$\lhd$ with a dot inside

Question

I want to make a new symbol which is $\lhd$ with a dot inside, similar to $\lessdot$.

I tried to follow some suggestions in this question: Horizontal equivalent to \raisebox?

e.g. \newcommand{\lhdot}[0]{\adjustbox{lap={\width}{0.em}}{$\cdot$}\lhd}

But the placement of the dot is really wired..

Any ideas would be appreciated!

Answer 1

Quite likely you don't want to use \lhd, which is classed a binary operator and not a relation, but instead \vartriangleleft.

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

\makeatletter
\DeclareRobustCommand{\dotvartriangleleft}{%
  \mathrel{%
    \vphantom{\vartriangleleft}%
    \mathpalette\dot@vartriangle{{\vartriangleleft}{5mu}}%
  }%
}
\DeclareRobustCommand{\dotvartriangleright}{%
  \mathrel{%
    \vphantom{\vartriangleleft}%
    \mathpalette\dot@vartriangle{{\vartriangleright}{-5mu}}%
  }%
}
\newcommand{\dot@vartriangle}[2]{%
  \dot@@vartriangle#1#2%
}
\newcommand{\dot@@vartriangle}[3]{%
  \ooalign{$\m@th#1#2$\cr\hidewidth$\m@th#1\mkern#3{\cdot}$\hidewidth\cr}%
}
\makeatother

\begin{document}

$a\vartriangleleft b\dotvartriangleleft c$

$a\vartriangleright b\dotvartriangleright c$

$\scriptstyle\dotvartriangleleft\dotvartriangleright$

\end{document}

Answer 2

Done with stacks; made to obey math styles.

\documentclass{article}
\usepackage{amsmath, amssymb,stackengine,scalerel}
\newcommand*{\lhdot}{\mathbin{\ThisStyle{\ensurestackMath{%
  \stackinset{c}{\dimexpr.8pt+.2\LMpt}{c}{\dimexpr.2pt-.5\LMpt}%
  {\SavedStyle\cdot}{\SavedStyle\lhd}}}}}
\begin{document}
\(a \lhd b\)

\(a \lhdot b~
\scriptstyle a \lhdot b~
\scriptscriptstyle a \lhdot b\)
\end{document}

Answer 3

A simple solution is to stack \lessdot itself on top of \lhd.

\documentclass{article}
\usepackage{amsmath, amssymb}
\newcommand*{\lhdot}{\mathbin{\ooalign{$\lhd$\cr$\lessdot$}}}
\begin{document}
\(a \lhd b\)

\(a \lhdot b\)
\end{document}