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I am working on my notes for my classes, and decided that it would be useful to have a list of theorems and definitions for my algebra section. The theorem environments themselves are working fine, but when I use the \listoftheorems command from the thmtools package, there is a page with the heading, but no entries on it.

I am using TexMaker. My code is below.

\documentclass[oneside]{book}
\usepackage[utf8]{inputenc}
\usepackage{amsthm,amsmath,amssymb,amsfonts}
\usepackage{graphicx}
\usepackage[svgnames]{xcolor}
\usepackage{thmtools}
\usepackage{subfiles}
\usepackage{indentfirst}
\usepackage{array}
\usepackage[hidelinks]{hyperref}
\usepackage[left=1in,right=1in,top=1in,bottom=1in, bindingoffset=0in]{geometry}

\setcounter{tocdepth}{0}

\graphicspath{{Images/}{../Images/}} 

\declaretheorem{theorem}
\declaretheorem{example}
\declaretheorem{definition}

\renewcommand{\listtheoremname}{List of Theorems and Definitions}

\begin{document}

\listoftheorems[ignoreall, show={theorem,definition}]

\chapter{Group Theory}
\section{30,000ft View}
Group theory is the theory of reversible action. More formally, it is the theory of symmetry and transformation. 

\begin{example}
Suppose that you have a square,and wish to perform. There are a couple different flips one could do, vertically or horizontally. We could name these as function f,g, which act on the vertex set $V=\{A,B,C,D\}$ like so:

$$f:V\rightarrow V \text{defined by}$$ $$f:A\mapsto B, f:B\mapsto A, f:D\mapsto C, \text{and } f:C\mapsto D$$

$$g:V\rightarrow V \text{defined by}$$ $$g:A\mapsto D, g:D\mapsto A, g:B\mapsto C, \text{and } g:C\mapsto B$$

or the identity function 1, which corresponds to doing nothing.

$$1:V\rightarrow V \text{defined by}$$
$$1:A\mapsto A, 1:B\mapsto B, 1:C\mapsto C,\text{and } 1:D\mapsto D$$.

Since these are functions, we may compose them. In this case, this corresponds to multiple flips. The group G is (informally) the set of all things I can do the the vertex set under these functions.

$$G=\{1,f,g,g\circ f, f\circ g, f\circ g\circ f,...\}$$

This particular group is not actually infinite, because $f=f^{-1}$, $g=g^{-1}$, $f\circ 1=1\circ f= f$ and$g\circ 1=1\circ g= g$. The full group is then just 

$$G=\{1,f,g,f\circ g\}$$

and we can prove this with the multiplication table.

\begin{example} 
Suppose we have two binary switches. Define l=flip the left switch, and r=flip the right switch, and 1=do nothing. These are function of the set of states $S=\{UU,UD,DU,DD\}$. Thus, the group is 

$$H=\{1,l,r,lr\}$$ This is the entire group for the same reasons as in the previous example.
\end{example}

Since H and G have the same multiplication table, up to re-labeling, we say these groups are isomorphic.

Both of these examples were groups of functions, and these were all bijective functions. In the context of group theory, this is called a permutation.

\begin{definition}
Suppose $S$ is a set. Then $f:S\rightarrow S$ is a permutation iff $f$ is a bijection.
\end{definition}

\begin{definition}
For $n\in \mathbb{Z}$, $S_n$ is the set of permutations of the set $\{1,...,n\}$.
\end{definition}

Summary: With $S_n$, we have an associative binary operation such that there exists a two-sided identity, and two sided inverses. This makes it a group.

\end{document}

1 Answer 1

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You've set the tocdepth to 0, so nothing will be displayed. You need to set it to at least 1 to have anything displayed.

\setcounter{tocdepth}{1}

If you want to have a tocdepth for the main TOC at zero, just set the counter to 0 before the main TOC (or in the preamble) and then set it to 1 just before the \listoftheorems.

enter image description here

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  • Is there a way to keep the counter at zero for the actual table of contents, but not the list? Jan 24, 2018 at 1:34
  • @BrandonMyers Yes, see edit.
    – Alan Munn
    Jan 24, 2018 at 1:41

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