\USym X to \USymX for X=G,H,f. Unresolved: \US f, \Usym composite

fingp.tex

@@ -11,16 +11,16 @@ \chapter{Finite groups}
  The orbit-stabilizer theorem \cref{sec:orbit-stabilizer-theorem} is at the basis of this analysis: if $G$ is a group and $X:\BG\to\Set$ is a $G$-set, then
 $$X(\shape_G)\simeq \coprod_{x:X/G}\mathcal O_x$$
 and each orbit set $\mathcal O_x$ is equivalent to the cokernel of the inclusion $G_x\subseteq G$ of the stabilizer subgroup of $x$.
-Consequently, if $X(\shape_G)$ is a finite set, then its cardinality is the sum of the cardinality of these cokernels.  If also the set $\USym G$ is finite much more can be said and simple arithmetical considerations often allow us to deduce deep statements like the size of a certain subset of $X(\shape_G)$ and in particular whether or not there are any fixed points.
+Consequently, if $X(\shape_G)$ is a finite set, then its cardinality is the sum of the cardinality of these cokernels.  If also the set $\USymG$ is finite much more can be said and simple arithmetical considerations often allow us to deduce deep statements like the size of a certain subset of $X(\shape_G)$ and in particular whether or not there are any fixed points.
 
 \begin{example}
   A typical application could go like this.
-If $X(\shape_G)$ is a finite set with $13$ elements and for some reason we know that all the orbits have cardinalities dividing $8$ -- which we'll see happens if $\USym G$ has $8$ elements -- then we must have that some orbits are singletons (for a sum of positive integers dividing $8$ to add up to $13$, some of them must be $1$).
+If $X(\shape_G)$ is a finite set with $13$ elements and for some reason we know that all the orbits have cardinalities dividing $8$ -- which we'll see happens if $\USymG$ has $8$ elements -- then we must have that some orbits are singletons (for a sum of positive integers dividing $8$ to add up to $13$, some of them must be $1$).
 That is, $X$ has fixed points.
 \end{example}
 
 The classical theory of finite groups is all about symmetries coupled with simple counting arguments.
-Lagrange's \cref{thm:lagrange} gives the first example: if $H$ is a subgroup of $G$, then the cardinality ``$|G|$'' of $\USym G$ is divisible by $|H|$, putting severe restrictions on the possible subgroups.  For instance, if $|G|$ is a prime number, then $G$ has no nontrivial proper subgroups! (actually, $G$ is necessarily a cyclic group).  To prove this result we interpret $G$ as an $H$-set.
+Lagrange's \cref{thm:lagrange} gives the first example: if $H$ is a subgroup of $G$, then the cardinality ``$|G|$'' of $\USymG$ is divisible by $|H|$, putting severe restrictions on the possible subgroups.  For instance, if $|G|$ is a prime number, then $G$ has no nontrivial proper subgroups! (actually, $G$ is necessarily a cyclic group).  To prove this result we interpret $G$ as an $H$-set.
 
 
 Further examples come from considering the $G$-set $\typesubgroup_G$ of subgroups of $G$ from \cref{sec:subgroups}.  Knowledge about the $G$-set of subgroups is of vital importance for many applications and Sylow's theorems in  \cref{sec:sylow} give the first restriction on what subgroups are possible and how they can interact.  The first step is Cauchy's \cref{thm:cauchys} which says that if $|G|$ is divisible by a prime $p$, then $G$ contains a cyclic subgroup of order $p$.  Sylow's theorems goes further, analyzing subgroups that have cardinality powers of $p$, culminating in very detailed and useful information about the structure of the subgroups with cardinality the maximal possible power of $p$.
@@ -54,7 +54,7 @@ \section{Lagrange's theorem, counting version}
 We start our investigation by giving the version of Lagrange's theorem which has to do with counting, but first we pin down some language.
 \begin{definition}
   \label{def:finitegrd}
-A \emph{finite group}\index{finite group} is a group such that the set $\USym G$ is finite.   If $G$ is a finite group, then the \emph{\gporder}\index{\gporder} $|G|$ is the cardinality of the finite set $\USymG$ (\ie $\USymG:\conncomp\FinSet{|G|}$).
+A \emph{finite group}\index{finite group} is a group such that the set $\USymG$ is finite.   If $G$ is a finite group, then the \emph{\gporder}\index{\gporder} $|G|$ is the cardinality of the finite set $\USymG$ (\ie $\USymG:\conncomp\FinSet{|G|}$).
 \end{definition}
 \begin{example}
   The trivial group has \gporder $1$, the cyclic group $C_n$ of order $n$ has \gporder $n$ %(which is good)
@@ -73,7 +73,7 @@ \section{Lagrange's theorem, counting version}
 If $|H|=|G|$, then $H=G$ (as subgroups of $G$).
 \end{lemma}
 \begin{proof}
-  Consider the $H$ action of $H$ on $G$, \ie the $H$-set $i^*G:\BH\to\Set$ with $i^*G(x)\defequi(\shape_G=\Bi(x))$, so that $G/H$ is just another name for the orbits $i^*G/H\defequi \sum_{x:\BH}i^*G(x)$.  Note that composing with the structure identity $p_i:\shape_G=\Bi(\shape_H)$ gives an equivalence $i^*G(\shape_H)\equiv \USym G$, so that $|i^*G(\shape_H)|= |G|$.
+  Consider the $H$ action of $H$ on $G$, \ie the $H$-set $i^*G:\BH\to\Set$ with $i^*G(x)\defequi(\shape_G=\Bi(x))$, so that $G/H$ is just another name for the orbits $i^*G/H\defequi \sum_{x:\BH}i^*G(x)$.  Note that composing with the structure identity $p_i:\shape_G=\Bi(\shape_H)$ gives an equivalence $i^*G(\shape_H)\equiv \USymG$, so that $|i^*G(\shape_H)|= |G|$.
 
