4
\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}

Is there a way to left justify the first line of the equation, but then keep the = signs aligned in the subsequent lines? I've tried many combinations of \begin{align} etc with no luck. (It appears that the \begin{equation} favors center alignment.)

3 Answers 3

6

You can do it with aligned. I also provide what I firmly believe is a better version.

\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}

enter image description here

8
  • The second sample is missing a line break before the "Now"? Commented Feb 20, 2017 at 22:40
  • I must agree, I like this better than the \noindent clause of @Zarko's solution Commented Feb 20, 2017 at 22:42
  • 1
    @AthenaWidget: In my eyes, the second sample looks far better!
    – AboAmmar
    Commented Feb 20, 2017 at 22:42
  • 1
    @AthenaWidget There's no reason for starting a new line. I understand it's a stylistic issue, but it's just writing in plain English. Also the \therefore symbol should be avoided in such contexts: writing the full adverb is better for understanding.
    – egreg
    Commented Feb 20, 2017 at 22:56
  • 1
    @egreg. Agreed about the therefore symbol. Thx. Commented Feb 20, 2017 at 23:30
4

Another solution consists in nesting align* in the fleqn environment from nccmath:

\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} 

enter image description here

1
  • It is an interesting extension for another feature: it defines commands for medium-sized formulae (it happens from time to time that one would like to have a size between \textstyle and \displaystyle).
    – Bernard
    Commented Feb 20, 2017 at 22:57
4

Like this?

enter image description here

\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}
3
  • Beat me by 4 min. Commented Feb 20, 2017 at 22:36
  • Perfect. I had this fixation that I had to use {equation}. Thx! Commented Feb 20, 2017 at 22:37
  • @AthenaWidget, I add proof environment. Is now better?
    – Zarko
    Commented Feb 20, 2017 at 22:53

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