This might be a stupid question but if I have a subgroup $H$ of $G$, and I have $h\in H$ and $g\notin H$, then $gh\notin H$ for if $gh\in H$ then $g\in Hh^{-1}=H$ is a contradiction. But what if I have $x\notin H$, what is my argument for why $gx\notin H$?
Product of two elements not in subgroup is not in subgroup
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abstract-algebra
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0Let your $G = \mathbb{Z}$, and let $H = 4\mathbb{Z}$. Let $g = x = 2$. Then $g$ and $x$ are both not in $H$, but what's $g + x$? (I love being able to reduce a question to this identity.) – 2012-06-14