在证明中左对齐方程但保持= 对齐

Question

\begin{proof}
Let $g_1, g_2 \in G$. Then $\theta(g_1) = Ng_1$ and $\theta(g_2) = Ng_2$. \\
\begin{equation*}
\begin{split}
\text{ Now, } \theta(g_1g_2) &= Ng_1g_2 \\
& = Ng_1 Ng_2 \\
& = \theta(g_1) \theta(g_2)\\
\end{split}
\end{equation*}
$\therefore \theta$ is a group homomorphism. $\theta$ is clearly onto.
\end{proof}

有没有办法左边证明第一的等式行，但随后的行中 = 符号保持对齐？我尝试了 \begin{align} 等的多种组合，但都没有成功。（看来 \begin{equation} 更倾向于居中对齐。）

Answer 1

你可以用 来完成aligned。我还提供了我坚信更好的版本。

\documentclass{article}
\usepackage{amsmath,amssymb,amsthm}

\begin{document}

\begin{proof}
Let $g_1, g_2 \in G$. Then $\theta(g_1) = Ng_1$ and $\theta(g_2) = Ng_2$. \\
$\begin{aligned}
\text{Now, } \theta(g_1g_2) &= Ng_1g_2 \\
& = Ng_1 Ng_2 \\
& = \theta(g_1) \theta(g_2)\\
\end{aligned}$\\
$\therefore \theta$ is a group homomorphism. $\theta$ is clearly onto.
\end{proof}

\begin{proof}
Let $g_1, g_2 \in G$. Then $\theta(g_1) = Ng_1$ and $\theta(g_2) = Ng_2$. Now
\begin{align*}
\theta(g_1g_2) &= Ng_1g_2 \\
& = Ng_1 Ng_2 \\
& = \theta(g_1) \theta(g_2)
\end{align*}
Therefore $\theta$ is a group homomorphism. $\theta$ is clearly onto.
\end{proof}

\end{document}

Answer 2

另一种解决方案是在环境align*中嵌套：fleqnnccmath

\documentclass{article}
\usepackage{amsthm, amssymb}
\usepackage{mathtools, nccmath}

\begin{document}
\begin{proof}
Let $g_1, g_2 \in G$. Then $\theta(g_1) = Ng_1$ and $\theta(g_2) = Ng_2$. \\
\begin{fleqn}
\begin{align*}
\text{Now, } \theta(g_1g_2) &= Ng_1g_2 \\
& = Ng_1 Ng_2 \\
& = \theta(g_1) \theta(g_2)
\end{align*}
\end{fleqn}
$\therefore \theta$ is a group homomorphism. $\theta$ is clearly onto.
\end{proof}

\end{document} 

Answer 3

像这样？

\documentclass[12pt,a4paper]{article}
\usepackage{mathtools,amssymb}

\begin{document}
\begin{proof}
Let $g_1, g_2 \in G$. Then $\theta(g_1) = Ng_1$ and $\theta(g_2) = Ng_2$. 
    \begin{flalign*}
\text{Now, } \theta(g_1g_2) 
    & = Ng_1g_2                 &&  \\
    & = Ng_1 Ng_2               &&  \\
   & = \theta(g_1) \theta(g_2)
    \end{flalign*}
$\therefore\theta$ is a group homomorphism. $\theta$ is clearly onto.
\end{proof}
\end{document}