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Let $H$ a proper subgroup then $G=$.

I do not know how to prove this problem, if $H$ is a proper subgroup then H is not G, but how generates G from $G-H$?

2 Answers 2

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All we need is that also $H\subseteq \langle G-H\rangle$. So given $h\in H$, pick any $g\in G-H$ (which is possible because $H$ is proper). Then $hg^{-1}\notin H$ and so $h=hg^{-1}\cdot g\in \langle G-H\rangle$.

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    why you have to see that $H \subseteq $? – 2017-02-12
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This follows also from the fact that a group cannot be the union of two proper subgroups.

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    Sorry but I do not see that this implies my fact? – 2017-02-12
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    Note that $G=H \cup (G-H)$ as sets, whence certainly $G=H \cup \langle G-H \rangle$ as groups. Since $G-H$ is not empty, $\langle G-H \rangle$ cannot be proper and must be the whole group $G$. – 2017-02-13