbb{Z}_{2n}\times\mathbb{Z}_2$}. Now consider the noncommutative case {$za=az^{-1}$}. Applying this repeatedly, we have {$z^ia=az^{-i}$}. But {$z^i=a^2$} so {$z^i$} commutes with {$a$} and we have {$az^i=az^{-i}$} and {$z^i=z^{-i}$} so {$z^{2i}=1$}. Thus {$2i\equiv 0 \mod 2n$}. This means that {$i=0$} or {$i=n$}, which is to say, {$a^2=1$} or {$a^2=z^n$}. The dicyclic group is the noncommutative group of order {$4n$} generated by {$z,a$} where {$z^{2n}=1, a^2=z^n, za=az^{-1}$}. The dihedral group is the noncommutative group of order {$4n$} generated by {$z,a$} where {$z^{2n}=1, a^2=1, za=az^{-1}$}. The dicyclic group {$\textrm{Dic}_n$} is a subgroup of {$\textrm{Pin}_-(2)$} for which {$a^2=z^n$}, {$a^4=1$}. Note that {$a\neq a^{-1}$} so {$az\neq za$}. {$\mathbb{Z}_{2n}$} is not in the center of {$X$} and this is not a central extension. The dihedral group {$\textrm{Dih}_{2n}$} is a subgroup of {$\textrm{Pin}_+(2)$} for which {$a^2=1$}. Note that {$a=a^{-1}$} so {$az=za$}. {$\mathbb{Z}_{2n}$} is in the center of {$X$}, which is to say, this is a central extension. Group Extension Example: Extending {$\mathbb{Z}_2$} by {$U(1)$}
Consider the group extension {$1\rightarrow U(1)\overset{f}{\rightarrow} G\overset{g}{\rightarrow}\mathbb{Z}_2\rightarrow 1$} This is a left split short exact sequence if there exists a morphism {$t:G\rightarrow U(1)$} such that {$tf$} is the identity on {$U(1)$}. This is a right split short exact sequence if there exists a morphism {$u:\mathbb{Z}_2\rightarrow G$} such that {$gu$} is the identity on {$\mathbb{Z}_2$}. The injectivity of {$f$} means that {$N=f(U(1))\cong U(1)$}. {$N$} is a normal subgroup of {$G$}. {$G/N\cong\mathbb{Z}_2$}. Define {$\mathbb{Z}_2=\{1,\bar{a}\}$}. The surjectivity of {$g$} implies there exists an {$a\in G$} such that {$g(a)=\bar{a}, g(a^2)=\bar{a}^2=1, a^2\in f(N)$}. Thus there exists {$\theta\in U(1)$} such that {$a^2=\theta$}. Note that {$a\notin \ker g=N$}. We have {$G=N\sqcup aN$}.
The map {$\phi_a(z)=aza^{-1}$} defined for all {$z\in N\cong U(1)$} is an automorphism. But there are only two automorphisms of {$U(1)$}: {$z\rightarrow z$} and {$z\rightarrow z^{-1}$}. This fact is proven here: How to prove the group of automorphisms of {$S_1$} as a topological group is {$\mathbb{Z}_2$}? Informally, one way to prove it is to note that the continuous homomorphisms from a circle into a circle have to wind around the circle, forwards or backwards, an integer number of times. Of these homomorphisms, only two are invertible, namely, the identity homomorphism but also the inverse homomorphism. This means that we have two cases for the map {$\phi_a(z)=aza^{-1}$}. Either {$aza^{-1}=z,\forall z\in N$} or {$aza^{-1}=z^{-1},\forall z\in N$}.
