我不知道为什么证明的结尾显示在下一行，我希望它出现在(iii)
\documentclass[11pt, a4paper]{report}
% \usepackage{eurosym}% you probably don't need this (most fonts have euro)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% \usepackage{amsmath} % you load this below
%%\usepackage{amsfonts} % you load this below
\usepackage{bm}
\usepackage{amsfonts, graphicx, verbatim, amsmath,amssymb}
\usepackage{color}
% \usepackage{lipsum} % only for demos
\usepackage{array}
\usepackage{setspace}% if you must (for double spacing thesis)
\usepackage{fancyhdr}
\usepackage{enumitem}
\usepackage{tikz}
%\setcounter{MaxMatrixCols}{10} 10 is teh default anyway
%TCIDATA{OutputFilter=Latex.dll}
%TCIDATA{Version=5.50.0.2953}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{BibliographyScheme=Manual}
%TCIDATA{LastRevised=Sunday, November 26, 2017 16:01:29}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
% These are sort of OK, but better to use geometry package
% to set a consistent set of page dimensions
\setlength{\textheight}{22cm}\setlength{\textwidth}{16cm}
\setlength{\topmargin}{-1.5cm}
\setlength{\oddsidemargin}{-0.5cm}\setlength{\evensidemargin}{-0.5cm}
\setlength{\textheight}{24cm}\setlength{\textwidth}{16.5cm}
\setlength{\topmargin}{-1.5cm}
\setlength{\oddsidemargin}{0.5cm}\setlength{\evensidemargin}{0.5cm}
% This discards its argument, is that intended?
% \U{wibble} is same as \U{zzzzz}
\providecommand{\U}[1]{\protect\rule{.1in}{.1in}}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{Theorem}[theorem]{Theorem}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{Fact}[theorem]{Fact}
% Why all these variant corollary forms?
\newtheorem{corollary}[theorem]{Corollary}
%\newtheorem{corol}[theorem]{Corollary}
%\newtheorem{Corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
% why variant definition forms?
\newtheorem{definition}[theorem]{Definition}
%\newtheorem{Definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
% as above
\newtheorem{lemma}[theorem]{Lemma}
%\newtheorem{Lemma}[theorem]{Lemma}
\newtheorem{fact}[theorem]{Fact}
\newtheorem{lma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
% as above
\newtheorem{proposition}[theorem]{Proposition}
%\newtheorem{prop}[theorem]{Proposition}
%as above
\newtheorem{Property}[theorem]{Property}
%\newtheorem{property}[theorem]{Property}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{Comment}[theorem]{Comment}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
% it would be better to use amsthm package for the theorem definitions
% That defines a more extensive proof environment
\newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\newcommand{\ve}{\varepsilon}
%Better to use \mathrm than \text so it uses the same font in all contexts
\newcommand{\cvgpr}{\xrightarrow{\text{\upshape\tiny P}}}
\newcommand{\cvgdist}{\xrightarrow{\mathrm{d}}}
\newcommand{\G}{{\mathcal{G}}}
% \mathcal{K} not {\cal K} %\cal not defined by default since 1993
%\newcommand{\Kx}{{\cal K}}
%\newcommand{\tod}{\to^{\cal D}}
\newcommand{\ls}{\limsup_{n\to\infty}}
\newcommand{\rE}{\mathbb{E}}
\newcommand{\A}{{\mathcal{A}}}
\newcommand{\rP}{\mathbb{P}}
\newcommand{\p}{{\mathbb{P}}}
\newcommand{\Z}{{\mathbb{Z}}}
% \mathrm{Be} not {\rm BeK} %\cal not defined by default since 1993
%\newcommand{\Be}{{\rm Be}}
\newcommand{\re}{\mathrm{e}}
\newcommand{\ep}{\varepsilon}
%\newcommand{\Bin}{{\rm Bin}}
\newcommand{\qand}{\quad\mbox{and}\quad}
\newcommand{\quso}{\quad\mbox{so}\quad}
%\newcommand{\Nn}{{\bf N}}
%\newcommand{\St}{\underline{\rm S}}
%\newcommand{\Rt}{\underline{\rm R}}
%\newcommand{\It}{\underline{\rm I}}
%\newcommand{\one}{{\bf 1}}
\newcommand{\Ups}{{\Upsilon}}
\newcommand{\iu}{{i\mkern1mu}}
\newcommand{\II}{{\mathcal{I}}}
%\newcommand{\Var}{{\rm Var}}
%\newcommand{\var}{{\rm Var}}
%\newcommand{\Cov}{{\rm cov}}
