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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 $gx\notin H$?

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    Let 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

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It is not true that:
$x\notin H$ and $g\notin H$ $\implies$ $xg\notin H$.
(If I understand your question, you wanted to justify this claim.)

Counterexample:
Just take $H=\{e_G\}$ and $x=g^{-1}$ for some $g\ne e_G$.

E.g. $H=\{0\}$ and $g=1$, $x=-1$ in $(\mathbb Z,+)$.

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    @JackSchmidt: Of course! thank you!2012-06-14