Suppose {$aza^{-1}=z$} for all {$z\in N$}. Then {$az=za$} for all {$z\in N$}. We know that {$N$} and {$H$} are commutative and that the elements of {$N$} commute with those of {$H$} and vice versa. It follows that {$G$} is a commutative group. We know that {$a^2=\theta\in N$}. There exists {$\theta^{\frac{1}{2}}\in N$} such that {$(\theta^{\frac{1}{2})^2}=\theta$}. Consider the element {$b=a \theta^{-\frac{1}{2}}\notin N$}. We have that {$b^2=(a \theta^{-\frac{1}{2}})^2=a^2\theta^{-1}=1$}. Define {$H=\{1,b\}$}. {$G=N\sqcup Na = N\sqcup Nb=NH$}. For any two elements {$g_1,g_2\in G$}, there is a unique decomposition {$g_1=z_1h_1, g_2=z_2hz_2, g_1g_2=z_1z_2h_1h_2$} where {$z_1,z_2\in Z, h_1,h_2\in H$}. So {$G\cong N\times H \cong U(1)\times\mathbb{Z}_2$}.
For all {$z\in N$}, {$aza^{-1}=z^{-1}$} and {$az=z^{-1}a$}. We know that {$a^2=\theta\in N$}. Thus {$a\theta=\theta^{-1}a, a^3=a^{-1},a^4=1. This means that {$\atheta^2=1$}, so {$theta=-1$} or {$theta=1$}, which is to say, {$a^2=1$} or {$a^2=-1$} (the rotation halfway round the circle).
Define the map {$u:\mathbb{Z}_2\rightarrow G$} as follows. {$u(1)=1, u(\bar{a})=a$}. Then {$gu(1)=1, gu(\bar{a})=\bar{a}$}. This is the requisite morphism and so the sequence is right split. This means that we have a semidirect product {$G=N\ltimes H$} where {$H=\{1,a\}$}. Recall that {$a\notin N$} and {$N\cap H=1$}.
There does not exist a morphism {$t:G\rightarrow U(1)$} such that {$tf$} is the identity on {$U(1)$}. The semi-direct product defines the group homomorphism {$\phi :\mathbb{Z}_2\rightarrow \textrm{Aut}(U(1))$} given by {$\phi: h\rightarrow \phi_h$} where {$\phi_h(n)=hnh^{-1}$}. Since {$\textrm{Aut}(U(1))\cong U(1)$}, we are looking at embeddings {$φ:\mathbb{Z}_2\rightarrow U(1)$}.
If it's left split, since it is also right split, we have a direct product {$G\cong U(1)\times \mathbb{Z}_2=t(G)\times u(G)$}.
Thus there exists a {$z'\in f(N)$} such that {$za=az'$}, where {$z'\neq 1$}, {$z'\in U(1)$}. We see that {$z^na=a(z')^n$} for all {$n\in\mathbb{Z}$}. In general, every element in {$U(1)$} can be written as {$z^{\lambda}$} where {$\lambda\in\mathbb{R}$}. This map {$z\rightarrow z'$} implies the continuous homomorphism {$z^r a\rightarrow az'^r$} from {$U(1)$} to {$U(1)$}. Topologically speaking, the homology group {$H_1(U(1))=\mathbb{Z}$} is given by the winding number. Now likewise, there exists a {$z''\in f(N)$} such that {$z'a=az''$}. Consequently, {$z^2=za^2=az'a=a^2z''=zz''$} and {$z=z''$}. This implies that our map is an automorphism. The only inverses in {$\mathbb{Z}$} are {$\pm 1$}. Thus the only possible maps here are {$z\rightarrow z$} and {$z\rightarrow z^{-1}$}, which means that either we have commutativity {$za=az$} or we have {$za=az^{-1}$}.
We can represent {$U(1)$} in terms of {$2\times 2$} matrices of real numbers. Suppose that we can represent all of {$G$} likewise and with {$U(1)$} thus embedded. We represent {$z\in U(1)$} with the matrix {$\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{pmatrix}$}. Let {$a$} be represented as {$\begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{pmatrix}$}.