%\newcommand{\cov}{{\rm cov}}
%\newcommand{\corr}{{\rm corr}}
%\newcommand{\lhs}{{\rm lhs}}
%\newcommand{\rhs}{{\rm rhs}}
\newcommand{\ra}{\rightarrow}
\newcommand{\I}{{\mathbf 1}}
\newcommand{\R}{{\mathbb R}}
\newcommand{\N}{{\mathbb N}}
\newcommand{\LL}{{\mathbb L}}
\newcommand{\E}{{\mathbb{E}}}
%\newcommand{\bin}{{\rm Bin}}
%\newcommand{\Pois}{{\rm Pois}}
%\newcommand{\Po}{{\rm Pois}}
%\newcommand{\Bi}{{\cal B}}
\newcommand{\ri}{\mathrm{i}}
\newcommand{\rd}{\mathrm{d}}
\newcommand{\XXi}{\Xi_{k,m}^{(n)}}
\newcommand{\xxi}{\bar{\xi}}
\newcommand{\qedhere}{{\diamond}}
\newcommand{\eqdef}{\stackrel{\mathrm{def}}{=}}
\newcommand{\eqdist}{\stackrel{\mathrm{D}}{=}}
\newcommand{\braket}[2]{{\langle{#1|#2}\rangle}}
\newcommand{\independent}{\perp}
% use amsmath package (that you have loaded) align environment, not eqnarray
%\newcommand{\bb}{\begin{eqnarray*}}
%\newcommand{\ee}{\end{eqnarray*}}
%\newcommand{\bbb}{\begin{eqnarray}}
%\newcommand{\eee}{\end{eqnarray}}
\newcommand{\F}{{\mathcal{F}}}
\newcommand{\qed}{$\diamond$}
\newcommand{\cross}{\mathbin{\tikz [x=1.4ex,y=1.4ex,line width=.075ex] \draw (0,0) -- (1,1) (0,1) -- (1,0);}}%
% \parindent 0pt % this is just low level version of following line
% \setlength{\parindent}{0pt}%
% use parskip package (if you must) to stop indent and put vertical space betwen paragraphs
% although most documents lok better with traditional typesetting with indentation and no vertical space
\usepackage{parskip}
%\newcommand{\forceindent}{\leavevmode{\parindent=3em\indent}}%eek
%\input{tcilatex}
\begin{document}
\begin{proposition}
If $\chi$ is the character of a representation $\rho$ of degree $n$, then
\begin{enumerate}[label=(\roman*)]
\item $\chi(1) = n$
\item $\chi(s^{-1}) = \chi(s)^*\quad for s \in G$ \qquad (where $*$ denotes complex conjugation)
\item $\chi(tst^{-1})=\chi(s) \quad for s,t \in G$
\end{enumerate}
\end{proposition}
\begin{proof}
\begin{enumerate}[label=(\roman*)]
\item We have $\rho(1) = 1$, and $Tr(1)=n$ since $V$ has dimension $n$.
\item $\rho(s^\alpha)=I$ for $\alpha$ large enough. It follows that the eigenvalues $\lambda_i$ of $\rho_s$ are roots of unity. Now with $*$ complex conjugation we have:
\[
\chi(s)^* = Tr(\rho_s)^* = \sum \lambda_i^* = \sum \lambda^{-1} = Tr(\rho_s^{-1}) = Tr(\rho_{s^{-1}}) = \chi(s^{-1})
\]
\item Using the widely known formula, $Tr(ab)=Tr(ba)$ the result follows.
\end{enumerate}
\end{proof}
\end{document}
编辑：使用amsthm包：
\documentclass[11pt, a4paper]{report}
% \usepackage{eurosym}% you probably don't need this (most fonts have euro)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% \usepackage{amsmath} % you load this below
%%\usepackage{amsfonts} % you load this below
\usepackage{bm}
\usepackage{amsfonts, graphicx, verbatim, amsmath,amssymb, amsthm}
\usepackage{color}
% \usepackage{lipsum} % only for demos
\usepackage{array}
\usepackage{setspace}% if you must (for double spacing thesis)
\usepackage{fancyhdr}
\usepackage{enumitem}
\usepackage{tikz}
\usepackage{parskip}
%\setcounter{MaxMatrixCols}{10} 10 is teh default anyway
%TCIDATA{OutputFilter=Latex.dll}
%TCIDATA{Version=5.50.0.2953}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{BibliographyScheme=Manual}
%TCIDATA{LastRevised=Sunday, November 26, 2017 16:01:29}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
% These are sort of OK, but better to use geometry package
% to set a consistent set of page dimensions
\setlength{\textheight}{22cm}\setlength{\textwidth}{16cm}
\setlength{\topmargin}{-1.5cm}
\setlength{\oddsidemargin}{-0.5cm}\setlength{\evensidemargin}{-0.5cm}
\setlength{\textheight}{24cm}\setlength{\textwidth}{16.5cm}
\setlength{\topmargin}{-1.5cm}
\setlength{\oddsidemargin}{-0.1cm}\setlength{\evensidemargin}{-0.1cm}
% This discards its argument, is that intended?
