Rewrite symmetric groups

src/Advanced/Symmetric Groups/parts/03 subgroup.typ

@@ -0,0 +1,172 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.4.2"
+#import "../macros.typ": *
+
+= Subgroups
+
+#problem(label: "s2s3share")
+What elements do $S_2$ and $S_3$ share?
+
+#v(2cm)
+
+
+Consider the sets $\{1, 2\}$ and $\{1,2,3\}$. Clearly, $\{1, 2\} subset \{1, 2, 3\}$. \
+Can we say something similar about $S_2$ and $S_3$?
+
+#v(2mm)
+
+Looking at @s2s3share, we may want to say that $S_2 subset S_3$ since every element of $S_2$ is in $S_3$. \
+This however, isn't as interesting as it could be. Remember that $S_2$ and $S_3$ are _groups_, not _sets_: \
+their elements come with structure, which the "subset" relation does not capture.
+
+#v(2mm)
+
+To account for this, we'll define a similar relation: subgroups.
+
+#definition("Subgroup")
+Let $G$ and $G'$ be groups. We say $G'$ is a _subgroup_ of $G$ (and write $G' subset G$) if the following are true:\
+(Note that $x, y$ are elements of $G$, and $x y$ is multiplication in $G$)
+- the set of elements in $G'$ is a subset of the set of elements in $G$.
+- the identity of $G$ is in $G'$
+- $x,y in G' => x y in G'$
+- $x in G' => x^(-1) in G'$
+
+The above definition may look fairly scary, but the idea behind a subgroup is simple. \
+Consider $S_3$ and $S_4$, the groups of permutations of $3$ and $4$ elements. \
+
+#v(2mm)
+
+Say we have a set of four elements and only look at the first three. \
+$S_3$ fully describes all the ways we can arrange those three elements:
+
+#table(
+  columns: (1fr,),
+  align: center,
+  stroke: none,
+  align(center, cetz.canvas({
+    import cetz.draw: *
+
+    let s = 0.7
+
+    set-style(content: (frame: "rect", stroke: none, fill: white, padding: .1))
+    content((0 * s, 0.5 * s), $1$, name: "1a")
+    content((1 * s, 0.5 * s), $2$, name: "2a")
+    content((2 * s, 0.5 * s), $3$, name: "3a")
+    content((3 * s, 0.5 * s), $4$, name: "4a")
+
+    content((0 * s, -2 * s), $2$, name: "2b")
+    content((1 * s, -2 * s), $3$, name: "3b")
+    content((2 * s, -2 * s), $1$, name: "1b")
+    content((3 * s, -2 * s), $4$, name: "4b")
+
+
+    // These arrows are wrong,
+    // but create a symmetric picture
+    markline(s, "1a", "1b")
+    markline(s, "2a", "3b")
+    markline(s, "3a", "2b")
+    markline(s, "4a", "4b", c: ogreen)
+
+    content(
+      (1 * s, -0.55 * s),
+      $S_3$,
+      fill: white,
+      stroke: oblue + 0.6mm,
+      padding: 1.3mm,
+    )
+  })),
+)
+
+
+#problem()
+Show that $S_3$ is a subgroup of $S_4$.
+
+#v(1fr)
+#pagebreak()
+
+
+
+#definition("Isomorphism")
+Let $G$ and $H$ be groups. We say that $G$ and $H$ are _isomorphic_ (and write $G tilde.equiv H$) \
+if there is a bijection $f: G -> H$ with the following properties:
+- $f(e_G) = e_H$, where $e_G$ is the identity in $G$
+- $f(x^(-1)) = f(x)^(-1)$ for all $x$ in $G$
+- $f(x y) = f(x) f(y)$ for all $x, y$ in $G$
+
+Intuitively, you can think of isomorphism as a form of equivalence. \
+If two groups are isomorphic, they only differ by the names of their elements. \
+The function $f$ above tells us how to map one set of labels to the other.
+
+
+
+#problem()
+Show that $ZZ_7^times$ and $ZZ_9^times$ are isomorphic.
+#hint[
+  Build a bijection with the above properties. \
+  Remember that a group is fully defined by its multiplication table.
+]
+#v(1fr)
+
+
+#problem()
+Show that $ZZ_10^times$, $ZZ_5^times$, and $ZZ_4$ are isomorphic.
+#hint[
+  Build a bijection with the above properties. \
+  Remember that a group is fully defined by its multiplication table.
+]
+
+
+#v(1fr)
+
+#problem()
+Show that isomorphism is transitive. \
+That is, if $A tilde.equiv B$ and $B tilde.equiv C$, then $A tilde.equiv C$.
+
+#v(1fr)
+#pagebreak()
+
+
+#problem(label: "firstindex")
+How many subgroups of $S_4$ are isomorphic to $S_3$? \
+
+#v(1fr)
+
+
+
+
+#problem()
+What are the orders of $S_3$ and $S_4$? \
+How is this related to @firstindex?
+
+#solution[
+  $|S_4| = |S_3| times [S_4 : S_3]$
+
+  #v(2mm)
+
+  This solution is written using index notation, \
+  but the class doesn't need to know what it means yet.
+]
+
+#v(1fr)
+
+#problem()
+$S_4$ also has $S_2$ and the trivial group as subgroups. \
+How many instances of each does $S_4$ contain?
+
+#v(1fr)
+
+
+#problem()
+$(ZZ_4, +)$ is also a subgroup of $S_4$. Find it! \
+How many subgroups of $ZZ_4$ are isomorphic to $S_4$?
+
+#solution[
+  A good hint is "look at generators."
+
+  #v(4mm)
+
+  There are four instances of $ZZ_4$ in $S_4$, each of which is generated by a 4-cycle of $S_n$. \
+  (i.e, the group generated by $(1234)$ is isomorphic to $ZZ_4$)
+]
+
+#v(1fr)