Then {$$\begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{pmatrix}\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{pmatrix}=\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{pmatrix}\begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{pmatrix}$$} {$a_{11}\cos\theta + a_{12}\sin\theta = a_{11}\cos\theta - a_{21}\sin\theta$} implies {$a_{12}=-a_{21}$} or {$\sin\theta = 0$} {$-a_{11}\sin\theta + a_{12}\cos\theta = a_{12}\cos\theta + a_{11}\sin\theta$} implies {$a_{11}=0$} or {$\sin\theta = 0$}. {$a_{21}\cos\theta + a_{22}\sin\theta = a_{11}\sin\theta + a_{21}\cos\theta$} implies {$a_{22}=a_{11}$} or {$\sin\theta = 0$}. {$-a_{21}\sin\theta -a_{22}\cos\theta = a_{12}\sin\theta -a_{22}\cos\theta$} implies {$a_{12}=-a_{21}$} or {$\sin\theta = 0$}. Note furthermore that {$a^2=z$} and {$\det z=1$}. This means that {$\det a=\pm 1$}. Thus, if {$\sin\theta\neq 0$}, then {$a$} is represented by {$\pm\begin{pmatrix}0 & -1 \\ 1 & 0 \\ \end{pmatrix}$}, but then {$z$} is represented by {$\begin{pmatrix}-1 & 0 \\ 0 & -1 \\ \end{pmatrix}$} with a contradiction. Thus {$\sin\theta = 0$} and {$z$} is represented by {$\pm\begin{pmatrix}1 & 0 \\ 0 & 1 \\ \end{pmatrix}$}.
{$$\begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{pmatrix}\begin{pmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \\ \end{pmatrix}=\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{pmatrix}\begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{pmatrix}$$} {$a_{11}\cos\theta - a_{12}\sin\theta = a_{11}\cos\theta - a_{21}\sin\theta$} implies {$a_{12}=a_{21}$} or {$\sin\theta = 0$} {$a_{11}\sin\theta + a_{12}\cos\theta = a_{12}\cos\theta - a_{22}\sin\theta$} implies {$a_{11}=-a_{22}$} or {$\sin\theta = 0$}. {$a_{21}\cos\theta - a_{22}\sin\theta = a_{11}\sin\theta + a_{21}\cos\theta$} implies {$a_{11}=-a_{22}$} or {$\sin\theta = 0$}. {$a_{21}\sin\theta +a_{22}\cos\theta = a_{12}\sin\theta +a_{22}\cos\theta$} implies {$a_{12}=a_{21}$} or {$\sin\theta = 0$}. Thus {$a$} is represented by {$\begin{pmatrix}a_{11} & a_{12} \\ a_{12} & -a_{11} \\ \end{pmatrix}$} Again, {$a^2=z$} and {$\det z = 1$} implies {$\det a=\pm 1$}. Thus {$-a_{11}^2-a_{12}^2=-1$} so {$a$} is represented by a matrix of the form {$\begin{pmatrix}\cos\psi & \sin\psi \\ \sin\psi & -\cos\psi \\ \end{pmatrix}$}. Then {$a^2=z$} means that {$\begin{pmatrix}\cos\psi & \sin\psi \\ \sin\psi & -\cos\psi \\ \end{pmatrix}\begin{pmatrix}\cos\psi & \sin\psi \\ \sin\psi & -\cos\psi \\ \end{pmatrix}=\begin{pmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \\ \end{pmatrix}$} {$\cos^2\psi + \sin^2\psi=\cos\theta$} implies {$\cos\theta = 1$} {$\cos\psi\sin\psi - \cos\psi\sin\psi = -\sin\theta$} implies {$\sin\theta=0$} Thus {$z$} is represented by {$\begin{pmatrix}1 & 0 \\ 0 & 1 \\ \end{pmatrix}$}, which is to say, {$z=1$}. Whereas {$a$} can be represented by any matrix of the form below, which is a rotation times a reflection. {$$\begin{pmatrix}\cos\psi & \sin\psi \\ \sin\psi & -\cos\psi \\ \end{pmatrix}=\begin{pmatrix}\cos\psi & -\sin\psi \\ \sin\psi & \cos\psi \\ \end{pmatrix}\begin{pmatrix}1 & 0 \\ 0 & -1 \\ \end{pmatrix}$$} |

This page was last changed on October 28, 2024, at 08:12 PM