From 2ba5d67adce0d704112c385c5bebd59fe242b39a Mon Sep 17 00:00:00 2001
From: Robert <9291792+bushshrub@users.noreply.github.com>
Date: Sat, 7 Sep 2024 23:26:51 -0400
Subject: [PATCH] whoops, made a typo. also added a new problem

---
 ...ter-group-presentations-and-generators.tex |  8 +++++++
 chapter-lagrange-theorem.tex                  | 22 ++++++++++++++-----
 2 files changed, 25 insertions(+), 5 deletions(-)

diff --git a/chapter-group-presentations-and-generators.tex b/chapter-group-presentations-and-generators.tex
index b9a82e0..b5ac099 100644
--- a/chapter-group-presentations-and-generators.tex
+++ b/chapter-group-presentations-and-generators.tex
@@ -243,4 +243,12 @@ \subsection{Problems and Exercises}
     \end{enumerate}
 \end{exercise}
 
+\begin{prob}    
+    Let $G$ be a finitely generated group, and suppose that $[G:H]$ is finite.
+    Prove that $H$ is finitely generated. \textit{(You might need the content
+    from \cref{chapter:lagrange-theorem} to do this.)}
+
+    \textit{Bonus: Try to find a proof of this fact using algebraic topology}
+\end{prob}
+
 \end{document}
diff --git a/chapter-lagrange-theorem.tex b/chapter-lagrange-theorem.tex
index 9bc7a8f..8d91af3 100644
--- a/chapter-lagrange-theorem.tex
+++ b/chapter-lagrange-theorem.tex
@@ -138,7 +138,12 @@
 We again direct our attention to the power of definitions. Having the correct
 choice of equivalence relation made the proof of Lagrange's Theorem very easy.
 As such, it would do a lot of good to understand how such an equivalence
-relation was chosen.
+relation was chosen. Lagrange's theorem now motivates the following definition:
+the \emph{index of a subgroup}. If $H$ is a subgroup of $G$, then we let $[G:H]$
+denote the number of left cosets of $H$. This is called the \emph{index of $H$
+in $G$}. We leave it to the reader to verify that $[G:H]$ is the same number if
+we used right cosets instead of left. If there are infinitely many cosets, we
+write $[G:H] = \infty$.
 
 We now state some corollaries of Lagrange's Theorem. While
 obvious, they are still good to mention.
@@ -199,7 +204,8 @@
     So if $t = h\inv h' = k {k'}\inv$, then it all works out. This shows that
     every element in $HK$ is represented by precisely $\abs{H \cap K}$ products.
 \end{proof}
-The proof here actually leads to a proof of a more general fact, which is outlined in \cref{ex:generalization-of-hk-theorem}.
+The proof here actually leads to a proof of a more general fact, which is
+outlined in \cref{ex:generalization-of-hk-theorem}.
 
 
 Let us now see another application of Lagrange's theorem. This time, we classify
@@ -234,6 +240,11 @@
 
 \subsection{Exercises and Problems}
 
+\begin{exercise}
+    Suppose that $G$ is finite. Let $H \leq G$ and $K \leq H$. Show that $[G:K]
+    = [G:H][H:K]$.
+\end{exercise}
+
 \begin{exercise}
 \label{ex:prove-thm-classifying-order-2p}
     Prove \cref{thm:groups-order-2p} by following the steps below.
@@ -260,9 +271,10 @@ \subsection{Exercises and Problems}
 \begin{exercise}[Generalization of $HK$ theorem]
 \label{ex:generalization-of-hk-theorem}
     Let $H, K$ be subgroups of $G$, and $\alpha: H \times K \to G$ be the map
-    defined by $\alpha(h, k) = hk$. Prove that $\alpha\inv \paren{hk} = \set{(ht, t\inv k): t \in H \cap K}$,
-    and that additionally the cardinality of $\alpha\inv\paren{hk}$ equals to the cardinality of $H \cap K$. 
-    Conclude that if $HK$ has finite cardinality then $\abs{HK} = \abs H \abs K/(\abs{H \cap K})$.
+    defined by $\alpha(h, k) = hk$. Prove that $\alpha\inv \paren{hk} =
+    \set{(ht, t\inv k): t \in H \cap K}$, and that additionally the cardinality
+    of $\alpha\inv\paren{hk}$ equals to the cardinality of $H \cap K$. Conclude
+    that if $H, K$ are finite, then $\abs{HK} = \abs H \abs K/(\abs{H \cap K})$.
 
     See \autocite[Exercise~9,\pno~58]{Jacobson_2009}
 \end{exercise}