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booktabs 包中的脚注

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booktabs 包中的脚注

我想在表格中的某个位置添加脚注。我正在使用booktabs包。我使用编写了脚注\footnote。它在一个小框内显示符号 1，但没有在任何地方显示脚注。

\documentclass[12pt,a4paper]{article}
\usepackage{amsmath,amsthm,amsfonts,latexsym,amscd,amssymb,mathrsfs,hyperref,textcomp,booktabs}
We have already discussed about completeness.   
\begin{table}[ht]
\begin{minipage}{\linewidth}
\def\arraystretch{1.4}\tabcolsep=2pt\small\centering
\begin{tabular}{@{} l l l l l @{}}\toprule
Order&Group $G$&Gap Id&Presentation&$\Gamma_d(G)$\footnote{We shall exclude the cases $d=1,|G|$, since $\Gamma_1(G)=K_1=\Gamma_{|G|}(G)$}\\ \midrule
1&$C_1$&1(1)&$\langle a;a^1\rangle$&$\Gamma_1(C_1)=K_1$ \\
2&$C_2$&2(1)&$\langle a;a^2\rangle$&$\Gamma_1(C_2)=\Gamma_2(C_2)=K_1$ \\ 
3&$C_3$&3(1)&$\langle a;a^3\rangle$&$\Gamma_1(C_3)=\Gamma_3(C_3)=K_1$ \\ 
4&$C_4$&4(1)&$\langle a;a^4\rangle$&$\Gamma_1(C_4)=\Gamma_2(C_4)=\Gamma_4(C_4)=K_1$\\ 
&$C_2\times C_2$&2(2)&$\langle a,b;a^2,b^2,aba^{-1}b^{-1}\rangle$&$\Gamma_1(G)=\Gamma_4(G)=K_1,\Gamma_2(G)=K_3$\\ 
5&$C_5$&5(1)&$\langle a;a^5\rangle$&$\Gamma_1(C_5)=\Gamma_5(C_5)=K_1$\\ \bottomrule
\end{tabular}
\end{minipage}
\caption{Abelian groups}
\end{table}
This table shows $\Gamma_d(G)$ for some abelian groups.

\end{document}

这显示了表格前的行This table shows $\Gamma_d(G)$ for some abelian groups.，表格显示在下一页。另外，我想要符号*或\dagger位于 的位置a，而我不想要脚注前的行。

答案1

使用minipage表示表格。但是，您的表格太宽，当然，您不应该使用太多行。没有它会使它更具可读性：

\documentclass[12pt,a4paper]{article}
\usepackage{amsmath,amsthm,amsfonts,latexsym,amscd,amssymb,mathrsfs,hyperref,textcomp,booktabs}
\hypersetup{colorlinks}
\begin{document}

\begin{table}[ht]
\centering
\begin{minipage}{\linewidth}
\begin{tabular}{|p{0.9cm}|p{1.6cm}|p{1.6cm}|p{4cm}|p{4.5cm}|}
\hline
Order&Group $G$&Gap Id&Presentation&$\Gamma_d(G)$\footnote{We shall exclude the cases $d=1,|G|$, since $\Gamma_1(G)=K_1=\Gamma_{|G|}(G)$}\\ \hline
1&$C_1$&1(1)&$\langle a;a^1\rangle$&$\Gamma_1(C_1)=K_1$ \\ \hline
2&$C_2$&2(1)&$\langle a;a^2\rangle$&$\Gamma_1(C_2)=\Gamma_2(C_2)=K_1$ \\ \hline
3&$C_3$&3(1)&$\langle a;a^3\rangle$&$\Gamma_1(C_3)=\Gamma_3(C_3)=K_1$ \\ \hline
4&$C_4$&4(1)&$\langle a;a^4\rangle$&$\Gamma_1(C_4)=\Gamma_2(C_4)=\Gamma_4(C_4)=K_1$\\ \cline{2-5}
&$C_2\times C_2$&2(2)&$\langle a,b;a^2,b^2,aba^{-1}b^{-1}\rangle$&$\Gamma_1(G)=\Gamma_4(G)=K_1,\Gamma_2(G)=K_3$\\ \hline
5.&$C_5$&5(1)&$\langle a;a^5\rangle$&$\Gamma_1(C_5)=\Gamma_5(C_5)=K_1$\\ \hline
\end{tabular}
\end{minipage}
\caption{Abelian groups}
\end{table}

\begin{table}[ht]
\begin{minipage}{\linewidth}
\def\arraystretch{1.4}\tabcolsep=2pt\small\centering
\begin{tabular}{@{} l l l l l @{}}\toprule
Order&Group $G$&Gap Id&Presentation&$\Gamma_d(G)$\footnote{We shall exclude the cases $d=1,|G|$, since $\Gamma_1(G)=K_1=\Gamma_{|G|}(G)$}\\ \midrule
1&$C_1$&1(1)&$\langle a;a^1\rangle$&$\Gamma_1(C_1)=K_1$ \\
2&$C_2$&2(1)&$\langle a;a^2\rangle$&$\Gamma_1(C_2)=\Gamma_2(C_2)=K_1$ \\ 
3&$C_3$&3(1)&$\langle a;a^3\rangle$&$\Gamma_1(C_3)=\Gamma_3(C_3)=K_1$ \\ 
4&$C_4$&4(1)&$\langle a;a^4\rangle$&$\Gamma_1(C_4)=\Gamma_2(C_4)=\Gamma_4(C_4)=K_1$\\ 
&$C_2\times C_2$&2(2)&$\langle a,b;a^2,b^2,aba^{-1}b^{-1}\rangle$&$\Gamma_1(G)=\Gamma_4(G)=K_1,\Gamma_2(G)=K_3$\\ 
5.&$C_5$&5(1)&$\langle a;a^5\rangle$&$\Gamma_1(C_5)=\Gamma_5(C_5)=K_1$\\ \bottomrule
\end{tabular}
\end{minipage}
\caption{Abelian groups}
\end{table}

\end{document}

在此处输入图片描述

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