% \U{wibble} is same as \U{zzzzz}
\providecommand{\U}[1]{\protect\rule{.1in}{.1in}}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{Theorem}[theorem]{Theorem}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{Fact}[theorem]{Fact}
% Why all these variant corollary forms?
\newtheorem{corollary}[theorem]{Corollary}
%\newtheorem{corol}[theorem]{Corollary}
%\newtheorem{Corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
% why variant definition forms?
\newtheorem{definition}[theorem]{Definition}
%\newtheorem{Definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
% as above
\newtheorem{lemma}[theorem]{Lemma}
%\newtheorem{Lemma}[theorem]{Lemma}
\newtheorem{fact}[theorem]{Fact}
\newtheorem{lma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
% as above
\newtheorem{proposition}[theorem]{Proposition}
%\newtheorem{prop}[theorem]{Proposition}
%as above
\newtheorem{Property}[theorem]{Property}
%\newtheorem{property}[theorem]{Property}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{Comment}[theorem]{Comment}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
% it would be better to use amsthm package for the theorem definitions
% That defines a more extensive proof environment
\newcommand{\ve}{\varepsilon}
%Better to use \mathrm than \text so it uses the same font in all contexts
\newcommand{\cvgpr}{\xrightarrow{\text{\upshape\tiny P}}}
\newcommand{\cvgdist}{\xrightarrow{\mathrm{d}}}
\newcommand{\G}{{\mathcal{G}}}
% \mathcal{K} not {\cal K} %\cal not defined by default since 1993
%\newcommand{\Kx}{{\cal K}}
%\newcommand{\tod}{\to^{\cal D}}
\newcommand{\ls}{\limsup_{n\to\infty}}
\newcommand{\rE}{\mathbb{E}}
\newcommand{\A}{{\mathcal{A}}}
\newcommand{\rP}{\mathbb{P}}
\newcommand{\p}{{\mathbb{P}}}
\newcommand{\Z}{{\mathbb{Z}}}
% \mathrm{Be} not {\rm BeK} %\cal not defined by default since 1993
%\newcommand{\Be}{{\rm Be}}
\newcommand{\re}{\mathrm{e}}
\newcommand{\ep}{\varepsilon}
%\newcommand{\Bin}{{\rm Bin}}
\newcommand{\qand}{\quad\mbox{and}\quad}
\newcommand{\quso}{\quad\mbox{so}\quad}
%\newcommand{\Nn}{{\bf N}}
%\newcommand{\St}{\underline{\rm S}}
%\newcommand{\Rt}{\underline{\rm R}}
%\newcommand{\It}{\underline{\rm I}}
%\newcommand{\one}{{\bf 1}}
\newcommand{\Ups}{{\Upsilon}}
\newcommand{\iu}{{i\mkern1mu}}
\newcommand{\II}{{\mathcal{I}}}
%\newcommand{\Var}{{\rm Var}}
%\newcommand{\var}{{\rm Var}}
%\newcommand{\Cov}{{\rm cov}}
%\newcommand{\cov}{{\rm cov}}
%\newcommand{\corr}{{\rm corr}}
%\newcommand{\lhs}{{\rm lhs}}
%\newcommand{\rhs}{{\rm rhs}}
\newcommand{\ra}{\rightarrow}
\newcommand{\I}{{\mathbf 1}}
\newcommand{\R}{{\mathbb R}}
\newcommand{\N}{{\mathbb N}}
\newcommand{\LL}{{\mathbb L}}
\newcommand{\E}{{\mathbb{E}}}
%\newcommand{\bin}{{\rm Bin}}
%\newcommand{\Pois}{{\rm Pois}}
%\newcommand{\Po}{{\rm Pois}}
%\newcommand{\Bi}{{\cal B}}
\newcommand{\ri}{\mathrm{i}}
\newcommand{\rd}{\mathrm{d}}
\newcommand{\XXi}{\Xi_{k,m}^{(n)}}
\newcommand{\xxi}{\bar{\xi}}
\newcommand{\eqdef}{\stackrel{\mathrm{def}}{=}}
\newcommand{\eqdist}{\stackrel{\mathrm{D}}{=}}
\newcommand{\braket}[2]{{\langle{#1|#2}\rangle}}
\newcommand{\independent}{\perp}
% use amsmath package (that you have loaded) align environment, not eqnarray
%\newcommand{\bb}{\begin{eqnarray*}}
%\newcommand{\ee}{\end{eqnarray*}}
%\newcommand{\bbb}{\begin{eqnarray}}
%\newcommand{\eee}{\end{eqnarray}}
\newcommand{\F}{{\mathcal{F}}}
\newcommand{\cross}{\mathbin{\tikz [x=1.4ex,y=1.4ex,line width=.075ex] \draw (0,0) -- (1,1) (0,1) -- (1,0);}}%
% \parindent 0pt % this is just low level version of following line
% \setlength{\parindent}{0pt}%
% use parskip package (if you must) to stop indent and put vertical space betwen paragraphs
% although most documents lok better with traditional typesetting with indentation and no vertical space
%\newcommand{\forceindent}{\leavevmode{\parindent=3em\indent}}%eek
%\input{tcilatex}
\begin{document}
\newpage
\pagestyle{fancy}
\fancyhf{}
\fancyhead[EL]{\nouppercase\leftmark}
\fancyhead[OR]{\nouppercase\rightmark}
\fancyhead[ER,OL]{\thepage}
\maketitle
\chapter{Introduction}
%\bigskip
%\smallskip
\section{Definitions and prerequisites}
Let us recall some basic knowledge, we begin by giving some definitions.
