confirm the bug fixed

This commit is contained in:
EthanDeng
2022-04-18 18:41:14 +08:00
parent 083483f7b5
commit b3108df9c2
2 changed files with 6 additions and 2 deletions

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@@ -1,4 +1,4 @@
\documentclass[lang=cn,10pt,founder]{elegantbook} \documentclass[lang=cn,10pt]{elegantbook}
\title{ElegantBook优美的 \LaTeX{} 书籍模板} \title{ElegantBook优美的 \LaTeX{} 书籍模板}
\subtitle{Elegant\LaTeX{} 经典之作} \subtitle{Elegant\LaTeX{} 经典之作}

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@@ -535,7 +535,8 @@ Note that a subgroup~$H$ of a group $G$ is itself a left coset of $H$ in $G$.
Let $G$ be a finite group, and let $H$ be a subgroup of $G$. Then the order of $H$ divides the order of $G$. Let $G$ be a finite group, and let $H$ be a subgroup of $G$. Then the order of $H$ divides the order of $G$.
\end{theorem} \end{theorem}
\ref{thm:lg} As theorem \ref{thm:lg} refered.
\lipsum[3] \lipsum[3]
@@ -543,6 +544,9 @@ Let $G$ be a finite group, and let $H$ be a subgroup of $G$. Then the order of $
The content of theorem. The content of theorem.
\end{theorem} \end{theorem}
we can refer this theorem as \ref{thm:label text}.
\begin{proposition}[Size of Left Coset] \begin{proposition}[Size of Left Coset]
Let $H$ be a finite subgroup of a group $G$. Then each left coset of $H$ in $G$ has the same number of elements as $H$. Let $H$ be a finite subgroup of a group $G$. Then each left coset of $H$ in $G$ has the same number of elements as $H$.
\end{proposition} \end{proposition}