   Lagrange's \cref{thm:lagrange} says that $i^*G$ is a free $H$-set \footnote{\cref{thm:lagrange} doesn't say this at present: fix it} and so all orbits $\mathcal O_x$ are equivalent to the $H$-set $\tilde H(x)=(\shape_H=x$).
 Consequently, the equivalence
@@ -134,34 +134,34 @@ \section{Cauchy's theorem}
 Informally, $\B\CG_p$ consists of pairs $(S,j)$, where $S$ is a set of cardinality $p$ and $j:S\eqto S$ is a cyclic permutation in the sense that for $0<k<p$ we have that $j^k$ is not $\refl{}$ while $j^p=\refl{}$.
 % Note also that $j^?:\bn p\to ((S,j)=(S,j))$ given by $j^?(k)=j^k$ is an equivalence (just as for the integers, a symmetry of $(S,j)$, \ie an $f:S=S$ so that $fj=jf$, must be $j^k$ for some $k:\bn p$, and if $k\neq l$, then $j^k\neq j^l$)
 % If $(S,j):BC_p$ let
-% $$A(S,j)\defequi ((S,j)=(S,j)\to \USym G).$$  Since we have an equivalence $j^?:\bn p\to ((S,j)=(S,j))$ we get that $J:A(S,j)\to \prod_{\bn p}\USym G$ given by $J(g)=(g_{j^0},g_{j_1},\dots,g_{j^{p-1}})$ is an equivalence.
+% $$A(S,j)\defequi ((S,j)=(S,j)\to \USymG).$$  Since we have an equivalence $j^?:\bn p\to ((S,j)=(S,j))$ we get that $J:A(S,j)\to \prod_{\bn p}\USymG$ given by $J(g)=(g_{j^0},g_{j_1},\dots,g_{j^{p-1}})$ is an equivalence.
 Given a set $A$, a function $a : \bn p \to A$ is an ordered $p$-tuple of elements of $A$: it suffices to write $a_i$ for $a(i)$ to retrieve the
 usual notations for tuples. Given $(S,j) : \B\CG_p$ however, functions $S \to A$ cannot really be thought the same because $S$ is not
 explicitely enumerated. But as soon as we are given $q : \zet/p \eqto (S,j)$, then functions $S \to A$ are just as good to model ordered
-$p$-tuples of $A$ (just by precomposing with the first projection of $q$). With this in mind, define $\mu_p : (\bn p \to \USym G) \to \USym G$
+$p$-tuples of $A$ (just by precomposing with the first projection of $q$). With this in mind, define $\mu_p : (\bn p \to \USymG) \to \USymG$
 to be the $p$-ary multiplication, meaning $\mu_p (g) \defequi g_0g_1\ldots g_{p-1}$. Then, one can define
-$\mu:\prod_{(S,j):\B\CG_p}(\zet/p \eqto (S,j)) \to (S\to\USym G)\to \USym G$ by $\mu_{(S,j)}(q)(g)\defequi (gq)_{0}\cdot\dots\cdot (gq)_{p-1}$
+$\mu:\prod_{(S,j):\B\CG_p}(\zet/p \eqto (S,j)) \to (S\to\USymG)\to \USymG$ by $\mu_{(S,j)}(q)(g)\defequi (gq)_{0}\cdot\dots\cdot (gq)_{p-1}$
 (where we use $gq$ abusively to denote the composition of $g$ with the equivalence given by applying the first projection to the identification
 $q$). We can now define the $\CG_p$-set $X:\B\CG_p\to\Set$ as:
-$$X(S,j)\defequi\sum_{g:S \to \USym G}\prod_{q : \zet/p \eqto (S,j)}\mu_{(S,j)}(q)(g) = e_G.$$
-% The map from $X(S,j)$ to the $p-1$-fold product of $\USym G$ with itself sending $(g,!)$ to $(g_{j^1},\dots,g_{j^{p-1}})$ is an equivalence ($\mu_{(S,j)}g=e_G$ says exactly that $g_{j^0}$ can be reconstructed as $(g_{j^1}\cdot\dots\cdot g_{j^{p-1}})^{-1}$), so $X(S,j)$ is a set of cardinality $p-1$ times the \gporder of $G$.  In particular, $p$ divides the \gporder of $X(S,j)$.
+$$X(S,j)\defequi\sum_{g:S \to \USymG}\prod_{q : \zet/p \eqto (S,j)}\mu_{(S,j)}(q)(g) = e_G.$$
+% The map from $X(S,j)$ to the $p-1$-fold product of $\USymG$ with itself sending $(g,!)$ to $(g_{j^1},\dots,g_{j^{p-1}})$ is an equivalence ($\mu_{(S,j)}g=e_G$ says exactly that $g_{j^0}$ can be reconstructed as $(g_{j^1}\cdot\dots\cdot g_{j^{p-1}})^{-1}$), so $X(S,j)$ is a set of cardinality $p-1$ times the \gporder of $G$.  In particular, $p$ divides the \gporder of $X(S,j)$.
 