%\medskip
\begin{definition}
A group is a set $G$ together with a binary operation $*$ on $G$ satisfying
the following properties:
\begin{enumerate}[label=(G\arabic*),series=group]
\item Closure: $\forall x,y \in G, x * y \in G$.
\item Associativity: $\forall x,y, z \in G, (x * y) * z = x * (y * z)$.
\item Identity: There is an element $e \in G$ such that $e * x = x * e = x$ for all $x \in G$.
\item Inverses: For any $x \in G$ there is an element $y \in G$ such that $x * y = y * x = e$.
\end{enumerate}
\end{definition}
\begin{definition}
A group $G$ is called an abelian group if the following axiom is satisfied:
\begin{enumerate}[label=(G\arabic*),resume=group]
\item Commutativity: $\forall x,y \in G, x * y = y * x$.
\end{enumerate}
\end{definition}
\end{document}
使用amsthm，间距分解
答案1
请从工作示例中移除所有不必要的杂乱内容，以便获得 MWE。要获得正确结果，请执行以下操作：
- 加载
amsthm包 - 消除
\newcommand{\qed}{...} - 消除
\newcommand{\qedhere}{...} - 消除
\newcommand{\qed}{...} - 消除
\newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
我可以随意删除文档中未使用的一些命令，因此整体代码如下所示：
\documentclass[11pt, a4paper]{report}
\usepackage{bm}
\usepackage{amsfonts, graphicx, verbatim, amsmath,amssymb, amsthm}
\usepackage{color}
\usepackage{array}
\usepackage{setspace}% if you must (for double spacing thesis)
\usepackage{fancyhdr}
\usepackage{enumitem}
\usepackage{tikz}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\usepackage{parskip}
\setlength{\textheight}{22cm}\setlength{\textwidth}{16cm}
\setlength{\topmargin}{-1.5cm}
\setlength{\oddsidemargin}{-0.5cm}\setlength{\evensidemargin}{-0.5cm}
\setlength{\textheight}{24cm}\setlength{\textwidth}{16.5cm}
\setlength{\topmargin}{-1.5cm}
\setlength{\oddsidemargin}{0.5cm}\setlength{\evensidemargin}{0.5cm}
\begin{document}
\begin{proposition}
If $\chi$ is the character of a representation $\rho$ of degree $n$, then
\begin{enumerate}[label=(\roman*)]
\item $\chi(1) = n$
\item $\chi(s^{-1}) = \chi(s)^*\quad for s \in G$ \qquad (where $*$ denotes complex conjugation)
\item $\chi(tst^{-1})=\chi(s) \quad for s,t \in G$
\end{enumerate}
\end{proposition}
\begin{proof}
\begin{enumerate}[label=(\roman*)]
\item We have $\rho(1) = 1$, and $Tr(1)=n$ since $V$ has dimension $n$.
\item $\rho(s^\alpha)=I$ for $\alpha$ large enough. It follows that the eigenvalues $\lambda_i$ of $\rho_s$ are roots of unity. Now with $*$ complex conjugation we have:
\[
\chi(s)^* = Tr(\rho_s)^* = \sum \lambda_i^* = \sum \lambda^{-1} = Tr(\rho_s^{-1}) = Tr(\rho_{s^{-1}}) = \chi(s^{-1})
\]
\item Using the widely known formula, $Tr(ab)=Tr(ba)$ the result follows.
\qedhere
\end{enumerate}
\end{proof}
\end{document}
请注意如何使用\qedhere您所要求的内容。