答案1
\left ... \right
使用对和的快速方法array
(感谢@AndrewSwannA \middle
):
\documentclass[]{article}
\usepackage[]{amsmath}
\begin{document}
\[
B = \left\langle \sigma_1,\dots,\sigma_{n-1} \middle\vert\,
\begin{array}{@{}cc@{}}
\sigma_i \sigma_j = \sigma_j \sigma_i,
& \lvert i - j \rvert > 1 \\
\sigma_i \sigma_j \sigma_i = \sigma_j \sigma_i \sigma_j,
& \lvert i - j \rvert = 1
\end{array}
\right\rangle
\]
\end{document}
答案2
您可以使用\middle\vert
以下规则和alignedat
条件:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation*}
B_{n} = \left\langle \sigma_{1},\dots,\sigma_{n-1} \;\middle\vert\;
\begin{alignedat}{3}
\sigma_{i}\sigma_{j} &=\sigma_{j}\sigma_{i},&\ &\lvert i-j \rvert
> 1,\\
\sigma_{i}\sigma_{j}\sigma_{i} &=
\sigma_{j}\sigma_{i}\sigma_{j},&&
\lvert i-j \rvert = 1
\end{alignedat}
\right\rangle
\end{equation*}
\end{document}