 % Specializing to $(S,j)$ being $\zet/p$ and allowing to index the elements in $A(\zet/p)$ with $i:\bn p$ (instead of the very awkward ``$(\sqrt[p]\id)^i$'' as purism would dictate) we proceed as follows.
 In particular, an element of $X(\zet/p)$ is a tuple $(g_0,\dots,g_{p-1})$ satisfying that $g_{\sigma 0}\dots g_{\sigma (p-1)} = e_G$ for every
 $\sigma : \USym \CG_p$. Note that this is equivalent to the set of tuples $(g_0,\dots,g_{p-1})$ satisfying that $g_{0}\dots g_{(p-1)} =
-e_G$. So, the map $X(\zet/p)\to \USym G^{p-1}$ that send an element $(g_0,\dots,g_{p-1})$ to $(g_1,\dots,g_{p-1})$ is an equivalence (the
+e_G$. So, the map $X(\zet/p)\to \USymG^{p-1}$ that send an element $(g_0,\dots,g_{p-1})$ to $(g_1,\dots,g_{p-1})$ is an equivalence (the
 condition $g_{0}\dots g_{(p-1)} = e_G$ says exactly that we can reconstruct $g_0$ from $(g_1,\dots,g_{p-1})$). In particular, $p$ divides the
 \gporder of $X(\zet/p)$.%
 
 Now, a $\CG_p$-fixed point of $X$, that is an element $f: \prod_{(S,j):\B\CG_p}X(S,j)$, will have $f_{\zet/p}$ being an element
 $(g_0,\dots,g_{p-1})$ of $X(\zet/p)$ that satisfies (in particular) $(g_0,\dots,g_{p-1})=(g_1,\dots,g_{p-1},g_0)$, \ie{} such that
 $g_0=g_1=g_2=\dots=g_{p-1}$. In other words, a fixed point $f$ is such that $f_{\zet/p}: X(\zet/p)$ is of the form $(g,\dots,g)$ where $g$
-satisfies $g^p=e_G$. So, there is a map $\ev : X^{\CG_p} \to \sum_{g:\USym G}g^p=e_G$ simply given by evaluation at $\zet/p$. This map is an
-equivalence. Indeed, each fiber of $\ev$ is already a proposition, and we only need to show that each is inhabited. Given any $g : \USym G$ such
-that $g^p = e_G$, and given $(S,j) : \B\CG_p$, one can consider the constant function $\hat g : S \to \USym G$ given by $\hat g (s) = g$ for all
+satisfies $g^p=e_G$. So, there is a map $\ev : X^{\CG_p} \to \sum_{g:\USymG}g^p=e_G$ simply given by evaluation at $\zet/p$. This map is an
+equivalence. Indeed, each fiber of $\ev$ is already a proposition, and we only need to show that each is inhabited. Given any $g : \USymG$ such
+that $g^p = e_G$, and given $(S,j) : \B\CG_p$, one can consider the constant function $\hat g : S \to \USymG$ given by $\hat g (s) = g$ for all
 $s:S$. Then, for all $q : \zet/p \eqto (S,j)$, $\hat g q$ is the tuple $(g,\dots,g)$, so that we have $(\hat g, !) : X(S,j)$. In other words, we
 just constructed a fixed point of $X$ whose image through $\ev$ is $g$, that is an element of the fiber of $\ev$ at $g$. In particular,
-$X^{\CG_p}$ is not empty as it is equivalent to $\sum_{g:\USym G}g^p=e_G$, which contains at least $e_G$.
+$X^{\CG_p}$ is not empty as it is equivalent to $\sum_{g:\USymG}g^p=e_G$, which contains at least $e_G$.
 
