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\begindocument
\beginsolution Let $[G:H] = 2$, so $H$ has exactly two left cosets: $H$ and $gH$ for any $g \notin H$. Similarly, the right cosets are $H$ and $Hg$. For any $g \notin H$, we have $gH = G \setminus H = Hg$. Thus left and right cosets coincide, so $H \trianglelefteq G$. \endsolution
% Solution environment \newtcolorboxsolution colback=gray!5, colframe=blue!30!black, arc=2mm, title=Solution, fonttitle=\bfseries
\begindocument
\beginsolution Let $[G:H] = 2$, so $H$ has exactly two left cosets: $H$ and $gH$ for any $g \notin H$. Similarly, the right cosets are $H$ and $Hg$. For any $g \notin H$, we have $gH = G \setminus H = Hg$. Thus left and right cosets coincide, so $H \trianglelefteq G$. \endsolution Dummit And Foote Solutions Chapter 4 Overleaf High Quality
% Solution environment \newtcolorboxsolution colback=gray!5, colframe=blue!30!black, arc=2mm, title=Solution, fonttitle=\bfseries \begindocument \beginsolution Let $[G:H] = 2$, so $H$