 Now, \cref{lem:fixedptsize} claims that $p$ divides the cardinality of $X^{\CG_p}$, and since there \emph{are} fixed points, there must be
 at least $p$ fixed points.  One of them is the trivial one (given by $g\defeq e_G$ above), but the others are nontrivial.
@@ -172,7 +172,7 @@ \section{Cauchy's theorem}
 
 \footnote{Two slight variations commented away.  Have to choose one.  The first needs some background essentially boiling down to $BC_n$ being the truncation of the $n$th Moore space.
 % ALTERNATIVELY:
-%   Consider  the $p-1$-fold product $(\bn{(p-1)}\to \USym G)$ of $\USym G$ with itself.  We give this set the structure of a $\ZZ/p$-set as follows: ((here it is convenient to say that a $\ZZ/p$-set is the same as a $\ZZ$-set commuting with the $p$-fold cover)) define $X:S^1\to\Set$ by $X(\base)\defequi (\bn{(p-1)}\to \USym G)$ and by setting $X(\Sloop)$ to be the element in $X(\base)=(\base)$ sending $(g_1,\dots,g_{p-1})$ to $(g_2,\dots,g_{p-1},(g_1\cdots g_{p-1})^{-1})$.  Note that
+%   Consider  the $p-1$-fold product $(\bn{(p-1)}\to \USymG)$ of $\USymG$ with itself.  We give this set the structure of a $\ZZ/p$-set as follows: ((here it is convenient to say that a $\ZZ/p$-set is the same as a $\ZZ$-set commuting with the $p$-fold cover)) define $X:S^1\to\Set$ by $X(\base)\defequi (\bn{(p-1)}\to \USymG)$ and by setting $X(\Sloop)$ to be the element in $X(\base)=(\base)$ sending $(g_1,\dots,g_{p-1})$ to $(g_2,\dots,g_{p-1},(g_1\cdots g_{p-1})^{-1})$.  Note that
 % $$\xymatrix{S^1\ar[d]_{(-)^p}\ar[dr]^X&\\S^1\ar[r]_X&\Set}$$
 % commutes and so we get a $\ZZ/p$-action ((expand)).  A fixed point of this action is an element of $X(\base)$ of the form $(g,g,\dots,g)$ such that $g^{p-1}=g^{-1}$  (expand)). The choice $g=e$ always gives such a fixed point, but if there is any other point ... (do you need decidability here?), the pointed map $S^1\to \BG$ given by sending $\Sloop$ to $g$ gives the desired subgroup. Hence we need to show that $X$ has fixed points.
 
@@ -184,16 +184,16 @@ \section{Cauchy's theorem}
 % ALTERNATIVELY:
 % Recall the cyclic group of order $p$.  For the sake of convenience, we identify $\bn p\times\bn 1$ and $\bn p$, so that elements will be denoted $0,1,2,\dots$ and not $(0,0), (1,0),(2,0)\dots$, and we also write $s$ instead of $\sqrt[p]{\id}$ for the element in $\bn p=\bn p$ that shifts to the successor (mod $p$).  In other words we choose an identification $\zet/p=(\bn p,s)$ once and for all.  If
 % $$(S,j):BC_p\defequi\sum_{S:\UU}\sum_{j:X=X}||(S,j)=(\bn p,s)||,$$ consider the set
-% $$X(S,j)\defequi\sum_{g:S\to \USym G}||\sum_{a:(S,j)=(\bn p,s)}g_{a^{-1}(0)}\cdot\dots\cdot g_{a^{-1}(p-1)}=e_G||.$$
-% Note that if $g:\bn p\to\USym G$, then we have an equality of propositions
+% $$X(S,j)\defequi\sum_{g:S\to \USymG}||\sum_{a:(S,j)=(\bn p,s)}g_{a^{-1}(0)}\cdot\dots\cdot g_{a^{-1}(p-1)}=e_G||.$$
+% Note that if $g:\bn p\to\USymG$, then we have an equality of propositions
 % $$||\sum_{a:(\bn p,s)=(\bn p,s)}g_{a^{-1}(0)}\cdot\dots\cdot g_{a^{-1}(p-1)}=e_G||=g_{0}\cdot\dots\cdot g_{p-1}=e_G$$
 % since $(g_1\cdot\dots\cdot g_{p-1}=e_G)=(g_2\cdot\dots\cdot g_{p-1}\cdot g_1=e_G)$,
 % and so
-% $$X(\bn p,s)=\sum_{g:\bn p\to \USym G}g_{0}\cdot\dots\cdot g_{p-1}=e_G$$
-% which is equivalent to the $p-1$-fold product of $\USym G$ with itself ($g_0=(g_1\cdot\dots\cdot g_{p-1})^{-1}$, but all the other $g_i$s are then chosen arbitrarily).  Consequently, the number of elements in $X(\bn p,s)$ is divisible by $p$.
+% $$X(\bn p,s)=\sum_{g:\bn p\to \USymG}g_{0}\cdot\dots\cdot g_{p-1}=e_G$$
+% which is equivalent to the $p-1$-fold product of $\USymG$ with itself ($g_0=(g_1\cdot\dots\cdot g_{p-1})^{-1}$, but all the other $g_i$s are then chosen arbitrarily).  Consequently, the number of elements in $X(\bn p,s)$ is divisible by $p$.
 
 % Now, a $C_p$-fixed point of $X(\bn p,s)$ is an element $(g_0,\dots,g_{p-1},!)$ such that $(g_0,\dots,g_{p-1},!)=(g_1,\dots,g_{p-1},g_0,!)$, \ie $g_0=g_1=g_2=\dots=g_{p-1}$\footnote{if I am allowed to write that}.  In other words, a fixed point is of the form $(g,\dots,g,!)$, where $!$ expresses that $g^p=e_G$:
-% $$X(\bn p,s)^{C_p}=\sum_{g:\USym G}g^p=e_G.$$  If we can show that $(g,!):X(\bn p,s)^{C_p}$ is nonempty, we'd have established an abstract cyclic subgroup consisting of the powers of $g$.  Of course, setting $g=e_G$ will give us such a fixed point.
+% $$X(\bn p,s)^{C_p}=\sum_{g:\USymG}g^p=e_G.$$  If we can show that $(g,!):X(\bn p,s)^{C_p}$ is nonempty, we'd have established an abstract cyclic subgroup consisting of the powers of $g$.  Of course, setting $g=e_G$ will give us such a fixed point.
 
 %  Now, $X(\bn p,s)$ splits as a disjoint union of its orbits.  Since $p$ is prime, the group $C_p$ is simple ((where proved?)), and so the orbits are either singletons (fixed points) or free orbits:
 % $$X(\bn p,s)= X(\bn p,s)^{C_p}+(\text{free part of }X(\bn p,s)).$$
@@ -209,7 +209,7 @@ \section{Cauchy's theorem}
 \begin{proof}
   Recall the $G$-set $\Ad_G:\BG\to\Set$ given by $\Ad_G(z)=(z=z)$.
 Then the map
-$$\ev_{\shape_G}:\prod_{z:\BG}(z=z)\to\USym G,\quad \ev_G(f)=f(\shape_G)$$
+$$\ev_{\shape_G}:\prod_{z:\BG}(z=z)\to\USymG,\quad \ev_G(f)=f(\shape_G)$$
 has the structure of a (n abstract) inclusion of a subgroup; namely the inclusion of the center $Z(G)$ in $G$.
 The center thus represents the fixed points of the $G$-set $\Ad_G$.
 Since $G$ has \gporder a power of $p$, all orbits but the fixed points have cardinality divisible by $p$.
@@ -221,7 +221,7 @@ \section{Cauchy's theorem}
 \end{corollary}
 \begin{proof}
   The center $Z(G)$ is by \cref{lem:nontrivcenter} of \gporder $p$ or $p^2$.
-  Since $G$ is not cyclic we have that $g^p=e_G$ for all $g:\USym G$.
+  Since $G$ is not cyclic we have that $g^p=e_G$ for all $g:\USymG$.
 \footnote{((To be continued: the classical proof involves choosing nontrivial elements  -- see what can be done about that.  At present this corollary is not used anywhere))}
 \end{proof}
 \section{Sylow's Theorems}